# Prove that x is irrational unless it is an integer.

1. Feb 12, 2012

### LaMantequilla

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

This is taken from an answer book that I have. I don't understand the bolded step. Can someone explain it to me?

Suppose x = p/q where p and q are natural numbers with no common factor. Then:

pn/qn + an-1pn-1/qn-1 + ... + ao = 0

and multiplying both sides by qn gives

pn + an-1pn-1q + ... + aoqn = 0

Now if q ≠ ±1 then q has some prime number as a factor. This prime number divides every term of the second equation other than pn, so it must divide pn also. Therefore it divides p, a contradiction. So q = ±1, which means that x is an integer.

Once again, it's the bolded step that I don't understand. Why must it divide pn?

2. Feb 12, 2012

### SammyS

Staff Emeritus
Looks like you're asking us to pick things up in the middle of a proof, without letting us know what it is that is being proved.

I'm guessing that you're looking at a solution for the following.
Prove that any root of the following polynomial of degree, n, with integer coefficients:
xn + an-1 xn-1 + an-2 xn-2 + an-3 xn-3 + an-4 xn-4 + … + a0
is either an integer, or the root is irrational.​
The proof is by contradiction, and done by assuming that there is a rational, non-integer root.

Rewrite that second equation of yours as:

an-1pn-1q + ... + aoqn = -pn

So, q divides the left hand side. Therefore, it must divide the right hand side.

3. Feb 12, 2012

### LaMantequilla

Thank you so much! I can't believe I missed that! Thanks!