Math Induction Proof

[yojoe]

Prove that if n is a positive integer, then (7^n)-1 is divisible by 6.

Step one:
Show statement is true for some value of n; n=1
(7^1)-1=6 ---> 6/6 = 1

Step two:
Assume (7^k)-1

Step Three:
Having trouble here:
((7^k+1)-1)/6
Then I don't know what to do following this part.
Would I start plugging in x1,x2,x3...x(-1),x(-2) and so on to prove?

Thanks

 

[Gene]

Watch your ()s
7^(k+1) - 1 =
7^k * 7 - 1 =
7*7^k-7 + 6 =
7(7^k-1) + 6 but
7^k-1 is a multiple of 6
Call it 6x so
7(7^k-1)+6 =
7(6x)+6 =
6(7x+1) which is obviously a multiple of 6
QED

 

[yojoe]

Hi Gene-

Thanks for responding so quick! I am desperately wanting to understand math induction. All of the people around me seem to get it but me. It probably is simple. I follow the logic thru step 2, but then wonder where the:

7*7^k-7 + 6 comes from? (esp k-7 part)
then it turns into k-1?

I understand how the last lines are multiples of six, but cannot understand why:
7(6x)+6 turns into:
6(7x+1)

 

[yojoe]

Oh, I see the last part, factoring out a six, the parentheses confused me

 

[Gene]

Yup, ya gotta watch them ()s.
We had 7*7^k-1 but to get the 7^k-1 used in step 2 the 7 had to be factored so I made -1 into -7+6 which still = -1, but now I can get 7(7^k-1)+6. The 7^k-1 is vital to the induction 'cause we are assuming it is 6x.
7*6x=42x=6*7x to get the 6 isolated.
You do understand the basic idea of induction, do you not?
Step 2 is a general restatement of step 1 which showed that it works when k=1. Next we have to prove that IF it works for k then it also works for k+1. After that you can step thru all the k's in the universe (1,2,3...) and know that it will work for each one.