# Need help with proving a number is irrational

1. Sep 25, 2011

### thenthanable

Prove that if x satisfies
'xn +an-1xn-1+ ... a0=0'​
for some integers an-1,..., a0, then x is irrational unless x is an integer.

My main question is that I don't quite understand what the question is trying to ask me prove. I'm fairly new with this so pardon me if this question is really basic. Am I suppose to prove that x is irrational or an integer? The 'unless' in the question really throws me off.

Nonetheless, I came up with a few ideas because I must say I really suck at proofs so I always begin by writing down things I notice about the equations given.

I wrote that the above equation is the binomial expansion of (x+(ao)1/n)n = 0, and if ao gives an integer solution, then x= -(ao)1/n so x is an integer? But then again, I'm not very sure what the question really wants. :S This question is madly confusing to me! :(

2. Sep 25, 2011

### D H

Staff Emeritus
Consider x2-1=0. This has solutions at x=1 and x=-1. Now consider x2-2=0. This is very similar in form to the previous equation, but now the solutions are irrational.

You are supposed to prove that the real solutions are either irrational or are integral. Another way to put it: There are no rational, non-integral solutions to such equations. This rephrasing suggests an attack: Assume a rational, non-integral solution exists show that this results in a contradiction.

Edit
Note that you can get rational, non-integral solutions if the leading coefficient is a non-zero integer other than 1 or -1 (and there's no reason for a leading coefficient of -1; it easily converts to the canonical form without loss of generality). Example: 2x-1=0.

Last edited: Sep 25, 2011
3. Sep 25, 2011

### thenthanable

Oh I think I sorta get it. I thought 'unless' might imply something that I might have missed, because in my mind, if 'unless' could simply be replaced by 'or', then they would have written 'or' in the first place. Ah, this Maths stuff is killing me. Thanks a lot! :)