real analysis hw1

7/30/2019 Real Analysis HW1

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Seung min Oh

Real Analysis

HW#1

Problem A. In no more than four sentences, describe the mainthemes and concepts of Chapter 1. The first sentence or two

should be at a level that your parent could understand even if

they never went to college. The other sentences should be

understandable by any college student.

A number system allows us to perform basic arithmetic such as

addition and multiplication and compare two elements in the number

system and conclude their relative size. In addition, for some

collection of numbers, there is a number which is greater than or less

than any chosen number from the collection, which we call upper

bounds and lower bounds respectively. Rational numbers, real

numbers and complex numbers all fulfill this requirement.

Problem B. Recall that in class, we defined a rational numberm/n to be an equivalence class of pairs (m,n) under an

equivalence relation. Check that this equivalence relation is

transitive: if (p,q)~(m,n) and (m,n)~(a,b), then (p,q)~(a,b).

Recall that transitivity implies if aRb is true and bRc is true that aRc is

true. In addition, we have defined rational numbers as the set of m/n

in which m,n are integers and n does not equal 0 and the equivalence

relation for (p,q)~(m,n) as pn=qm.

Therefore in order to demonstrate transitivity in this relationship, we

must show that (p,q)~(m,n) and (m,n)~(a,b), then (p,q)~(a,b).


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By the defined equivalence relation, pn=qm (1) and ma=nb. (2)

Recall that for integers for c that does not equal 0, ca=cb implies a=b.

Multiplying a to (1), we obtain pan=qam. Similarly multiplying q to (2),

we obtain qma=qnb. Hence, pan=qam=qnb.

Since n does not equal 0, we can divide by n for pan=qnb and obtain

pa=qb which is the definition of (p,q)~(a,b).

Therefore, we have demonstrated transitivity for rational number pairs

(m,n)

Problem C. We defined addition of rational numbers in terms ofrepresentatives: a/b + c/d = [ad+bc]/[bd]. Show that the

addition of rational numbers is well-defined.

In order to show that a/b + c/d = [ad+bc]/[bd] is well-defined,

Let (a,b) ~ (a,b) and (c,d) ~ (c,d). Then We want to show that

(ad+bc, bd) ~ (ad+bc, bd).

Since a/b + c/d = (ad+bc)/bd = a/b + c/d = (ad+bc)/bd.

Hence choosing any elements from the classes of a/b and c/d and

adding them together leads to the same results.


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Problem D. Define a multiplication of rational numbers, and show this

multiplication is well-defined.

In order to show that multiplication is well-defined, we will assume

(a,b)~(a,b) and (c,d)~(c,d) and show that (ac,bd)~(ac,bd).

Since a/b * c/d = ac/bd = a/b * c/d = ac/bd.

Hence, multiplication is well-defined.

Do also Chapter 1 ( 1, 2, 3a ). Rudin 1.1 If r is rational (r0) and x is irrational, prove that r+x

and rx are irrational.

Proof (by contradiction)

Suppose r+x and rx are rational.

Recall that addition of Q is well-defined which means that addition of

rational numbers can be defined as rational numbers.

By the addition axioms, r+x-x = r is rational which contradicts our

assumption that r is rational.

Similarly, suppose rx is rational.

Recall that multicplication of Q is well-defined meaning that

multiplication of rational numbers can be defined as rational numbers.

By the multiplication axioms, rx * 1/x = r is rational which contradicts

our assumption that r is rational


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2. Prove that there is no rational number whose square is 12.

Proof by contradiction

Suppose there is a rational number a/b (a,b are integers and b does not equal 0) whose square is

12 and a,b have no common factors.

Then (a/b)^2 = 12 and thus a^2 = 12b^2.

Since a^2 is a multiple of 12, a is a multiple of 12. (if a is not a multiple of 12, then a^2 is not a

multiple of 12).

Hence we can write a^2=144k^2 for some k leading to k^2 = 12b^2 and thus b is a multiple of

12.

Since a and b are both multiples of 12, the assumption that a,b have no common factors is not

true.

Hence, there is no rational number whose square is 12.

3. Prove Proposition 1.15.

(a) By multiplication axioms, y= 1 * y = x *1/x * y = 1/x * x * y = 1/x * (xy) = 1/x * (xz) = 1/x * xz

= z.

(b) Set z=1 and apply (a).

(c) Set z=1/x and apply (a)

(d) By M5, set x as 1/x and we obtain (d).

real analysis hw1

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