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Finger [1]
1 year ago
7

Using principle of mathematical induction prove that 6^-1 divisble by 5 .​

Mathematics
1 answer:
salantis [7]1 year ago
5 0

I suppose the claim is 5 \mid 6^n - 1 for n\in\Bbb N.

When n=1, we have 6^1 - 1 = 6 - 1 = 5, and of course 5 divides 5.

Assume the claim holds for n=k, that 5 \mid 6^k - 1. We want to use this to show it holds for n=k+1, that 5 \mid 6^{k+1} - 1.

We have

6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5

Since 5 \mid 6^k - 1, we can write 6^k - 1 = 5\ell for some integer \ell. Then

6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)

which is clearly divisible by 5. QED

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