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Math Help - Proof on Powers of Numbers Greater Than 1

  1. #1
    Newbie eikou's Avatar
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    Proof on Powers of Numbers Greater Than 1

    I'm working on a proof and I'm stuck at what seems to me a severe case of mathematician's block (which is sort of like writer's block). I need to prove that when you take a number greater than 1 and raise it to any power the result is going to be larger than the original number. In other words:

    If x \ge 1 and n is a natural number, then x^n \ge x.
    Last edited by eikou; January 11th 2010 at 01:01 PM.
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  2. #2
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    Quote Originally Posted by eikou View Post
    If x > 1 and n is a natural number, then x^n > x.
    Well of course it should be x^n\ge x. What if n=1?

    Is this true x^{n-1}\ge 1?
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  3. #3
    Newbie eikou's Avatar
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    Right right, greater than or equal to. With the natural numbers, there will be no x^{n-1} though (beneath 1), because they start at 1 and count upwards by integers. A condition on the proof was that the exponent belong to the natural numbers. Therefore, by consequence, x^{n-1} \ge 1 will always hold true since n-1 \ge 1. Otherwise the conditions of the proof are violated and it's... what's that called... vacuously true I think.
    Last edited by eikou; January 11th 2010 at 01:06 PM.
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    MHF Contributor Drexel28's Avatar
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    Let f:[1,\infty)\mapsto\mathbb{R} be given by f(x)=x^n-n,\text{ }n\in\mathbb{N}. We have that f(1)=0 and f'(x)=nx^{n-1}-1\geqslant n-1\geqslant 1-1=0. Where it follows that f(x)\geqslant f(0)=0 for all x\in[1,\infty)
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    MHF Contributor Drexel28's Avatar
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    Alternatively, x^n-x==x\left(x^{n-1}-1\right)\geqslant x\left(1-1\right)=0.
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    MHF Contributor Drexel28's Avatar
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    You could also do induction. The possibilities are endless.
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  7. #7
    Newbie eikou's Avatar
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    I actually tried induction first, epic flop. Not to say you can't, just that it was beyond my ability. I tend to run into difficulty frequently with induction proofs. I always get stuck in vicious circles.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by eikou View Post
    I actually tried induction first, epic flop. Not to say you can't, just that it was beyond my ability. I tend to run into difficulty frequently with induction proofs. I always get stuck in vicious circles.
    Really? Seriously? Are you sure that you didn't make a careless clerical error?

    Assume that x^n\geqslant x for all x\geqslant1 then ]x^{n+1}\geqslant x^2\geqslant x. Induction. Over.
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