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TAF 3023 - Discrete Mathematic
Pang Kok An 032195
Siti Aminah Bt Ahmad Sahrel 032806
Nurul Fasihah Bt Che Azmi 032220
Hasyazid b Osman 033147
Muhd. Faiz Fahimi Bin Razali 033225
Nurul Nabilah Bt Azman 032236
Roshatul Hidaya Bt Rosdin 032558
Nusaibah Bt Yahaya 032357
Group 1
1. What do you show in the basis step and what do you show in the inductive step when you use (ordinary) mathematical induction to prove that a property involving an integer n is true for all integers greater than or equal to some initial integer?
Mathematical Induction involves 2 steps:
- Basis step - Inductive step
In the basis step, we take a value arbitrarily to test its correctness.
Mathematical Induction
For example: If x ≥ 4, then 2x ≥ x2
Prove this statement’s validity
Since x ≥ 4, we take any number which is greater or equal to 4. For example, we take 4.
Mathematical Induction
24 ≥ 42
16 ≥ 16 (proved)Since we have proved x = 4, we should prove the statement
If x ≥ 4, then 2x ≥ x2
is true for other values of x greater than 4.
Basis step
Substitute x as x + 1, as to determine thatIf x is true, then x + 1 is also true.
2x ≥ x2
2x+1 ≥ (x+1)2
2 2∙ x ≥ (x+1)2 ①
2 2∙ x ≥ 2x2 ②
Inductive step
Compare ① and ②2 2∙ x ≥ 2x2 ≥ (x+1)2
2x2 ≥ (x+1)2
2x2 ≥ x2 + 2x + 12x2 ≥ x2 + 2x + 1
x2 ≥ 2x + 1x ≥ 2 +
Inductive step
x ≥ 2 +
When x is bigger, becomes smaller.
If x ≥ 4, then 2x ≥ x2 is proved.
Inductive step
2. What is the inductive hypothesis in a proof by (ordinary) mathematical induction?
In inductive step, we proof a statement by applying the method
If P(x) is true, then P(x + 1) is true.
Before we can proof that P(x + 1) is true, we have to proof the inductive hypothesis is true.
The assumption that if P(x) is true in the statement is called the inductive hypothesis
Inductive hypothesis
3. Are you able to use (ordinary) mathematical induction to construct proofs involving various kinds of statements such as formulas, divisibility properties and inequalities?
We can use mathematical induction to prove the validity of a formula.
For example:
a + (a + d) + (a + 2d) + … + [a + (n – 1)d] = n [ a + (n – 1)]for n N∈
Prove the above formula.
Using Induction in Proving Formula
Test the validity of the statement using an initial number: For example, we take 1.
a = 1 [ a + (1 – 1)]
a = a (proved)
We have proved that the statement is true for n = 1. Now, we should prove the validity of the statement with
any general values of n
Basis step
Substitute n with n + 1a + (a + d) + … + [a + (n – 1)d] + [a + nd] = (n + 1) [ a + n]
n [ a + (n – 1)] + [a + nd] = (n + 1) [ a + n]
na + n(n – 1) + a + nd = na + n2 + a + n
2na + dn(n – 1) + 2a + 2nd = 2na + dn2 + 2a + dn2na + dn2 – dn + 2a + 2nd = 2na + dn2 + 2a + dn
Inductive step
Eliminate both sides2na + dn2 – dn + 2a + 2nd = 2na + dn2 + 2a + dn
2dn – dn = dn dn = dn (proved)
Therefore, a + (a + d) + (a + 2d) + … + [a + (n – 1)d] = n [ a + (n – 1)]
is true for all n N∈
Inductive step
We can use mathematical induction to proof whether a statement is divisible by a certain integer.
For example:
Proof that n3 – n is divisible by 3.
Using Induction in Proving Divisibility
We take n = 113 – 1 = 0
0 is divisible by 3.
Therefore , the statement, n3 – n is divisible by 3 for n = 1.
Basis step
Substitute n with n + 1
Since n3 – n and 3(n2 + n) is divisible by 3, therefore, n3 – n is divisible by 3
Inductive step
(n + 1)3 – (n + 1) = n3 + 3n2 + 3n + 1 – n – 1= n3 – n + 3n2 + 3n= n3 – n + 3(n2 + n)
We can also use mathematical induction to proof an inequality.
For example:
Proof that n < 2n for n N. ∈
Using Induction in Proving Inequalities
Take n = 1.1 < 21
1 < 2
n < 2n is true for n = 1
Basis step
Substitute n with n + 1n + 1 < 2n + 1
Go back to the original statement, n < 2n. Add 1 to both sides of the equation.
n + 1 < 2n + 1
Inductive step
n + 1 < 2n + 1
Therefore, we have proven that n < 2n is true for n N. ∈
Inductive step
n + 1 < 2n + 1 < 2n + 2n
n + 1 < 2n + 1 < 2 2∙ n
n + 1 < 2n + 1 < 2n + 1