# Thread: Principle of Induction Proof

1. ## Principle of Induction Proof

Prove that 7^(2n-1) + 13^(2n-1) is divisible by 10 for all n in N. I know this is done using algebra and followed by the definition of divisibility. I just got a bit lost in the algebra.

2. The induction step:
Assume that 10 divides $\displaystyle 7^{2n-1}+13^{2n-1}$.
Want to show for n+1, 1.e want to show that 10 divides
$\displaystyle 7^{2(n+1)-1}+13^{2(n+1)-1}=7^{2n+1}+13^{2n+1}=49.7^{2n-1}+169.13^{2n-1}$

Let see if you can go from here............

3. shouldn't that n-1 you inserted be n+1 ??

4. Originally Posted by Caity
shouldn't that n-1 you inserted be n+1 ??
Yes I have fixed it now.

5. Also the 49 and 169 be 7 and 13 since they were factored out...??? Sorry I get confused easily..

6. never mind you factored out 2 of each of them I see...

7. Thanks I think I got it... I let the original eqation equal 10j j is in N. and ended up with 10(40j + 12 times 13^(2n-1)) which is in N so original is divisible by N. Thanks again...

8. Here's another one and not sure what to do with the exponents on this... Prove the if 0<a<b then a^(1/n) < b^(1/n) for n in N. Or how to show for n + 1.

Oh nm this would be nth roots so prove by contradiction... sorry I answered my own question..

9. To make it work the statement should be for every 0<a<b any any n in N we have a^(1/n)<b^(1/n).

Now we want to show a^(1/(n+1))<b^(1/(n+1)). Can you find A and B such that a^(1/(n+1))=A^(1/n) and b^(1/(n+1))=B^(1/n) ?

10. Originally Posted by watchmath
To make it work the statement should be for every 0<a<b any any n in N we have a^(1/n)<b^(1/n).

Now we want to show a^(1/(n+1))<b^(1/(n+1)). Can you find A and B such that a^(1/(n+1))=A^(1/n) and b^(1/(n+1))=B^(1/n) ?
I know that is what I need to do. How does the algebra go on this? Do you think nth root contradiction proof would maybe be easier to show this?