Prove that x is irrational unless it is an integer.

[LaMantequilla]

Homework Statement

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 + a_n-1p^n-1/q^n-1 + ... + a_o = 0

and multiplying both sides by qⁿ gives

pⁿ + a_n-1p^n-1q + ... + a_oqⁿ = 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?
Thanks in advance.

 

[SammyS]

Homework Statement

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 + a_n-1p^n-1/q^n-1 + ... + a_o = 0

and multiplying both sides by qⁿ gives

pⁿ + a_n-1p^n-1q + ... + a_oqⁿ = 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?
Thanks in advance.

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 + a_n-1 x^n-1 + a_n-2 x^n-2 + a_n-3 x^n-3 + a_n-4 x^n-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.To answer your question:

Rewrite that second equation of yours as:

a_n-1p^n-1q + ... + a_oqⁿ = -pⁿ

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

 

[LaMantequilla]

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