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