# Mathematical Induction

1. Oct 19, 2005

### dglee

Show taht for every natural numbers n>=2 $$n \geq 2$$ the number 2^2^n - 6 $$2^{2^n} -6$$ is a multiple of 10. Using mathematical induction.
Okay i got no clue how to start this question. Ahhh Is there a series where the x^n is a series? Well this stuff really sucks.

Last edited: Oct 19, 2005
2. Oct 19, 2005

### Tide

HINT:

$$2^{2^{n+1}} = \left( 2^{2^n}\right)^2$$

3. Oct 19, 2005

### dglee

Oh WOw... that helped a lot.. now i can show that its inductive hmm but how would i show its a multiple of 10?
hmm maybe if i said 10^2 is 100 so thats a multiple of 10. $$10*n\leq2^{2^n}-6$$ if i proved that.. hmm i wonder if that would be right... $$10*(n+1)\leq2^{2^(n+1)}$$ hmm if i proved that would i have solved the question?
wow if hmm well ahaha thanks a LOT!!!!!!!! YOUR AWSOME. that little hint helped a lot but i got no clue if im actually doing it right.

Last edited: Oct 19, 2005
4. Oct 19, 2005

For Mathematical Induction, you assume that P_{k} is true, in this case for n greater or equal to 2. With this, you can immediately say that 2^(2^k) - 6 is equal to 10a, for some a, where a is a positive integer.
Can you then use this fact, and the hint provided, to prove the desired result for 2^(2^(k+1)) - 6?

5. Oct 19, 2005

### dglee

hmm should could i say that

$$10(a+x)\leq (\left2^{2^n}\right )^2 - 6$$ where x is some positive number and should that it is inductive to prove that 10 is a multiple?
soo
$$10(a)\leq2^{2^n} - 6$$ then
$$10(a+x)\leq(\left 2^{2^n}\right)^2 - 6$$
then show that
$$10(a+x)\leq2^{2^{n+1}} -6$$ ahhh im confusd now.. ahh

Last edited: Oct 19, 2005
6. Oct 19, 2005

### Tide

HINT: If M - 6 is a multiple of 10 then M ends in the digit 6! :)

7. Oct 19, 2005

Hmmm... Why are there so many inequalitites in your working? From what I know, the only inequality to appear in your solution should be the fact that n is greater than or equal to 2, but this is just a specification, and should not appear in your proof.

8. Oct 19, 2005

Steps in Mathematical Induction
1) Let $$P_{n}$$ be the statement $$2^{2^n}-6$$ is divisible by 10, for n$$\geq$$ 2.
2) Check that the result you want to prove is valid for n=2, so $$P_{2}$$ is true.
3) Assume $$P_{k}$$ is true, for some n $$\geq$$ 2. So, $$2^{2^k}-6$$ = 10a, for some a, which is a positive integer.
4) Using this result, you must somehow prove that $$2^{2^{k+1}}-6$$ is a multiple of 10. How would you go around doing it? Look at the first hint provided and observe... What has been done to the term $$2^{2^n}$$ ? USE BOTH THE RESULT FROM STEP 3 AND THE FIRST HINT

5) Once you have proven step 4, give a conclusion. "Since $$P_{2}$$ is true, and for some n$$\geq$$2, $$P_{k}$$ is true $$\Longrightarrow$$ $$P_{k+1}$$ is true. By Mathematical Induction, $$P_{n}$$ is true for all n$$\geq$$2."

Last edited: Oct 19, 2005
9. Oct 19, 2005

### dglee

Wow thanks a lot for your help. I will try to figure this out. You helped a lot pizzasky and Tide.