# P 2n-1 (a)=0 for some a, real.

1. Dec 8, 2012

### peripatein

P2n-1(a)=0 for some a, real.

Hi,
1. The problem statement, all variables and given/known data

I was wondering whether I could prove the following using induction? I have tried, and was successful, am simply now not sure whether employing induction in this case is just.
I am asked to show that for any odd-power polynomial P2n-1, there exists a number a, real, so that P2n-1(a)=0.
I first thought of using continuity, but wasn't sure how to proceed in that direction so managed to show it via induction. Is it valid?

2. Relevant equations

3. The attempt at a solution

2. Dec 8, 2012

### tiny-tim

hi peripatein!
induction should be valid (depending how you've done it! )

have you tried using compactness?

3. Dec 8, 2012

### peripatein

Re: P2n-1(a)=0 for some a, real.

Compactness?

4. Dec 8, 2012

5. Dec 8, 2012

### peripatein

Re: P2n-1(a)=0 for some a, real.

Thank you, tiny-tim!

I am afraid I have another question. Could you please tell me whether the following proof is lacking in any regard? I am trying to show that for continuous function f: [a,b] -> R, f([a,b]) is a closed interval.

My attempt:

Using Weierstrass, since f(x) is continuous in [a,b], f(x) must also be bounded in that interval and must have a minimum and maximum. Hence, m <= f([a,b]) <= M
Using Cauchy's mean value theorem, since f(x) is continuous in [a,b], and for every y in [m,M] -> m <= f(a) <= y <= f(b) <= M (or f(b) <= y <= f(a)), then there exists x in [a,b] so that f(x)=y.
I doubt this is complete, moreover I am not certain all of the above statements are indeed correct, or that this is indeed the way to prove it.