I've just bought "Inside Interesting Integrals" by Paul Nahin and I've come to an impasse after the first paragraph in the preface. He posed this problem as a test to see whether or not the reader should continue on or put down the book. Basically, if the reader was interested and already trying to figure it out, then the reader should continue. I fit that category, and tried a proof. However I feel like I'm cheating. So would you please tell me if this is a legal proof? Any insights are welcomed. If there is an entirely different way to do this please show, but I'm partial to something similar to this.(adsbygoogle = window.adsbygoogle || []).push({});

Without actually solving for x, prove the following: if x + 1/x = 1, then x^{7}+ 1/x^{7}= 1.

Assume that x^{7}+ 1/x^{7}= 0.

Then:

x^{7}= -1/x^{7}

x^{7}x^{7}= -1

x^{14}= -1.

Therefore x is an imaginary number.

Now by hypothesis we know that x + x^{-1}= 1.

Since x is imaginary, x^{-1}is also imaginary.

By the closure property of the set of imaginary numbers, x + x^{-1}must also be imaginary.

However, the hypothesis states that x + x^{-1}= 1, which is not imaginary. Thus by contradiction, if x + 1/x = 1, then x^{7}+ 1/x^{7}= 1.

Does this work at all? Not only am I not sure, I feel like I cheated by just stating that x^{7}+ 1/x^{7}= 0. What I was attempting to do was assume ~Q and then derive a contradiction using P, using the basic proof by contradiction technique. Saying it is equal to 0 is more specific than saying it is not equal to 1, but either way it's still stating it isn't equal to 1. Is this even a viable way to approach this proof?

Thanks!

EDIT- I should point out that the book actually does a direct proof using just a bit of recursive algebraic manipulation. I just want to see if this way is legal.

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# I Prove if x + (1/x) = 1 then x^7 + (1/x^7) = 1.

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