discreet structures

7/31/2019 Discreet Structures

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Chapter 1: Graphs

(VERY SHORT ANSWER TYPE QUESTIONS)

Q. 1. Define directed graph with example and draw a diagram?

Sol. Directed Graph: A directed graph is defined as an ordered pair (V, E) where V is aset and E is a binary relation on V. A directed graph can be represented geometrically asa set of marked points V with a set of arrows E between pairs of points. Also the

elements in V are called vertices. The ordered pairs in E are called edges. For example : The directed graph is given below. Here, vertices are a, b, d and edges are (a, b), (b, a), (b, d) (d, a).

Q. 2. Define undirected graph with example?

Sol. Undirected Graph: It consists of a set of vertices, V and a set of edges E. The edgeset contains the unordered pair of vertices. If (u , v) E E, then we say u and v are

connected by edge where u and v are vertices in the set V.

For example The undirected graphs are given below in Fig 1 and Fig. II.

We note that the edges x1 x2 x3 in Fig. II are associated with the unordered points (1,2), (2, 3) and (3, 1) respectively. In Fig. II node I is adjacent to node 2 and 3.

Q. 3. Define the following terms with example

Sol. (a) Simple Graph: If there is no more than one edge between a pair of vertices (nomore than one directed edge in the case of a directed graph) then such a graph is knownas simple graph.

(a) Simple graph

(b) Non-simple graph(c) Multi-graph

(d) Weighted graph


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(b) Non-Simple Graph: A graph which is not a simple is called a non-simple graph.

(c) Multigraph : A multigraph G = (V. E) consists of a set of vertices V and a set of edgesE such that edge set E may contain multiple edges and self-loops.

For example : Consider the following graphs.

In undirected multigraph, e4, and e5 are multiple edges and e6 is a self loop. In directed multigraph, edges e1, e2 and e4 e5 are multiple edges and e7, is a self-loop.(d) Weighted Graph : A graph G = (V, E) is called a weighted graph if each edge of

graph G is assigned a positive number w, called the weight of the edge e.

For Example : The fig I is a weighted graph in which the number of any edge show themultiplicity of the edge. We may regard the multiplicity as a weight assigned to an edge.

The Fig II is a weighted graph in which the numbers on the edge show the weight of theedges which are not necessarily integer.

Q. 4. How can you define degree of a graph?

Sol. Degree of a graph: Let v be a vertex of an undirected graph, the degree of v,denoted by d(v), is the number of edges that connects v to the other vertices in a graph.The degree of a graph cannot be negative.

Out degree : If v is a vertex of a directed graph, then the out degree of v, denoted by d(v), is the number of edges of the graph that initiate VIn degree : The indegree of v, denoted by indeg (v), is the number of edges thatterminate of v.

e.g. Consider the graph, we see that the degrees of A, B, C and D are 3, 3, 5 and 3

respectively.


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Q. 5. What are the conditions to make two graph isomorphic?

OrDefine isomorphic Graphs?

Sol. Two graphs G = [V1. E1] and C2 = [V2, E2] are called isomorphic graphs if there

is a one-to-one correspondence between their vertices and between their edges i.e. the

graphs have identical representation except that vertices may have different labels.Two groups G1 and G2 are said to be isomorphic if there exist a function

f:

such that

1. f is one-one and onto 2. f preserves adjancies i.e.

1. f preserves non-ad jancies

The function f is called isomorphism between G1 and G2 (i.e. between G1 and G2).Note : Two graphs are isomorphic, will have1. Same number of vertices. 2. Same number of edges

3. An equal number of vertices with given degree. e.g.


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These are isomorphic graphs.

Q. 6. Define the following terms

(a) Finite multigraph

(b) Trivial graph

Sol. Finite multigraph : A multigraph is said to be finite if it has a finite number of

vertices and a finite number of edges;

Trivial Graph : The finite graph with one vertex and no edge i.e. a single point is

called a trivial graph.

Q. 7. The sum of the degrees of the vertices of a undirected graph G is twice thenumber of edges in G.Or

Let G IV, E] be an undirected graph with e edges, then

Sol. If the edge e is a self loop at node x, it is counted twice in deg (x). If e is an edge

with distinct end nodes x and y, it is also counted twice, once in degree (x) and again indeg (y).The sum of the degrees of all vertices is equal to twice the number of edges in C.

Q. 8. Define the following terms:

(d)Cut point

(a) Connected graph

(b) Diameter of a connected graph

(c) Complete graph

Sol. (a) Connected Graph : A graph is connected if there is a path from vertex u to v

or vice-versa. If there is no path between any of its vertices are called disconnectedgraph.(b) Diameter of a connected Graph : Let G be a connected graph. The distance betweenvertices u and v in G written as d (u, v); is the length of the shortest path between u

and v. The diameter of G written as diag (G), is the maximum distance between any twopoints in G.


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(c) Complete Graph: A complete graph is a graph with the property that between anytwo distinct vertices there is exactly one edge.

(d) Cut-point : be a connected graph. If v be a vertex of G such that C-vis not connected then we call vertex v as a cut vertex. The subgraph G-v is obtained bydeleting the vertex v from the Graph G and also deleting all the edges incident on v.

In given Fig., v is a cut vertex of C.

Q. 9. Define the following:

1. Subgraph2. Spanning-subgraph

Sol. Subgraph: A Subgraph of graph G = (V. E) is a graph C = (V1 E) in which V V andE E and each edge of G has the same end vertices in G as in graph G. That is C is asubgraph of G, if1. All the vertices of G are in G.

2. All the edges of C are inC and

3. Each edge of C has same end vertices in G as in G.Note : Graph C is itself a subgraph of G. Spanning Subgraph: A graph G1 = (V1. E1) is called a spanning sub graph of C = (V. F)

if G1 contains all the vertices of G and

e.g.

Q. 10. Differentiate between path and circuit.

Sol. Path: A path of length n is a sequence of (n + 1) vertices of a graph in which eachpair of vertices is an edge of the graph. The path is called simple one if no edge isrepeated in the path i.e., all the vertices are distinct except that first vertex equal to lastvertex.

If first vertex equal to last vertex then path is called a closed path and if first vertex notequal to last vertex then path is called an open path.

If p is a path in a graph G then clearly


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1. i.e. number of vertices in P is a subset of number of vertices in C.

2. i.e. number of edges in P is a subset of number Of edges in G.

3. i.e.number of vertices in P lies between l and

n + 1 and number of edges in P lies between 1 and n.

Circuit: If G = [V. E] be a graph, then a path is a cycle (circuit) which originates andends in the same vertex. It is simple if its path is simple. Incircuit no. edge occurs morethan once in the path.

Note: 1. A self loop is also a circuit, but converse is not true

2. The degree of every vertex in a circuit is 2. e.g. In the graph of fig.

Q. 11. What properties a graph must have to call as strongly connected graph?

Sol. Strongly connected graph: A directed graph is called strongly connected if there is adirected path from any node u to v and vice-versa, for any pair of nodes of the graph.

In other words, two vertices u and v are strongly connected, if there is a path from uand v and v and u. That is if they are connected in both directions to one another andgraph is strongly connected if every pair of its vertices is strongly connected.

Q. 12. Define weekly connected?

Sol. Weekly Connected : A weekly connected graph is a directed graph, if itsundirected graph is connected i.e. the graph obtained after neglecting the direction.


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Q. 13. Define the following (a) Chromatic number(b) Chromatic number of chromatic polynomial.

Sol. Chromatic Number The minimum number of colours needed to produce a proper colouring of a graph is called chromatic number of G and is denoted by (G).

In the following figure minimum 3-.colourable. HenceSimilarly, for the complex graph K6, we need 6 colours to colur K6. Since every vertexis adjacent to every other vertex and we need a different colour for each vertex.

Therefore

Chromatic Polynomial : The problems of computing (G) is the problem of computing thetotal number of different proper colouring of a graph C using a set

Let G be a graph and x 0 is an integer. Let P (x) be the number of ways to properly colour G using x or fewer colours, as P (x) is a definite number for each x, we see that

PG is a function. That is P is a polynomial in x. We call P the chromatic polynomial in G.

Q. 14. Define Planner Graphs?

Sol. Planar Graph : A graph is said to be planar if it can be drawn in the plane with nointersecting edges. That is a graph G is said to be planar if it can be drawn on a plane

without crossovers and A graph G that cannot be drawn on a plane without cross overbetween its edges is called non-planar graph.


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Q. 14. Show that following graphs are planar.

Sol.

Q. 15. State Eulers formula for connected planar Graph.

Sol. Let C (V. E, R) be a planar graph with V, E and R be number of vertices, edges andregion, respectively then V - E + R = 2.

Q. 16. Define the following (a) Eulerian Path(b) Eulerian circiuts(c) Eulerian Graphs

Sol.Eulerian Path : An Eulerian path through a graph is a path whose edge list contains eachedge of the graph exactly once.

Eulerian Circuit: An Euler circuit is a path through a graph, in which the initial vertexappears 2nd time as the terminal vcrtex.Eulerian Graph: A graph C is called Eulerian graph which possesses an Euler circuit.

Euler circuit uses every edge exactly once but vertices may be repeated.

Q. 17. In the following graph the Euler Path is:

(i) abcdef(ii) abcf


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(iii) ftlceah

(iv) fdeabc

Sol. (iii) fdceah.

Q. 18. Write short Note on (a) Hamiltonian graphs (b) Hamiltonian path

(c) Hamiltonian circuit.

(d) Hamiltonian chain

Sol. (a) Hamiltonian graph : A Hamiltonian graph is a graph that possesses a Hamiltoniapath. A Hamiltonian path uses each vertex exactly once but edges may not be included. (b) Hamiltonian Path: A Hamiltonian path through a graph is a path whose vertex listcontain each vertex of the graph exactly once, except if path is a circuit.

(c) Hamiltonian Circuit: A Harniltonian circuit is a path in which the initial vertex appearsa second time as the terminal vertex.

(d) Hamiltonian Chain : It is a path through a graph whose vertex list contains eachvertex exactly once.

Q. 19. Give an example of a graph which has an Euler circuit but not Hamiltoncircuit. Also discuss converse with example. Give an example of a graph thathas neither an Euler nor Hamiltonian circuit.

Sol. An Euler circuit but not Hamiltonian

The Euler circuit V1. V5. V2, V5, V3, V4, V6. V3. V2, V1. There is no Hamiltonian circuit

A Hamiltonian circuit but not Eulerian circuit

Hamiltonian circuit is a, b, d, c, a. There is no Euler path.

Neither Euler nor Hamiltonian circuit


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It has neither Euler circuit nor Hamiltonian circui

Q. 20. Prove that there is always a Hamiltonian path in a directed complete

graph ?

Sol. Directed complete graph A simple directed graph in which there exist an edgebetween

every pair of vertices is called a directed complete

graph.

The path W = v, e10, v5, e6, v6, e9, v2, 2 v7, e, v8, e, v4, is a Hamiltonian path and itis calle

a directed complete graph.

Q. 21. What is the minimum number of vertices necessary for a graph withsix edges to be planar, with eleven edges.

Sol. Here


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[SHORT ANSWER TYPE QUESTIONS]

Q. 1. Suppose a graph G contains two distinct parts from a vertex u to a vertexv.Show that G has a cycle.

Sol. Consider two distinct path from u to e be

Now delete from the paths P1 and P2. all the initial edges, v art identical i.e. of we

have

We will delete the all first k edges of both the paths P1 and P2. Now, after delecting the

K edges be!h the paths start from the same vertex, (Let u1) and ends at v. Now, toconstruct a Je. start trom vertex u, and

follow the left over path of P1 until we first meet any vertex of the left over path of P2. Ifthis vertex is u2, then the remaining cycle is computed by following the left over path of

2 which starts from u2 and ends at v.

Q. 2. State and prove Eulers Theorem.

Sol. Statement

Let

be a connected graph and Let R be the number of regions defined by any

planar depiction of C. Then we have

Proof. We prove the result by induction on number of regions determined by C.

We first prove the result for K = 1

A tree determines above region.

Let us assume that result is true for all

Let C be a connected plane group determining regions. Remove anedgewhich is common to the boundary of two regions we obtain a graph G. If V, E, Rdenotes number of vertices, number of edges and regions of C rep. then


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Q. 3. If a graph C has more than two vertices odd degree, then prove that thereis

no Euler path in G.

Sol. Let V1, V2, V3 be vertices of odd degree. Any possible Euler path must leave (orarrive at) each of V1, V2. V3 with no. way to return (or leave) since each of these

vertices has odd degree. One vertex of these three vertices has odd degree. One vetexof these three vertices may be the beginning of the Euler path and another the end, butthis leaves the third vertex at end of an untravelled edge. Thus there is no Euler path.

Q. 4. If G is a connected graph and has Exactly two vertices of odd degree,there is an Euler path in G. Any Euler path in G must begin at vertex of odddegree and ends at other.

Sol. Let C has two vertices V1 and V2 of odd degree. By adding the edge [V1, V2] to Gwe can produce a connected graph say C all of whose vertices are of even degree. As G,is a connected graph and every vertex of G has even degree so we can determine an

Euler circuit C in G. Removing the edge [V1. V2] from the circuit C, produces an Eulerpath that begins at V1 (or V2) and ends at V2 (or V1).

Q. 5. An undirected graph possesses Eulerian path if it is connected and haseither zero or two vertices of odd degree. Give suitable example.

Sol. Let the undirected graph possesses an Eulerian path. Then, by definition, the graphmust be connected. Now, since the graph has Eulerian path, it means that every time

the path meets a vertex, it goes through two edges which are incident with the vertexand have not been traced before.Ihus, except for the two vertices at the ends of the path, the degree of all other

vertices in the graph must be even.

When the two vertices at the ends of the Eulerian path are distinct, then there areonly two vertices with odd degree.Converse : Let the undirected graph is connected and two of its vertices are of odd degree.


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To Show : Graph C possesses an Eulerian path. Since the graph is connected, so no edgeof the graph will be traced more than once. For a vertex of even degree, when the path

enters the vertex through an edge, it can always leave the vertex through anotheredge that has not been traced before.

Iherefore, when the construction is completed, we must have reached the other vertexof odd degree. Tracing all the edges in this way, we will get an Eulerian path. If not all of

the edges in the graph were traced, we shall remove those edges that have been tracedand obtained a subgraph formed by remaining edges. The degree of the vertices of thissubgraph will be even. Starting from one of these vertices, we can again construct apath that passes through the edges. Because the degrees of the vertices are all even,

this path must return to the vertex at which it starts. Combining this path with the pathwe have constructed to obtain one which starts and ends at the two vertices of odddegree, the path so obtained is Eulerian path.

For Example, consider the graph as shown below:

This graph has only two vertices of odd degree (by above theorem), it has an Eulerianpath.

Q. 6. An undirected graph possesses an Eulerian circuit iff it is connected andits vertices are all of even dLrce. Prove that with the help of suitable example.

Sol. Let the u ndirected graph OSSCS an Eulerian circuit. Then, by definition, the graphmust he conne ted. Now since the graph has Eulerian circuit, it means that every timethe graph nwe1 a erte\, it goes through edges which are incident with the vertex andhave not been ltaud helore. ihus, the degree of all vertices in the graph must be even.

Converse let the undiiected graph is connected and all its vertices are of even degree.We shov ha the i api possesses an Eulerian circuit. Since the graph is connected, no edgewIll be traced more than once. As all the vertices are of even degree, whenever the

circuit, cntcrd the vetv through an edge, it can always leave the vertex through

another edge that has not been traced before and also the circuit must return to thevertex at which it starts. I lence the circuit is an Eulerian circuit. E.g Consider the graph given below


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Ihe graph is conner ted and all the vertices are of even degree. Therefore, it has anEulerian circuit.

Q. 7. State and rmve Fuleria theorem on graph to show that konigsbergs graphis not proved to a solution.

Sol. The word Konigs berg is the name a town, situated on the bank of river, pregel inGermany. This city has seven briges. L. Euler, the father of graph theory, proved that itwas not possible to cross each of the seven bridges once and only once in a walking

tour. Euler replaced the islands and the two sides of the river by points and the bridgesby curves as shown in Fig. II. Fig II is a multigraph. A multigraph is said to betraversable if it can be drawn with any breaks in the curve and without repeating anyedges i.e. if there is path which includes all vertices and uses each edge exactly onceand such a path is called traversable trial.

According to Euler, the walk in Konigs berg is possible if the multigraph in Fig. 2 istraversable. But Euler proved that the multigraph in Fig. 2 is not traversable and hencethe walk in konigsberg is impossible. We prove it.

We know that a vertex is even or odd according as its degree is even or odd. Suppose amultigraph is traversable and that a traversable trial does not begin or end at a vertex,say P, we claim that P is an even vertex. For whenever the traversable trials enters P byan edge, there must always be an edge not previously used by which the trial can leaveP. Thus, the edges in the trial incident with P must appear in pairs and so P is an evenvertex. F urther, if a vertex Q, is odd the traversable trial must begin or ends at Q.

Hence, a multigraph with more then two odd vertices cannot he traversable.Now the multigraph corresponding to the Konigsberg bridge has four odd vertices.


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Thus, one cannot walk through Kongisberg so that each bridge is crossed exactly once.

Euler actually proved the converse of the above statement, which is contained in the

following theorem, called Euler theorem.

Q. 8. Using either breadth first search aigz. m or Dijkstrass a1gorithem findthe shortest path from s to tin the following weghd graph.

Sol. Let S be the set of vertices of the given weighted shown in Fig. below. graph i.e. S {s, a, b, c, d, t) as

Step I. Include the vertex s in S and eterm ne all the direct paths from to all other

vertices without going through any other vertex

Step II. Include the vertex in S which is nearest to s and determine shortest paths to allvertices through this vertex and update the values. The nearest vertex is C.

Step III. The vertex which is 2 nearest t. s is a. Include this vertex in S.

Step IV. The vertex which is 3rd nearest to s is h. Include this vertex in S.

Step V. The vertex which is next nearest to S is (d). Include this vertex in S.

Since n - 1 = S vertices are included in S. Hence we have found the shortest distancefrom s to all other vertices. Thus, the shortest distance between s and t is 63 and the

shortest path is s;a, b, t.


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Important Expected Questions


(a) Mixed Graph(b) Finite Graph(c) Linear Graph

(d) Discrete or Null Graph

Sol. Mixed Graph: A graph C = [V, E] in which some edges are directed and some areundirected is called a mixed graph. The graph shown in Fig. I is a mixed graph.Finite Graph: A Graph C = [V, E] is said to be finite if V and E are finite sets.Linear Graph : A graph G = [V1 El is said to be a linear graph, if its edges joiningvertices lie along a line.

Discrete or Null Graph: A graph containing only vertices and no edge is called a discreteor null graph. The set E of edges in a graph G [V. E] is empty in a discrete graph. Alsoeach vertex in a discrete graph is an isolated vertex.

Q. 2. Show that the sum of degree of all the vertices in a graph G, is even.OrThe sum of the degrees of the vertices of a graph G is equal to twice the no. ofedges in G.

Sol. Each edge contribute two degrees in a graph. Also each edge contributes onedegree to each of the vertices on which it is incident.Hence, if there are N edges in C, then we have

Q. 3. Prove that in any graph, there are an even number of vertices of odddegree.

Sol. Consider a graph having vertices of degree even and odd. Now, make two groups ofvertices. One with even degree of vertices v1, v2 ,Vk and other with odd degree of.vertices U1, U2, ....u

Now we know that sum of degree of all the vertices is even so, U + V is even


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Since, V is sum of k even numbers. Hence, it is even. But U is sum of n odd numbers.So, to be U an even number, n must be even. Hence proved.

Q. 4. A graph G has 21 edges, 3 vertices of degree 4 and other vertices are ofdegree 3. Find the number of vertices in G.

Sol. let n be the number of vertices in C. Since, sum of degree of all vertices is equal

to twice the number of edges i.e.

Q. 5. Show that the maximum number of edges in graph with ii vertices and no

multiple edges are

Sol. Let n be the number of vertices of a graph C.

Q. 6. Prove that in a simple graph with n vertices, each vertex has maximum

degree (n1).

Sol. A simple graph is a graph without parallel edges or self loops.If G = [V, F] is a simple graph with only one vertex, then number of edges in C is

:. maximum degree of a vertex in G = I I = 0If G = [V. F] he a simple graph with 2 vertices, then number of edges in G is I andtherefore degree of each vertex is (21).and so on

If C = [V F] he a simple graph with n vertices, then maximum degree of each vertex =n-1.


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Q. 7. A graph C has a Hamilton circuit if

where 11 is the number of vertices and e the number of edges in C.

Sol. Let if possible, the graph C is non-Hamilton and we show

The converse of above theorem, however, is not true.

Q. 8. A planar and connected graph has a vertex of degree less on equal to 5.

Sol. Let C be connected and planar and suppose, if possible degree of each vertex


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Sectionll (Trees)

Q. 1. Prove that A tree with n vertices has 11 1 edges?

Sol. Because a tree is connected, there must at least 11 1 edges to connect the nvertices. Suppose that there are more than it 1. Then either the root has in degree Ior some other vertex has in-degree at least 2. Which is impossible. Thus there areexactly i-i 1 edges.

Q. 2. Define the following terms:(1) Complete Binary tree

(2) Binary search tree

Sol. Complete Binary tree : Complete binary tree is a binary tree if all its levels, exceptpossibly the least, have the maximum number of possible nodes as for left as possible.The depth of complete binary tree having it nodes is log2 ii + i.For example : The tree given below is a complete binary tree.

Binary Search Tree : Binary search trees has the property that the node to the leftcontains a smaller value than the node pointing to it and the node to the right contains alarger value then the node pointing to it.

It is not necessary that a node in a binary search tree point to the nodes whose values

immediately precede and follow it.

Q. 3. What are the properties of minimum spanning tree.

Sol. A minimum spanning tree T of a laph C is a tree whose total weight is the smallest

among all the spanning trees of the graph C. It has following properties.1 The total weight of spanning tree is the sum of the weights of the edges of the

spanning trees.2. The minimum weight of the spanning tree is unique.a tree.

Q. 1. Let T be a binary tree with ii vertices. Determine the number of leaf nodesin

Sol. We shall prove it by induction.


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Hence the result.

Q. 2. Define a rooted tree with an example and show how it may be viewed asdirected graph.

Ans. Rooted tree: We first define the term directed tree. A directed graph is said to be a

directed tree if it becomes a tree when the directions of the edges are ignored.

A directed tree is called a rooted tree if there is exactly one vertex whose incoming

degree is 0 and incoming degree of all other vertices are 1. The vertex with incoming

degree 0 is called the roots of the rooted tree.

In a rooted tree, a vertex whose outgoing degree is 0 is called a. leaf or a terminal nodeand a vertex whose out going degree is non-zero, is called a branch node or internalnode.Rooted tree may be viewed as directed graph : We know that a tree is a graph which isconnected and without any cycles. A rooted tree T is a tree with a designated vertex r,

called the root of the tree. Since there is a unique simple path from the root r to anyother vertex a in 1, this determines a direction to the edge of T. Thus T may be viewedas a directed graph.


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Q. 3. Prove that a simple graph is connected if it has a unique spanning tree.

Sol. First, SUOSC that a simple graph G has a spanning tree T. T contains every vertexof C. let a and b he the vertices of G. Since a and h are also vertices of T and T is a tree,there is a path P between a and h. Since T is subgraph, P also serves as path between aand 17 in C. Hence C is connected.

Conversely, Suppose that C is connected, if G is not a tree, it must contain a simplecircuit. Remove an edge from one of these simple circuits. The resulting subgraph hasone fewer edge hut still contains all the vertices of G and is connected. If this subgraphis not, a tree, it has a simple circuit, so as before, remove an edge that is in a simple

circuit. Repeat this process until no simple circuit remain. This is possible because thereare only a finite number of edges in the graph, the process terminates when no simplecircui remain. Thus, we eventually produce an a cycle sub graph T which is a tree. The

tree is spanning tree since it contain every vertex of G.

Q. 4. Suppose S people enter a Badminton tournament use a rooted tree model

of tournament to determine how many games must be played to determine achampian if a player is eliminated after one loss.

Sol. We can show if there are n people enter in a Badminton tournament, then thereare (ii 1) matches to he played. Number of people = 8.

So there are 8 people enter in a Badminton tournament, then there are (8 1) = 7matches to he played.Here number of rooted tree and no. of match to be played is same as the no. ofinternal vertices.

Q.5 . What is minimum spanning tree of a graph ? Write down Prims and

Kruskals algorithms and execute them by taking a suitable example.

Sol. Minimum Spanning tree : Consider a connected weight graph C (V. F). Aminimal spanning tree T of the graph C is a tree whose total weight is smallest among all

the spanning trees of the graph C. The total weight of the spanning trees is the sum of

theweights of the edges of the spanning trees.The minimum weight of the spanning tree is unique but the spanning tree may not

be unique because more than one spanning tree are possible when more than one edges exist having the same weight.Algorithm for minimal spanning Trees There are several methods available for actually

finding a minimal spanning tree in a graph. Two algorithm due to prim andkruskal for finding a minimal spanning tree for a connected weighted graph where no- weight is negative. These algorithms are example of greedy algorithms. A greedyalgorithm

is a procedure that makes an optimal choice at each of its steps without regard toprevious

choices,Prims Algorithm Initially the algorithm starting at a designated vertex chooses an


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edge with minimum weight and considers this edge and associated vertices as part ofthe

desired tree. Then iterate, looking for an edge with minimum weight not yet selectedthathas one of its nodes in the tree while the other node is not. The process terminates whenn I edges have been selected from a graph of n nodes to form a minimal spanning

tree.The algorithm involves the following steps.Input: A connected weighted graph C.Output: A minimal spanning tree 1.

Step I. Select any vertex in C. Among all the edges incident with the selected vertex,

choose an edge of minimum weight. Include it in T.Step II. At each stage, choose an edge of smallest weight joining a vertex already

included in T and a vertex not yet included if it does not form a circuit with the edges in T include it in T.Step III. Repeat until all the vertices in G are included.

e.g. Find the minimal spanning tree of the weighted graph of fig. I using Prismsalgorithm.

Kruskals Algorithm : This algorithm provides an acyclic subgraph T of a connected

weighted graph C which is a minimal spanning tree of G. The algorithm involves thefollowing steps:Input: A connected weighted graph COutput: A minimal spanning tree T.

Step I. List all the edges (Which do not form a loop) of C which are non-decreasing orderof their weightsStep Ii. Select an edge of minimum weight (if more than one edge of minimumweight, arbitrarily choose one of them). This the first edge of T.

Step III. At each stage, select an edge of minimum weight from all the remaining edgesof G if it does not form a circuit with the previously selected edges in T. Add the edge to

TStep IV. Repeat step 3 until n 1 edges have been selected, when n is the number ofvertices in G.The following examples illustrate these stepse.g. Show how Kruskals algorithm find a minimal spanning tree for the graph of fig.

given belowStep I. List the edges in non-decreasing orders of their weights.

Step II. Select the edge (b, c) since it has the smallest weight, include it in T.

Step III. Select an edge with the next smallest weight (c, e) since it does not formcircuit with the existing edges in T. So include it in T.


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Step IV. Select an edge with the next smallest weight (c, d) since it does not form circuitwith the existing edges in T, so include it in T

Step 5. Select an edge with the next smallest weight (a, b) since it does not formcircuit with the existing edges in T. So include it in T.

Since G contain 5 vertices and we have chosen 4 edges, we stop the algorithm andthe minimal spanning tree is produced.

Chapter 2 : Combinational Mathematics

Q. 1. Define Basic principle of counting? Explain,

Sol. Basic principle of counting: If we have to perform n operations, and there are ndifferent ways to perform 1 operation, n2 different ways to perform 2 operationand soon then number of ways to perform n operations will be n1xn2x x n. This isknown as Basic counting Principle.There are mainly two counting principles namely (i) Sum Rule (ii) Product Rule.

Sum Rule : Suppose some event E can occur in iii ways and a second event F can occur in n ways, and suppose both events cannot occur simultaneously. Then E or F can occur in m + n ways.Mathematically, If E and F are disjoint sets then

more generally, suppose an event E1 can occur in .n1 ways, a second event E2 can

occur in n2 ways, a yd event E3 can occur in n3 ways and suppose no two of the eventscan occur at the same time, then one of the event can occur in n1 + + + ways.


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Product Rule (Principle of Association) Suppose there is an eventE which can occur in rnways and independent to this event there is a second event F which can occur in n ways.Then combinations of E and F can occur in rnn ways.Mathematically, Let E x F be carterisan product of sets E and F, then n(ExF)=n(E).n(F)


(a) Permutation(b) combination

Sol. Permutation: A permutation is an arrangement of a number of objects in somedefinite order taken some or all at a time.The total number of permutations of n distinct objects taken r at a time is denoted by

Combination: A combination is a selection of some or all, objects from a set of givenobjects, where order of the objects does not matter.The number of combinations of n objects, takes r at a time is represented by

Q. 3. What is a circular permutation of n objects and how many are there?

Sol. Circular permutation : The circular permutation are the permutations of the objectplaced in a circle. Therefore, there is no starting and ending in the circular permutation.We only consider the relative positions. When persons are sitting around a circular table,then there is not first and last person.Let us fix this position of one person. The remaining (nfl persons can be arranged intheremaining (ni) places in 1P,,1 ways i.e (nI) ways.

Q. 4. How many 4-digit numbers have one or more repeated digits?

Sol.For a four digit number, unit place can be filled in 10 ways, tens place can be filled

in 10 ways, hundreds place can be filled in 10 waysand Thousands place can be filled in 9 ways. [.. As first digit can not be zero] Total 4digit numbers 10x10x10x9 = 9000Total 4 digit numbers without any repetition

=9x9x8x7=4536Total no. of 4 digit number with repeated digit

9000-4536

= 4464 Ans.


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Q. 5. In how many ways can a president and vice president be choosen from astateof 30 candidates?Sol. Two candidates one is president and other is vice-president can be choosen out of

30 candidates

Then No. of ways for choosing president and vice president are

870 Ans.

Q. 6. Automobile license plates in Massa chusetts usually consists of 3 digitsfollowed by 3 letters. The first digit is never zero. How many different plates ofthis

type could be made.

Sol. Consider the digits0, 1, 2 , 9. These are 10

digits. Since first digit of

License plate is non-zero.First place can be filled up in 9 ways.Similarly 2nd place can be filled up in 10 ways.

and 3 places can be filled up in 10 ways.Required No. of using the digits = 9x10x10 = 9000.

Again, there are 26 Alphabets.

1 place can be filled up in 26 ways

2 place can be filled up in 26 ways3rd place can be filled up in 26 ways

Required number of using the English Alphabets= 26x26x26= 17576By product rule, total Number of making the plate

= 900x17576= 15818400 Ans.15818400 Ans.


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Q. 7. Find the number of permutations that can be formed from the letterELEVtN. TTjW many of them begin and end with E. How many of these have thethree Es together.

Sol.Total No. of letter are 6 have three EsTherefore, Required No. of ways

1. The first and last places are occupied by E i.e.

(2) Take three Es one word.Total No. of words = 4. [FEE L V N]Req. No. of ways =4!= 24Ans.

Q. 8. State Principle of Inclusive and Exclusive?

Sol. Let A and B be any finite sets, then

lhis principle holds for any number of sets.

In other words, to find the number of elements in the union AUB, we add n (A) and n (B)& then we subtract i.e. we include n (A) & n (B) and we exclude . This

principle is known as Inclusion-Exclusion principle.

Let P and Q be any two non-disjoint sets. Then

Q. 9. How many subsets of { 1, 2 , 9) contains at least 5 elements.

Sol.

Let A = {1, 2 , 9) i.e. A has 9 elements. There are Subsets each consisting of 5 ofthe 9 elements of the given set.

There are subsets each consisting of 6 of the 9 elements of the given set.


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There are Subsets each consisting of all the 9 elements of the given set. Totalnumber of Subsets containing at Least 5 elements.

(SHORT ANSWER TYPE QUESTIONS)

Q. 1. How many words can be obtained by arranging the letters of the wordUNIVERSAL in different ways. In how many of them (a) E, R, S occur together

(b) No two of the letters F, R, S occur together?

Sol. There are 9 Letters in the word UNIVERSAL which can be arranged among in 9!

Ways.

1. As E, R, S occur together taking ERS as one letter, we can arrange seven Lettersin 7! ways and three Letters E, R, S can be arranged among themselves in 3P3 =3! ways.

By Product Rule, required Number of words when F, R, S occur together=7! x3!31240Ans.

(ii) Since no two of E, R, S occur together. We first place six Letters U, N, I, V, A, L in =6! Ways

Now we place three Letters at place marked X so that No two F, R, S occur together.

We can place these three Letters at 7 places in

Required no. of ways =6! x2106 x5 x4 x3 x2 x1 x210 = 30 x 24 x 210151200 Ans.


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Q. 2. Suppose that there are n people in room, n>=1 and they all shake handswith one another. Show that n (n 1)/2 hands shaken will have occured.

Sol. When two persons shake hand, it is counted as one hand Shake. Total Number of hand Shakes = Number of ways of choosing 2 persons from

Q. 3. In a class of 60 boys, 45 boys play cards and 30 play carrom. How manyboys play both games. How many play cards ony and how many plays carrom

only?

Sol. Let n (A) denotes Number of boys who play cardsLet n (B) denotes number of boys who play carrom.

n(A)= 45 & n(B)=30n (AUB) =60

Since n (AUB) = n (A) + n (B) -

=45+3060=15

There are 15 boys who play both games.

Since ,

No. of boys who play cards only =

Q. 4. Find the number of Subsets of a set S containing n elements?

Sol. There are nC1 subsets each consisting of one of the n elements of the given set S.There will be nC2 subsets each consisting of any two of the n elements of the givenset S.

At last there will be nCn subsets each consisting of all n elements of given set S.

Also, there will be one set

Hence, total No. of subsets are =


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Q. 7. Suppose that 100 out of 120 mathematics students at a college takeatleast one of the language French, German and Russian. Also suppose 65study French, 45 study German, 42 Study Russian, 20 study French & German,25 study French & Russian, 15 study German & Russian. Find the number ofstudents studying all subjects. Also find the number of students studying takingexactly one subject.

Sol. n (F) denotes no. of students study Frencho (C) Germano (R) Russian

:. No. of Students studying all three langauges = 8 Ans.

Q. 8. Prove that

Sol . We know

Hence Proved.

Q. 9. Suppose that an urn contains 15 balls of which 8 are Red & 7 are Black Inhow many ways can 5 balls to be choosen so that 2 are red and 3 are Black.

Sol, Total Number of Balls = 15Number of Red Balls 8Number of Black Balls = 7


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Q. 10. If A and B are disjoint sets such that AUB finite, then n(AUB) = n(A) +n(B)

Sol. In counting the element of AUB, we first count those elements which are in A.1 hese are n (A). The only other elements of AUB are those elements which are in B, butnot in A.

Since given that A and B are disjoint

No element of B is in AHence there are n (B) elements which are in B.

n (AUB) = n (A) + n (B)(AUB)

Q. 11. In how many ways can 5 Gentleman & 5 ladies be seated round a tableso that no two ladies are together.

Sol. Let us first seat 5 gentlemen on the round table M1, M2,M3, M4, M5 and this can be done in (n 1)!

(51)!4! =24 ways

Since no two Ladies are to sit together, they occupy the places marked as L1, L2, L3, L4,L54 Ladies can be arranged in 5 places in 5C4

= 120 ways.Required No. of ways = 24 x 120= 2880 Ans.

Q. 12. Find the least number of cables required to connect 100 computers to20 printers to guarantee that 20 computers can directly access 20 differentprinters. Justify your Answer.

Sol. Here Printer are Pigeonholes i.e. m = 20and computers are pigeon i.e. n = 100

By extended principle of pigeonholes, one of the pigeonhole must contain atleast


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pigeons.

Least no. of cables required = 5 Ans.


Q. 1. Define Extended pigeonhole principle.

Sol. Extended pigeonhole Principle: If n pigeons are assigned to m pigeonholeswhere n is sufficiently large as compared to m, then one of the pigeonhole must contain

at least pigeons.

Q. 2. Show that if 9 colours are used to paint 1000 houses, at least 112 houseswill

be of the same order.

Sol. We denote the no. of given houses by n and no. of colours by m i.e. n =1000, m=9ie there are n pigeons and m pigeonholes.

using extended pigeonhole principle

Q. 3. Show that at least two people must have their birthday in the same monthif 13 people are assembled in a room.

Sol. Let each person be assigned the month of the year on which he has born. As there are 12 months in a yearBy pigeonhole principle,

There must be at Least two people assigned to the same month.

Q. 4. How many 6-digit numbers can be formed from the digits 0, 1, 2 , 6,7, if


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no digit is repeated.

Sol. Out of 8 Numbers, 6 to be selected.

Total number of 6-digit Numbers =

Q. 5. How many permutations can be made out of the letter of word

COMPUTER.

How many of these(i) begin with C and ends with R(ii) C and R occupy the end places.

Sol.There are 8 Letters in the word COMPUIER and all are distinct.

Total No. of permutations = 8 ! = 40320

(i)Permutation begin with C and ends with R : lhe first position can be filled in

only one way i.e., C and the last place can he filled in only one way ie R and Remaining 6letters can be arranged in 6 ! ways.The total number of permutations begin with C and ends with R is

= 1 x 6! x 1 720 Ans.(ii)Permutations in which C and R occupy end places : C and R occupy end positionsin 21 ways ie C,R and R,Cand the remaining 6 letters can be arranged in 6! ways.

The total number of permutations in which C and R occupy end places is2! x 6! = 1440 Ans.

Q. 6. How many 16-bit strings are there containing exactly five os?

Sol. A 16-bit string having exactly five zeros is determined if we tell which bits are

zeros.

This can be done inTherefore, the total number of 16-bit string is =

.

Q. 7. Define pigeonhole Principle?

Sol. Pigeonhole Principle : if n pigeons are assigned to in pigenholes and m


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Sol. Let J denote for JAVA

and 0 denote for ORACLE

n (J) = 25n (O) = 28

n (JUO) = 40-7= 33

20 knew both Languages.

Sectionll (Recurrence Relations and Generating Functions)

Q. 1. Define Recurrence Relation?

Sol. Recurrence Relation : Let S be a sequence of numbers. A recurrence relation onS is a formula that relates all, but a finite number of terms of s, to previous terms of s.

A sequence is called as a solutioi of a recurrence relation if its terms satisfy therecurrence relation. Recurrence relations arealso called difference equation. Recurrence relations are used to model variety of problems. Few problems areinherently recursive, viz., Towers of Hanoi, divide and computer compound interest,

conquer algorithms etc.

Q. 2. Define Linear recurrence Relation?

Sol. Linear Recurrence Relation : A recurrence relation with degree one is called Linear

recurrence relation.e.g. S (K) 2 S (Ki) 0 is Linear recurrence relation.Linear recurrence relation of order ii with constant coefficients : The general linear

recurrence relation of order n with constant coefficients, is given by

Where C1, C2.. Cn are constants.

Homogeneous Linear Recurrence relation of order n : A homogeneous Linear recurrence

relation of order n is an equation of the form

Characteristic Equation : The Characteristic equation of (1) is


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Q. 1. Solve the recurrence relation :

given that

Sol. Given recurrence relation can be written asS(r)7S(r1) + 10S(r2) =0 (1)Where S (0) = 0 and S (1) = 3

Its order = r - (r- 2) = 2The characteristic equation is

Therefore, the homogeneous solution of the equation

(2)Using,

Using,

Solving equations (3) and (4), for the value of

Then putting the value of c1 and c2 in (2), we get

Q. 2. Solve the recurrence relation

Sol. Given recurrence relation can be written as

Its characteristic equation is


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The solution of equation (1) is

using S (0) = 1, then equation (2) becomes

Also using S (1) =2, then n =(2) becomes

Q. 3. Solve the recurrence relation

Sol. Given equation is

Given recurrence relation can be written as

Its characteristic equation is

a = 2,4Solution is given by


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Q. 4. Discuss an algorithm on solving flth order linear homogeneousrecurrence relation?

Sol. Consider the linear homogenous recurrence relation of order n is

Step II: Find the roots of the characteristic equation obtained in Step I.

Q. 3. Find a close form expression (generating function) for the terms of

Fibonaccisequence.

Sol. Consider the Fibonacci sequence F (K), given by

F(K) = F(K-1) + F(K-2) ; K


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Its order = K - (K-2) -2

is the required generating function for the fibonacci sequence.


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Chapter 3 : Sets and Functions (Part 1)

[VERY SHORT ANSWER TYPE QUESTION]

Q. 1. Define following terms

(a) Set(b) Power set(c) Union and Inter section of two sets(d) Complement of a set

(e) Symmetric difference of two sets

Sol. (a) Set: A set is defined as a collection of distinct objects of same type or class ofobjects. The objects of a set are called element or members of the set. Objects can be

number, alphabets, names etc.Eg.A = {1,2,3,4,5}

A is a set of numbers containing element 1, 2,3, 4 and 5.(b) Power set : The power set of any given set A is the set of all subsets of A and isdenoted by P(A).

If A has n elements then P(A) has elements.

(c) Union of two sets: Union of two sets A and B is defined as the set of all thoseelements which belongs to A or B or both and is denoted by AUB. i.e.,

AUB = {x:x A or x B}eg.

Let A ={a, b, c, d) and B={c, d, e, f}

ThenAUB ={a, b, c, d, e ,f }

Intersection of two sets: Intersection of two sets A and B is the set of all those elements

which belongs to both A and B and is denoted by


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(d) Complement of a set : Let U be universal set and A be its subset of all those element

of U which do not belong to A, is called complement of A and is written as

Symbolically, ifand is called complement of A.

(e) Difference of two sets : The difference of two sets A and B is the set of all those

elements of A which do not belong to B and is denoted by A - B.

Symbolically,

The shaded area in Venns diagram represents A B.(f) Symmetric Difference of two sets : The set which contains the elements of eitherset A or set B but does not contain those elements which occur in both A and B iscalled symmetric difference of two sets A and B.

The symmetric difference of two sets A, B is written as AVB and symbolically written

as

Q. 2. Describe with example, the two basic ways to specify a particular set.

Sol. There are two ways of expressing a set.1. Roster form or tabular form : When the actual elements of the set are given, we

put them with in bracket { } and separate them with commas. e.g. The set A of oddnaturalnumbers 1,3,5, 7,9,11 is written as {1, 3,5,7,9, 11}


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2. Set Builder form or Defining property: When instead of actual elements, a rule isgiven with the help of which we can write the elements of the set, then it is said tobe set builder form. eg. A is the set of all those odd +ve integers which are less than12, gives us the same data in (1) only expressed in a different way. That is writtenas

T he dot : is read as such that.

Q. 3. Describe the following sets in set builder form:(a) A = {2, 3, 5, 7, 11, 13, 17, 19} (b) B = {3, 5, 7, 9 77, 79}

(c) C = The even Integer

Sol.

1. A = {x:x is a prime number}

2. B = {2K + 1, , 1


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Q. 6. If A = {1, 2,4, 5} and B = {a, b, c, f}, C = {a, 5} , Compute AUC and(AUC) x B.

Sol.AUC = {1,2,4,5,a}(AUC) x B {1, 2, 4, 5, a) x {a, b, c, f}

Q. 7. Let A = [+, -] and B = [00, 01, 10, 11] find A x B.

Sol,Ax B [+, ] x [00, 01, 10, Ii]= [+00, +01, +10, +11, 00, 01, 10,11].

Q. 8 Define comparable and non-comparable sets.

Sol. Two sets are said to be comparable if one of the two sets is a sub set of the other.

e.g. Let A ={2,4,6}B ={2,4,6,8}C = {1, 2}

Here A B

A and B are comparable sets

But A C and C A .

A and C are non-comparable sets.

Q. 9. What do you understand by fundamental product of sets

Sol. Consider n distinct sets

A fundamental product of the sets is a set of the

form

We Note,

(i) There are such fundamental products.

(ii) Any two such fundamental products are disjoint and

(iii) The universal set U is the union of all the fundamental products. There is ageometrical description of these sets which is given below e.g. consider three sets X, Yand Z. The following lists the eight fundamental product of three sets:


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These eight products correspond precisely to the eight disjoint regions in the venn -diagram of sets X, Y and Z in figure.

Q. 10. Let A and B be a set such that . Draw the venns

diagram.

Sol. Since given that then Venns diagram is

Q. 11. Given that AUB = AUC. Is it necessary thus B = C.

Sol. Let A {1, 2, 3, 4, 5}B = {2, 3}Here C = {1,4}

AUB = {1,2,3,4,5}AUC = {1, 2, 3, 4, 5}i.e. AUB = AUC

But B C. Therefore, it is not necessary to have B = C.

Q. 12. Is a statement


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Sol.A set with n elements has 2 subsets when n = 0 this statement is true because A

set has subset .

Q. 13. Differentiate between an ordered and unordered partition of a finite set.

Sol. Let S be a set with n distinct elements and f be a +ve integer. By ordered partition

of S, we mean a set of S , such that

A partition of a set A into a collection of subsets where the subsets are

unordered is called an unordered partition of the set A.

Q. 14. If A = {1, 2, 3}, find P(A).

Sol.A = {1, 2, 3}

P(A) = { , {1}, {2}, {3), {1,2}, {1,3}, {2,3}, {1,2,3}.

Q. 15. If A x B = B XA (where A and B are general matrices), then

(i) A =

(ii) A=B(iii) B = A(iv) A= B

Sol.(iii) B = A.

Q. 16. The description of the shaded region in the following figure using the operation on set is.

Sol.


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Q. 17. Let N = (1, 2, 3...) be ordered by divisibility which of following subset istotally ordered.

(i) (2, 6, 24)(ii) (2, 9, 16)(iii) (3, 5, 15)(iv) (4, 15, 30)

Sol. (i) (2, 6, 24), Since 2 divides 6 which further divide 24.

Q. 18. For A = {1, 2, {1, 3} } determine the following sets.

Sol.

(a) A - {1} =

(b) A - = A

(c) A - { } =

(d) A {1, 2} = {2, {1, 3}, }

Q. 19. If n is an integer and is odd, then n is(i) even(ii) even or odd(iii) odd

(iv} prime

Sol. (ii) Odd.

Q. 20. If A and B are two subsets of U then prove that

Sol.


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[ SHORT ANSWER TYPE QUESTION]

Q. 1. If A, B are any two sets then

Sol. To prove:

Q. 2. Let A,B,C be any three sets then prove that

Sol. Let

Q. 3. If A is any set, then (A) = A.

Sol. To prove

(A) =ALet

Now converselyLet


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from (1) and (2), we have(A) =A.

Q. 4. Prove that(a)

(b)

Sol. (a) Let

Therefore

Now converselyLet

Therefore ,

From (1) and (2), we have

Hence Proved.

(b)To prove:

Let

Now conversely, Let


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Find (1) and (2), we have

Hence Proved.

Q. 5. Prove that AU(B -A) = AUB.

Sol. Let

Hence,AU(B -A) = AUB

Hence proved.

Q. 6. Let A, B, C be arbitrary sets then show that (A-B) -C = A- (BUC).

Sol.

Let

Conversely Let ,

From (1) and (2), we have(A-B) - C = A - (BUC)

Hence proved.


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Q. 7. Prove that

Sol. Let

Now converselyLet

Therefore from (1) and (2), we have

Hence Proved.

Q. 8. Given that

is it necessary that Q = R ? Justify your Answer.

Sol. This is not necessarily true.

Since contains elements common to both P and Q and contains elements

common to both P and R. But it is not necessary that Q = R because set R can haveelements of Q which are elements of P as well, but it can also have element other thanthose in Q.

e.g.P = {1, 2, 3, 4}Q = {1, 2, 5}

R = {1, 2, 7}

Then

= {1 , 2}

= {1 , 2}


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Here


Q. 1. Prove thatA x (BUC) = (A x B)U(A x C)

Sol.Let

(x, y) A x (BUC)

X A and y BUCX A and (y B or y C)

Now converselyLet

(x, y) (A x B) U (A x C)

(x, y) (A X B) or (x, y) (AxC)

(x A and y B) or (x A and y C)

X A and y B or y C

X A and y BUC

= x A x (BUC)

(A x B) U (A x C) A x

(BUC) (2)

From (1) and (2), we get

A x (BUC) = (AXB) U(A X C)

Hence proved.

Q. 2. Prove that if A B then A x C B x C .

Sol. Let


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Therefore

Q. 3. Given that

Is it necessary that Q = R ? Justify your answer.

Sol. It is necessary that Q = R.

Because

=

i.e. this set contains element which are in sets P or Q but does not containelements common to both sets.

Similarly,

i.e. This set contains elements which are in sets P or R but does not contain

elements common to both sets.

As

R = Q

Q. 4. Two finite sets have in and ii elements. The total number of subsetsof the first set is 56 more than the total number of subsets of the secondset. Find the value of in and n.

Sol. Let A = first set n(A) = m

B = Second set n(B) = n

Then total no. of subsets of

Then total no. of subsets of

According to question

Hence n = 3 and m-n=3 m-3 = 3

M = 6 Ans.


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Q.5. IfA x B = B x A then A = B.

Sol.Let us assume that A = B.

To Prove

Now

conversely, let

To prove A = B

Let x be an arbitrary element of A, then

Let y be an arbitrary elements of B, then

from (1) and (2), we haveA = B Hence proved.

Q. 6. If A is any set then

Sol. (a) To prove:

We know that

Let


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from (1) and (2), we have

(b) To prove:

We know

Let

from (1) and (2), we haveAUA = A

(c)To prove:


1. To prove:

Clearly

Let


(e) To prove:

Clearly

Let



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1. Let

Let


1. To prove:

Let

1. To prove:

Let

1. To prove :

Let

Conversely ,


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1. To prove:

We have

Let

Hence Proved.

1. To prove:

We have

Let


Q.7. Prove that

Sol. Given:

To prove: A = B

Proof : Let


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Similary

Conversely : Let A = B

And

from these two results, we have

Q. 8. If then prove that .

Sol. Let

Then

Hence Proved.

Q. 9. If A and B are any two sets, then prove that

Sol. (i) Let

(ii) Let

(iii) Let


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SECTION II (RELATIONS)

Q. 1. Define the following terms.

(a) Relations(b) Inverse Relation

(c) Domain and Range of a Relation(d) Anti symmetric Relation(e) Partial order Relation

(f) Composition of Relation

(g) Equivalence Relation(h) Tarnary Relation

Sol. (a) Relation: Let U he the universal set and A and B are any two subsets of U, thena relation R in A x B is a subset of A x B.If

is related to element and we write

Such a relation is called binary Relation.

(b) Inverse Relation : Let R be a relation from a set A to another set B. Then, the inverserelation is a relation from B to A.

It is denoted by

(c) Domain and Range of a Relation: Let R be a relation from the set A to the set B.Then, the domain of R is the set of all first co-ordinates of the ordered pair belonging toR.

Range of R is the set of all second co-ordinates of the ordered pair belonging to R. e.g.A = {1, 3, 5, 7}B = {2, 4, 6, 8, 10}

And R = {(1, 8), (3, 6), (5, 2), (1, 4)} be a relation from A to WThen Domain of R = {1, 3, 5}Range of R = {2, 4, 6, 8}.

(d) Anti symmetric Relation: Let A be any set. A relation R on a set A is called an anti-

symmetric relation if

(e) Partial order Relation : A relation R on a set A is called partial order relation if it is:(i) Reflexive(ii) Anti-symmetric(iii) transitive.

(t) Composition of Relation : Let R and S be two relations from sets A to B and B to Crespectively. Then a relation SoR is called composite relation from A to C where (a,

c) S or if we can find b B S.t. (a, b) R and (h,c) S.

The relation SoR is read as composition of R and S.

Also

And


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e.g. LetA = {1 , 2 , 3}

B = {a , h, c}

C = {p, q, r} be three sets such that

R = ((1, a), (2, 5), (1, c)} is a relation from A to B S = {(a, p), (b, q), (c, r)} is a relation from B to C.

The relation SoR is a relation from A to C is given by

SoR = {(1, p), (2, r), (1, r)}.

(g) Equivalence Relation : Let A be any set. Then a relation R on A is said to be anequivalence relation on A if it satisfies the following (PTU May 2009)(i) Reflexive: A relation R on a set A is said to be a reflexive relation if every element ofa is related to itself. Thus, R is reflexive iff (x, x) ER V xEA.(ii) Symmetric : Let A be any set. Then a relation R on a set A is called symmetric if

(iii) Transitive: Let A be any set. Then a relation R on a set A is said to be transitive if

(h) Tarnary Relation : It is a set of ordered triples. In particular, if S is a set, then asubset of S x 5 x S is called a tarnary relation on S.If L is a line, then betweenness is a tarnary relation among points of L.

Q. 2. Let A = (1, 2, 3, 4 and Let r be the relation on A. Draw the digraph and theHasse diagram of r.

Sol. r {(1., 1), (1., 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}

Q. 3. Give an example that shows that R does not necessarily partition a set

of Mathematicians?

Sol. Here the set of mathematicians is given by the set of all pairs (a, h). Consider the collections of (a, h) as.

{(a, h) (h, a))Since both are not disjoint.They cannot form a partition of set of mathematicians.


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Q. 4. Let R be a relation on A = {2, 3, 4, 5, 6) defined by x is relative prime toy.

Write R as a set of ordered pair.

Sol. We know two integers x and y are said to be relative prime iff (x, t) = 1. i.e. g.c.dof(x,y)

The required set of ordered pair is given by R {(2, 3), (2, 5), (3, 4), (3, 5), (4, 5), (5, 6)).

Q. 5. Find the numbers of relation from A {a, b, c} to B = fi, 2).

Sol. We know that if a set A contain m elements and set B contains n elements, thentotal number of relations from A to B is 2imHere m = 3 and n = 2.

Required no. of relations from A to B = = 64 Ans.

Q. 6. A partial order relation is transitive reflexive

and

Sol. (i) Anti-symmetric.

Q. 7. A partial ordered relation is reflextive, antisymmetric and

Sol. (i) Transitive.

Q. 8. The relation {(1, 2), (1, 3), (3, 1), (1, 1), (3, 3), (3, 2), (1, 4), (4, 2), (3,4)} is

Sol. (ii) Transitive.

Sol. A = (0)

B = {1, 2}We know that if a set A contains m elements and set B contain n elements then total

no. of relations from A to B is

Here m = 1 and n = 2Then total No. of relations from A to B is

Ans.


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Q. 11. How many different reflexive, symmetric relation are there on a setwith three elements.

Sol. Let A = {a, b, c}Let r be a relation on A with adjancency matrix M.

Case I : When the diagonal entries of the matrix M are each equal to unity, then therelation is reflexive.Case II : When the three entries, above the principle diagonal are determined andalso the three entries below the principal diagonal are also determined then the relation

is symmetric. Combining the above two cases, total number of required relations

= = 8.

Q. 12. What are the properties for a relation to be equivalence relation?

Sol. Let A be any non-empty set A relation r on A is said to be an equivalence relation ifit satisfies the following properties

(1) r is reflexive is for

(ii) r is symmetric i.e.

(iii) r is transitive i.e.

Q. 1. Let be a one-one function where Z is a set of integers

and N is a set of natural numbers. Let R be a relation on A defined as under:

Prove that R is a partial order relation.

Sol. To prove: R is a partial order relation.Proff: For R to be partial order relation we have to prove that R is reflexive,

antisymmetric and transitive.

i. Reflexive: Now

f is reflexive.

i. Anti-symmetric : Let

From (1) and (2)


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i. Transitive : Let

R is transitive.

R is partial order relation on A.

Q. 2. Let A = 11, 2, 3, 4) and let R and S be the relation on a described by

using warshalls algorithm to compute the transitive closure of RUS.

Sol.


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Q. 3. Prove that following statement are equivalent. P1 : n is even Integersn -1 is odd integer

P3: n2 is even integer.

Sol.

Let Integers are

0, 1, 2, 3, 4even Integers are: 0, 2, 4.


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odd integers are : 1, 3, 5

Therefore ,given statements are equivalent.

Q.4 . In inclusion of a subset in another, in the context of a universal set anequivalence relation in the class of subset of sets ? Justify the answer.

Sol. The relations inclusion of a subset in another is not symmetric. As if A is a subsetof B, But B may not be a subset of A, so relation is not equivalent. No, inclusion is not

symmetric.

Q. 5. Prove that mod m relation is an equivalence relation.

Sol. (a) Reflexivity:

n n(mod m) as n-n = 0 m(b) Symmetry:

Ifn = k (mod m ), thenn - k = rm, for some integer r, thusk-n = -r(m) and k = n mod(m)(c) Transitive

Ifa = b (mod m) and b = c (mod m), thena - b = rm and b - c = s m, for some integers r and s, Thusac =a -b+b-c = (r+s)m

a = c(mod m)


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Q. 6. Suppose R and S are symmetric relation on a set show that is alsosymmetric.

Sol. Let

Then

As R and S are symmetric

Q. 7. If A is a set containing n elements. Find the number of relations from A to

A.

Sol. We know, if aset A has in elements and B has n elements, then total no. of

relations from A to B is 2.Therefore, total no. of relations from A to A is

Ans.

Q. 8. Prove that a relation R on a set A is symmetric if

Sol. Let R is symmetric on A. We show

Let

from (1) and (2)

Converse :

LetTo show R is symmetricLet


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Therefore , R is symmetric.

Q.9. Characterize the matrix of an irreflexive relation. What about that relationthat is neither reflexive nor irreflexive.

Sol.There should be only zeros on the diagonal of the matrix of an irreflexive relation.There should be at least one zero and atleast one I on the diagonal of the matrix of a

relation that is neither reflexive nor irreflexive.

Q. 10. If A = {a, b, c} and r, s are relation on A, whose matrices are :

Sol.

Sor = {(a, a),(a, c),(b, a),(b, b),(b, c),(c, b),(c, c)} Ans.


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Section III (FUNCTIONS)

Q. 1. Define functions.

Sol. Function: A function f from a set P into set Q is a relation from P to Q such that

each element of P is related to exactly one element of tile set Q. It is denoted by

Q. 2. Discuss the types of Function.

Sol. Types of Function

1. Injective (one to one) function.

2. Surjective (onto) function.3. Bijective (one to one onto) function. 4. Into function.5. One-one into function.

6. Many one function.7. Many one into function.8. Many one onto function.

Q. 3. Let

Sol. Let us assume the functions have same domain, they are equal if for each domain element, the image of that element under the two functions are equal. We want to provethat

Similarly,

from (1) and (2)

Q. 4. A function f : X -> Y will be invertible if f is one to one and onto.

Sol. Case I: When f: X -> Y is invertible.f possesses inverse say


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i. To prove f is one to one

Let

Therefore , f is one to one.

i. To prove f is onto:

Therefore , f is onto.

Q. 5. What are domain, co-domain and image of a function.

Sol. Domain of a function : Let f be a function from P to Q such that every element of iscalled the domain of function.Co-Domain of a function: Let f be a function from P to Q. The set Q is called codomain ofthe functionImage of a function: If the element x of P corresponds toy under function f then y is the

image of x under f and is written as f(x) = y.If fix) = y, then we say that x is a pre image of y.

Q. 6. Let A = B = (1, 2, 3, 4, 51. Define a function f: A B s.t. f is one-one and

onto function.

Sol. Let

f(1) = 1,f(2) = 2,f(3) = 3

f(4) =4,f(5)=5.

i.e. different element of A have different images in B. f is one-one.

Also each element of B is the image of some element of A. f is onto also.


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Q. 7. Find the formula for the inverse of

Sol.

exists if function is one-one and onto.One-one :Let

x = yTherefore, f is one-one.

Onto:Let

Therefore ,f is onto.

Now,

f (x) = y gives

Q. 8. Find the formula for inverse of

Sol. Let


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Q.9. Give an explicit formula for a function.(a) One-one not onto.

(b) onto not one-one(c) one-one and onto(d) neither one-one nor onto.

Sol. (a) One-one and not onto:

(b) Onto not one-one

(c) One-one and onto

(d) Neither one-one nor onto:

Q.10 . What are Types of Functions? Explain with example.

Sol. 1. Injective (one-to-one) Function :

Let f: X -> Y. The functionf is called one-to- one or injective if different element m Xhave different image in Y

e.g. Consider

X = {1, 2, 3, 4}and Y = {a, b, c, d}

and f is a function from X to Y st.f = {(1, a),(2, b),(3, c), (4, d)}


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The function f is one-to-one as every element of domain X has a unique image in the

co-domain Y

2. Surjective (onto) Functions: (PTU Dec. 2004)Let f: X> Y. The function f is called surjective function if each element in Y, is the

image of at least one element in X. In other words, the range off is equal to co-domain Yi.e.

e.g. Consider,X = {1, 2, 3, 4, 5}

Y = {a,b,c,d}f = {(1, a),(2, a),(3, b), (4, c), (5, d)}It is a onto function, as every element of Y is the image of some element of X.

3. Bijective (one-to-one onto) function: (Dec. 2004)A function which is both one-to-one and onto is called bijective (one-to-one onto) function.e.g. ConsiderA = {x, y, z}B = {a, b, c}and f: A -> B s.t. f= {(x, a), (y, b), (z, c)}The f is one-to-one and also it is onto. So it is a bijective function.

4. Into functions:Let f: X -> Y. The function f is called into function if the range of f is not equal to co-

domain of Y. Therefore, there must be an element by co-domain Y which is not theimage of any element of domain X.

e.g. consider X = (1, 2, 3}, Y = {a, b, c, d, e} and f: X -> Y s.t.f = {(1, a),(2, d),(3, e)}In the function, f the range i.e. (a, d, e} is not equal to co-domain of Y. Therefore, it isan into function.


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5. One-one into functions :Letf: X -> Y. The function f is called one-one Into function if different elements of X havedifferent unique image of Y.

X = {a, b, c}Y ={1, 2, 3, 4}And f: X -> Y such that f= {(a, 1), (b, 3), (c, 4)}The function f is one-one into function.

e.g. Consider

6. Many one functions :

Letf: X -> Y. The function f is said to be many one function if 3 two or more than twodifferent element in X having the same image in Y.

e.g. Consider X = {1, 2, 3, 4, 5) and Y (a, b, c} and f: X -> Y such thatf= {(1, a), (2, a), (3, a), (4, b), (5, c)} The function f is a many one function.


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Chapter 4 : Monoids and Groups

Q. 1. Define the following terms(a) Group(b) Subgroup

(c) Abelian group(d) Semi group

(e)Monoid

(f) Congruence Relation(g) Coset(h) Isomorphic

Sol. (a) Group Let us consider an algebraic system (G, *), where * is a binary operation C. Then the system (G, *) is said to be a group if it satisfies the following properties:(i) The operation * is a closed operation.

(ii) The operation * is an associative operation

(iii) There exist an identity element w. r. t the operation . (iv) For every aEG, there exist an element a1 E G s.t.

(b) Sub Group : Let G be a group and Let is called a subgroup iff it

satisfies the following propertries.(i) The operation * is closed operation on S.(ii) The operation * is an associative operation on S.(iii) As e is any identity element belonged to G. It must belong to the set S i.e. theidentity element of (G, *) must belong to (S, *).

(iv) For every element , a also belongs to S.

(c) Abelian Group : Let us consider, an algebraic system (C, *), where * is a binaryoperation on G. Then the system (G, *) is said to be an abelian group if it satisfies aBthe properties of the Group plus an additional following property i.e.(i) The operation * is cummulative i.e.

(d) Semi-Group : Let us consider an algebraic system (A, *), where * is a binary

operation on A. Then, the system (A, *) is said to be a semi-group if it satisfies thefollowing properties:(i) The operation * is a closed operation on Set A.(ii) The operation * is an associative operation.

(e) Monoid : Let us consider an algebraic system (A, 0) where 0 is a binary operation on

A. Then the system (A, 0) is called monoid if it satisfies the following properties:

(i) The operation0 is a closed operation on the Set A.

(ii) The operation 0 is an associative operation

(iii) There exist such that for all

(j) Congruence Relation: Let C be a semigroup and Let R be an equivalence relation onG. We know that the equivalence relation R inducesa partition of G into equivalenceclasses, where [x] represents the equivalence class containing the element x in C and

that the collection of equivalence classes is denoted by G/R. Let the equivalence relation R on C has following property:

Then R is called a congruence relation on C.

(g) Coset : Let H be a subgroup of G and Let be any element. Then called


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or define Right coset and left coset.

Let be a group and H be a subgroup of G. Let aEG. The tight coset of H w,r.ta

written as Ha is the set

Similarly the left coset of H w.r.t a is written as aH is the set

The subset H is itself a left and right coset since

When we use addition notation, then

(h) Isomorphic : The groups (G, o, e) and (H,*,i) are isomorphic if there is a bijection

The bijection f is called an isomorphism.

Q. 2. Show that the identity element on a group is unique.

Solution.Let us assume that there exist two identity element by C i.e. e and e.

We have

Similarly

we have

from (1) and (2), we have

Q. 3. Let (A, *) be an algebraic system s.t. for all

(a*b) * a = a ...(1) and(a*b)*b = (b*a) * a ---(2)

Show that

Sol. (i)a * (a * b) = [(a * b)*a] * (a * b) [using (ii)]= a * b(ii) Let

a*b =b*a (3)Asa = (a*b)*a= (b * a) * a [by (3)]

= (a * b) * b [using (2)]= (b * a) * b [using (3)]


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= b [using (1)]

Q. 4. Prove that for any commutative monoid (A, 0) the set of idempotent

element of A forms a sub monoid.

Sol. Let S be the set of idempotent elements

To prove S is a sub monoid

i.e. S is closed under operation 0.Also

i e S has an identity element S is a sub-Monoid.

Q. 5. Let s be a semi group with identity e and Let b and b are inverse of a.Show that b = b i.e. inverses are unique, if they exist.

Ans As S is a semi-group with identity e and b & bare inverses of a

b*a =e=a*h (1)&b*a=e=a* b (2)

As S is a semi-group under *, therefore associativity property holds in S under operation.*i.e.

b*(a*b) = (b*a)*bb*e = e* b [using (1) and (2)]b= bHence Proved.

Q. 6. Let denotes the set of integers {0, 1,2, , n 1) Let be the binary

operation on , s.t.

Show that is a semi-group for any n.

Sol. Since = {O, 1, 2,, n 1)

Let

Now

where r is the least + ye remainder when ab is divided by n. We shall show that r liesbetween 0 and n 1 i.e. 0 r < n I

If n does not divide a and n does not divide b, then dividing ab by n. Let q be the quotient and r, the remainder, where r < n


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Multiplication modulo n, is a binary operation.Associative:

as each will leave the same remainder when the ordinary product abc is divided by n.therefore, by the definition of semi-group

is a semi-group, under given binary operation.

Q. 7. A group is (a) Also a semi-group (b) also a ring (c) an Integral domain (d)

None of these.

Ans. (a) Also a semi-group.

Q. 8. Let (A, *) be a semi-group. Let a be an element in A. Consider a binary

operation on A, s.t. for every x and y in A.

Show that is an associative operation.

Sol. Consider

Consider

from (1) and (2), is an associative operation.

Q. 9. Let (A, x) be a commutative semi-group show that if a *a = a and b * b =bThen

Sol. Consider L.H.S.

Since associativity hold in a semi group then

Then from (1), we have


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Q.12.Ifa,b,c are elements of a group G and a*b =c*a. Then b = c? Explain youranswer.

Ans. Given that

Consider

We claim

given that

Hence the given statement is a false statement.

However, the given statement is a true statement if G is an abelian group. In this case

Left cancellation Law

Q. 13. Define Normal subgroup?

Sol. Normal Subgroup : A subgroup H of a group C is called normal subgroup of G if for

every

Q. 14. Define cyclic group?

Ans. Cyclic Group A group G is called cylic if for some aEG, every element xEG is of the

form for some The element a is called the generator of C.

If G is cyclic, we write G =

Q 15. Define Homomorphism of Groups.

Sol.A mapping from a group (G, ) in to a group is said to be a group

hornorphism if

Q. 16. Define Group Isomorphism.

Sol. A homorphism which is one-one and onto is called isomorphism and the group Cand G are called isomorphic, written


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A homomorphism which is onto is called epimorphism. A homomorphism which is one-one called monomorphism.

Q. 1. Prove that intersection of two normal subgroups is again a normalsubgroup.

Sol. Let H and K be two normal subgroups of G.

Since H and H are normal subgroup C then we shall prove that

is a normal subgroup of G.

Let x be any element of G and h be any element of

We have

Since H is a normal subgroup of G

Similarly K is normal subgroup of G,

From (1) and (2), we get

Thus we have,

Fleiice is a normal subgroup of G.

Q. 2. Let H and K be subgroup of group G. Neither of which contains the other.Show that there exists an element of belonging neither to H nor K.

Sol. Given H and K are subgroup of G and

To show:

Proof.

Now,and since H is a subgroup of G

But and since K is subgrouped of G.

from (1) and (2 , a contradiction.Hence the theorem is proved.


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Q. 3. If H and k are two subgroups of G, then is also a subgroup of G.

Sol. We know that a subset H of a group G is a subgroup of G iff

Let

To show:

Now

Also

Since H is sub group of C and a, b, H, then,

Also If K is a subgroup of G and a, b K, then


Hence is a sub group of G.

Q. 4. Prove that congruence modulo H, is an equivalence relationin G.

Sol. Let G be a group, H is a subgroup of G. For a, , we say a is congruent to bmode H, written as

An equivalence relation on a set A is a relation on A which satisfies all the properties.

1. Reflexivity:

Let

Then

Hence the relation is reflexive2. Symmetry :

We have

Since H is a subgroup of G.

There the relation is symmetric


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3. Transitivity:

Let

And

Then

Since H is a subgroup of G.H must be closed w.r.t the composition in G.

Hence the relation is transitive.

Therefore, the relation is an equivalence relation in G.

Q. 1. State and prove Lagranges Theorem.

Ans. Statement: If G is a finite group and H is a subgroup of G then o(H) or o(G)Proof Since H is a subgroup of finite group C:. H is also finite, say

Hence H a has distinct elements.Now G is finite

The number distinct right cosets of H in G is also finite, say, K Let

Q.2 . How Group theory is applied in according to Theory.

Sol. By using group code, we can find the solutions of many coding problems by using

Group Theory. A coding problem is a problem which is used to represent distinctmessages by mean of a sequence of letters from a given alphabet. A sequence of lettersfrom an alphabet is called a word. A code is a collection of words that is used to


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represent distinct messages. A word in a code is called code word. A code consisting ofwords that are of same length is called black code.

The concept Group Theory can be applied in many situations which arise in codingproblem which is clear from the following discussion.In error correction: Suppose a code word is transmitted from the origin to itsdestination. During the course of transmission, some of the information might cause

some of 1 in thecode word to be received as 0s and some of the os to be receivedas 1s during the course of transmission. (This may be due to some interferences suchas noises)

Let A denote the set of all binary sequence of length n, Let be a binary operation on

A such that for x, . Let denotes the sequences of length n that has ls when x

and y differ and 0s when x and y are same. We observe that is a group with allzero word as its identity and every word is its own inverse.

By employing the minimum distance coding criteria, it has been observed that a code ofdistance 2r + 1 can correct r or lesser transmission error.

Q. 3. Prove that the set G = {1, 2,.., 6} is a finite abelian group of order 6

with respect to multiplication modulo 7.

Sol. The order to prove C is finite albelian group of order 6 w.r.t. multiplication modulo7, Let us form the composition table.

We see that all the entries in the composition table are elements of the set G. Therefore

G is closed w.r.t multiplication modulo 7 i.e.

(ii)The composition X7 is associative. If a, b, c are any three elements of G

then= Least non-negative remainder when a(bc) is divided by 7.= Least non-negative remainder when (ab)c is divided by 7.

(iii) Existence of Identity : We have If a is any element G, then from thecomposition table. We see that

1 is the identity element.

(iv) Existence of Inverse From the table, we see that the inverse of 1, 2, 3, 4, 5, 6 are1, 4, 5, 2, 3, 6 respectively.

The composition is commutative as the corresponding rows and columns in thecomposition table are identical.

The set C has 6 elements. Hence is a finite abelian group of order 6.


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Chapter 5 : Rings and Boolean Algebra (Part 1)

SectionI (Rings)

Q. 1. Define Ring.

Sol. Let R be a non-empty set with two binary compositions, addition (+) and

multiplication (.). Then R is called a ring iff it satisfies the following.

I. R is an abelian group under + i.e.,

I. For each a, b R, a.b R i.e.,

R is closed under multiplication.

I. For a, b, c R, a.(b.c) = (a.b).c i.e.,

Associativity under multiplication holds in R.

I. For a, b, c R,

(i) a.(b + c) = a.b + a.c. (Left distribution Law)(ii) (a + b).c = a.c + b.c. (Right distribution Law)

Q. 2. What is a ring with identity.

Sol. A ring R is called ring with identity if for each x R,3 IER such that 1.x = x = x. 1.

The element 1 is called multiplicative identity of R.


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Q. 3. Define subring?

Sol. Let {R, +,} be a ring and S be a subset of R. Then S is called a subring of R iff Sis itself a ring under the operation of R.

Q. 4. Define Integral Domain.

Sol. Integral Domain : A commutative ring R is called an integral Domain if for every O

a, b R, ab=0

a=0

Orb=0Thus, a commutative ring R is called an integral domain if R has zero divisor.

Q. 5. Define field?

Sol. Field : A commutative ring with unity such that each non-zero element has a

multiplicative inverse is called field. It is denoted by F.

(a b)

Q. 6. Let D be the ring of all real 2 x 2 matrices of the form . Show that

D is isomorphic to the complex number C where D is a field.

Ans.

Let D is a ring of real 2 x 2 matrices of the form

Define

To show :f is homomorphism, one-one and onto.Let

Consider


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Thus f is a homomorphism.

Further, Let

Equating real and Imaginary parts, we get

Thus

Hence

defined by

Q. 7. Define Improper ideals.

Sol. A non-empty set k of a ring R is called an ideal (or two sided ideal) of R if k is bothleft ideal and ring ideal of R i.e.,

Every ideal other than and R are known as proper ideals. The field and Rare improper ideals of R.

Q. 8. Define Euclidean ring (Domain)

Sol. Let R be an integral domain and suppose that for every 0 a ER, there exists a

non-negative d(a) satisfying the following.

Then R is called Euclidean domain


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Q. 1. Define Quotient Ring ? Explain with suitable example.

Sol. Quotient Ring : Let R be a ring and I be an ideal of R. Define R/I, by

Then R/I is also a ring under the addition and multiplication defined by

The ring defined by R/I is known as quotient ring.

Example : Consider

Let

And

Also if

Hence H4 is an ideal of z.Further,

Thus

Q. 2. Show that relation of being associates is an equivalence relation in a ring

R.

Sol. Reflexive : For we can write x = x. 1 where 1 is a unit in R.

Hence reflexive.

Symmetric : Let there exist some unit such

discreet structures

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