ics 253 presents mathematical induction sultan almuhammadi email: muhamadi (@kfupm.edu.sa)
TRANSCRIPT
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ICS 253Presents
Mathematical Induction
Sultan AlmuhammadiEmail: muhamadi (@kfupm.edu.sa)
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Mathematical Induction
What is it?
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Mathematical Induction
Is it good?
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Mathematical Induction
Is it easy?
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Mathematical Induction
Is it complicated?
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Mathematical
Do you like it? Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
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Mathematical Induction
Dominos Effect
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Mathematical Induction
The First Principle of Induction:
Let P(n) be an open sentence about positive integers, 1,2,3,… and assume the following:
(a) P(1) is a true statement.
(b) k 1, if P(k) is true then P(k+1) is true.
Then, we conclude that P(n) is true for all n 1
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Mathematical Induction
So what?
We can prove statement P(n) is true for all n in simple two steps:
1. Basis step: P(1)
2. Inductive step: P(k) P(k+1)
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Mathematical Induction
Example:
What is the sum of the first n positive integers?
SUM(n) = 1+2+3+…+n = ?
What is SUM(3)?
SUM (3) = 1 + 2 + 3 = 6
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Mathematical Induction
Problem:
Find SUM(n) = ??
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Mathematical Induction
Solution:
SUM(n) = n(n+1) / 2
Prove it !!
Well, we know SUM(3) = 1+2+3 = 6
Try RHS, 3(3+1) / 2 = 3*4 / 2 = 6
OK ??
But this is not a proof for all n.
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Mathematical Induction
Problem:
Prove the statement P(n) for all n.
P(n): SUM(n) = n(n+1) / 2
Example, P(5) read:
The sum of the first five integers is 5(5+1)/2
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Mathematical Induction
OK, try this:
P(1): SUM(1) = 1, and 1(2)/2 = 1.
P(2): 1+2 = 3, and 2(3)/2 = 3.
P(3): 1+2+3=6, and 3(4)/2=6.
P(4): 1+2+3+4=10, and 4(5)/2=10.
P(5): 1+2+3+4+5=15, and 5(6)/2=15.
and so on… Therefore, P(n) is true for all n.
Do you buy this proof??!
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Mathematical Induction
Prove SUM(n) = n(n+1)/2
Observation:
SUM(k) = 1+2+3+…+ k
SUM(k+1) = 1+2+3+…+ k + (k+1)
So, SUM(k+1) = SUM(k) + (k+1)
There is a relation between P(k) and P(k+1)
P(k): SUM(k) = k(k+1)/2
P(k+1): SUM(k+1) = (k+1)(k+2)/2
Show that P(k) P(k+1)
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Prove SUM(n) = n(n+1)/2
SUM(k+1) = SUM(k) + (k+1)
P(k): SUM(k) = k(k+1)/2
P(k+1): SUM(k+1) = (k+1)(k+2)/2
Show that P(k) P(k+1)
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Prove SUM(n) = n(n+1)/2
SUM(k+1) = SUM(k) + (k+1)
P(k): SUM(k) = k(k+1)/2
P(k+1): SUM(k+1) = (k+1)(k+2)/2
Show that P(k) P(k+1)
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Prove SUM(n) = n(n+1)/2
SUM(k+1) = SUM(k) + (k+1)
P(k): SUM(k) = k(k+1)/2
P(k+1): SUM(k+1) = (k+1)(k+2)/2
Show that P(k) P(k+1)
SUM(k+1) = SUM(k) + (k+1)
= k(k+1)/2 + (k+1)
= k(k+1)/2 + 2(k+1)/2
= [ k(k+1) + 2(k+1) ] /2
= [ k(k+1) + 2(k+1) ] /2
= [ (k+1) (k+2) ] /2
Therefore, if SUM(k) = k(k+1)/2 then SUM(k+1) = (k+1)(k+2)/2
So, P(k) P(k+1)
We know that P(1) is true. What else can we imply?
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Prove SUM(n) = n(n+1)/2
SUM(k+1) = SUM(k) + (k+1)
P(k): SUM(k) = k(k+1)/2
P(k+1): SUM(k+1) = (k+1)(k+2)/2
Show that P(k) P(k+1)
SUM(k+1) = SUM(k) + (k+1)
= k(k+1)/2 + (k+1)
= k(k+1)/2 + 2(k+1)/2
= [ k(k+1) + 2(k+1) ] /2
= [ k(k+1) + 2(k+1) ] /2
= [ (k+1) (k+2) ] /2
Therefore, if SUM(k) = k(k+1)/2 then SUM(k+1) = (k+1)(k+2)/2
So, P(k) P(k+1) (This is called the inductive step)
We know that P(1) is true. What else can we imply?
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Mathematical Induction
To prove statement P(n) for n = 1,2,3,... Basis step: Show P(1) is true. Inductive Step:
Assume P(k) is true. Show that P(k+1) is true based on the assumption.
This is called the first principle of induction. In the second principle of induction (or strong induction), we
assume that P(j) is true for all j k and then we prove P(k+1).
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Exam Tip:Prove by induction (First Principle)
Show that 1+2+ … + n = n(n+1)/2 using induction. Solution Template P(n): 1+2+ … + n = n(n+1)/2 Basis Step: P(1) is true because … (~25%)Inductive Step: Assume P(k) is true, 1+2+…+k = k(k+1)/2 Now, we show that P(k+1) is true. (~25%)
In this part you must use the assumption.
By induction principle, P(n) is true for all n 1.
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Exam Tip:Prove by induction (First Principle)
Show that 1+2+ … + n = n(n+1)/2 using induction. Solution Template (First Principle ONLY)P(n): 1+2+ … + n = n(n+1)/2 Basis Step: P(1) is true because … (~25%)Inductive Step: Assume P(k) is true, 1+2+…+k = k(k+1)/2 Now, we show that P(k+1) is true. (~25%)
In this part you must use the assumption.
By induction principle, P(n) is true for all n 1.
![Page 31: ICS 253 Presents Mathematical Induction Sultan Almuhammadi Email: muhamadi (@kfupm.edu.sa)](https://reader036.vdocuments.site/reader036/viewer/2022062314/56649e1b5503460f94b09dd8/html5/thumbnails/31.jpg)
Example:
Prove 11n – 6 is divisible by 5, for n=1,2,3…P(n): 11n – 6 is divisible by 5Basis Step: for n = 1,P(1): 111 – 6 = 5 which is divisible by 5.So, P(1) is true.Inductive Step:Assume P(k) is true, 11k – 6 is divisible by 5.We show that P(k+1) is also true. i.e. 11k+1 – 6 is also divisible by 5 ---------- ~50% is obtained upto this line --------- (Remember: use the assumption here)
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Example: P(n): 11n – 6 is divisible by 5
Inductive Step:Assume P(k) is true, 11k – 6 is divisible by 5.We show that S(k+1):11k+1 – 6 is divisible by 5 11k+1 – 6 = 11 . 11k – 6
= (10+1) . 11k – 6= 10 . 11k + 11k – 6 = 10 . 11k + 11k – 6
But 10 . 11k is divisible by 5, and 11k – 6 is divisible by 5 (by assumption).Therefore, 11k+1 – 6 is divisible by 5, and hence P(k+1) is true. By mathematical induction, P(n) is true for all n 1
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Exam Tip:Prove by induction (Second Principle) E.g. Show that the statement is true for all n=1,2,3,…
Solution Template P(n): put the statementBasis Step: Show that P(1), P(2),.. are true by testing/substitution … (~25%)Inductive Step: Assume P(j) is true for all j kNow, we show that P(k+1) is true. (~25%)
In this part you must use the assumption.
By induction principle P(n) is true for all n 1.
![Page 34: ICS 253 Presents Mathematical Induction Sultan Almuhammadi Email: muhamadi (@kfupm.edu.sa)](https://reader036.vdocuments.site/reader036/viewer/2022062314/56649e1b5503460f94b09dd8/html5/thumbnails/34.jpg)
Prove that any integer n > 1 can be written as the product of primes
P(n): n can be written as the product of primes.
Basis Step: P(2) is true since 2 is prime
Inductive Step: Assume P(j) is true for all j k
Example on Second Principle.
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Example on Second Principle.
Want to show: P(k+1) Case 1:
if k+1 is prime, then P(k+1) is true and we are done. Case 2:
if k+1 is composite, then there are two integers, a, b, such that k+1 = a.b and 2 a b < k+1
By assumption, P(a) and P(b) are true, which implies that a and b can be written as the products of prime.Therefore, a.b = k+1 can be written as the product of primes.
By induction principle, all n 2 can be written as the product of primes.
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Mathematical Induction
Like it or not.. Can you do it?
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Mathematical Induction
First step
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Mathematical Induction
Inductive Step
First step
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Mathematical Induction
Inductive Step
First step
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Mathematical Induction
Inductive Step
First step
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Mathematical Inductionyeah!
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Mathematical Induction
Remember…
50% ??!!
At least you get half of the points almost free !!
Just work on the other half..
Good luck