Prove x^n > x for all x > 1.

**450081592** · November 21st 2009, 07:04 PM

Originally Posted by lvleph

Yes, because the hint was to prove $f(x) > g(x)$ and we wanted to prove $x^n > x$ so it makes sense to assume $g(x)=x$ .

how about I do this

Factorise! $x^n-x=x(x^{n-1}-1)=x(x-1)(1+x+x^2+\cdots+x^{n-2}),$ the product of three terms which are obviously positive for $x\,>\,1.$

does this make sense?

**Moo** · November 22nd 2009, 06:27 AM

Yes, and it's way better.

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