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Page 1: Mathematical Induction Examples - mrskerbitz.commrskerbitz.com/InductDiv.pdf · Mathematical Induction Examples One important observation One fact that will prove useful in divisibility

Mathematical Induction Examples

One important observation One fact that will prove useful in divisibility problems is this ...

If each of a, b, and c are divisible by p, then a+ b+ c must also be divisible by p.

This allows us to break divisibility problems into pieces - hopefully this will make sense as youstudy the examples below.

Prove: 32n � 1 is divisible by 8 for all positive integers.

Starts: 32(1) � 1 = 8, which of course is divisible by 8. This is the "base case."

Continues: We assume 32k � 1 is divisible by 8 (this is the "induction hypothesis), and we mustshow that this forces 32(k+1) � 1 = 32k+2 � 1 to be divisible by 8.

Note that 32k+2 � 1 = 32�32k � 1

�+ 8: Then

32 ��32k � 1

�| {z }Div. by 8 (ind. hyp.)

+ 8|{z}Div. by 8

;

therefore, 32k+2 � 1 is divisible by 8

Prove n3 + 5n+ 6 is divisible by 6 for all positive integers.

Starts: (1)3 + 5 (1) + 6 = 12, which is divisible by 6. This is the "base case."

Continues: We assume k3 + 5k + 6 is divisible by 6 (this is the "induction hypothesis), and wemust show that this forces (k + 1)3 + 5 (k + 1) + 6 to be divisible by 6.

Note that

(k + 1)3+ 5 (k + 1) + 6 = k3 + 3k2 + 3k + 1 + 5k + 5 + 6 (expanded)

= k3 + 5k + 6 + 3k2 + 3k + 6 (rearranged terms)

In the second line above, we know already that k3 + 5k + 6 is divisible by 6 by the inductionhypothesis. What about 3k2 + 3k + 6? Note that

3k2 + 3k + 6 = 3k (k + 1) + 6.

The "6" is divisible by 6, so all that remains is to prove that 3k (k + 1) is also divisible by 6. Let�sthink about that for a moment ... note that k and k + 1 are consecutive integers, which meansone is even and one is odd. If we multiply an even and an odd number together, our answer isalways even. This means that 3k (k + 1) is 3 times an even number, and that even number byde�nition must contain a factor of 2. Since the number 3k (k + 1) contains a factor of 2 as wellas a factor of 3, it must be divisible by 6. Therefore, we have

k3 + 5k + 6| {z }Div. by 6 by the ind. hyp.

+ 3k2 + 3k + 6| {z }Div. by 6 by the above argument.

is divisible by 6.

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