Upload
vutruc
View
217
Download
0
Embed Size (px)
Citation preview
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.