Finding $n$th root of 2 is irrational using given polynomial

Question

The polynomial $f(x)$ is defined by
$f(x)=x^n + a_{n-1}x^{n-1}+ \cdots + a_{2}x^2+a_1x+a_0$
where $n \geq 2$ and the coefficients $a_0, \cdots, a_{n-1}$ are integers, with $a_0 \neq 0$.

Suppose that the equation $f(x) = 0$ has a rational root $\frac p q$ where $p$ and $q$ are integers with no common factor greater than $1$, and $q>0$. By considering $q ^{n-1}f(p/q)$, we know that the value of $q=1$ and the rational root of the equation $f(x) = 0$ must be an integer.

1. Show that the $n$th root of 2 is irrational for $n \geq 2$
2. Show that the cubic equation $x^3 - x +1 =0$ has no rational roots

I solved the first question without using the given polynomial. ie. setting $n$th root rational and finding contradiction.

But the hint in the solution paper says to use $f(x) = x^n - 2$ and evaluate $f(1)$ and $f(2)$, then apply the stem of the equation.
How does this work?

I am unable to do second part.

Many thanks in advance

Answer 1

Took me a second to realize what the question was even asking you to do -- it could stand to be expressed more clearly -- so I'll just walk through what they're suggesting.

Since $f(1) = 1 - 2 = -1 < 0$ and $f(2) = 2^n - 2 > 0$ for $n \geq 2$, we have $f(1) < 0 < f(2)$ so there is a root of $f(x)$ between 1 and 2 by the intermediate value theorem.

On the other hand, since $x^n - 2$ is a monic polynomial, any rational root it has is necessarily an integer by the rational roots theorem. (What they've written in the second paragraph is a sketch of the key step in the proof of the rational roots theorem).

Since there are no integers between 1 and 2, the root of x^n - 2 between 1 and 2 is irrational.

(I guess you could worry that $x^n - 2$ might have some other root, but it's an increasing function for positive values of $x$ so any other roots would have to be either negative or complex.)

Presumably you can use a similar technique to do the second part.

Answer 2

(1) Consider $f(x)=x^n-2$ then we'll show that it have no rational root.

suppose $x=\frac{p}{q} ~with~gcd(p,q)=1$ be a root of $f(x)$.

$\implies (\frac{p}{q})^n-2=0$

$\implies p^n-2q^n=0$

$\implies p^n=2q^n$

$\implies q\mid p^n$ but $gcd(p,q)=1$

and therefore it shows that $q=1$

So we have $p^n=2~for~ all~ n\geq2$.

Which is a absurd since 2 is prime.

And Hence $n^{th}$ root of 2 is a irrational number.

(2) If possible suppose that $f(x)=x^3-x+1=90$ have rational solution and call it $p\in \mathbb{Z}$, then we have,

$p^3-p+1=90$

$\implies p^3-p=89$

$\implies (p-1)p(p+1)=89$

Which is absurd since 89 is a prime number and a prime number can not be written like this.