# Rational roots of monic integer polynomials are integers.
We make the following claim.
> Suppose we have some integer $N$, whose square root $\sqrt{N}$ is some rational number. Then we must have $\sqrt{N}$ is an integer.
An immediate consequence is that $\sqrt{2}$ is irrational. If it is rational, then by our claim $\sqrt{2}$ is some integer $k$. But there is no integer whose square is 2, contradiction.
A common proof. Suppose $\sqrt{N} = \frac{a}{b}$ for some integers $a,b >0$ with $\gcd(a,b) = 1$. If $b = 1$ then we are done. Otherwise $b$ has some prime factor $p$ that is not a factor of $a$. Then we have $a^{2} = N b^{2}$, which shows $p$ divides $a^{2}$ and hence $p$ divides $a$, a contradiction.
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Conway's calculus proof. Take $A$ to be the closest integer to $\sqrt{N}$. Then the quantity $|A-\sqrt{N}| < 1$. So the limit $|A- \sqrt{N}|^{n} \to 0$ as $n\to \infty$. Suppose $\sqrt{N} = \frac{a}{b}$ some positive integers $a,b$. Then at some $n$ we have $|A-\sqrt{N}|^{n} < \frac{1}{b}$. But $|A - \sqrt{N}|^{n} = P + Q \sqrt{N} = P + Q \frac{a}{b}= \frac{M}{b}$, for some integers $P,Q,M$, where $M \ge 0$. But the only nonnegative $M$ such that $M / b < 1 / b$ is $M =0$. This implies $|A-\sqrt{N}|^{n} = 0$, or $\sqrt{N} = A$, an integer.
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A proof by Bezout's. Suppose $\sqrt{N} = r = \frac{a}{ b}$ with $\gcd(a,b)=1$. Then by Bezout's lemma, there exists integers $x,y$ such that $ax + by = 1$. So $$\begin{align*}
0 &= (rb-a)(y-rx) \\ &= rby-r^{2}bx-ay+arx \\& = r(by+ax)-Nbx-ay \\ &= r-Nbx-ay \\ \implies r &= Nbx-ay \in \mathbf{Z}.
\end{align*}$$
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However, these are all specific cases of the following result,
> If a monic polynomial $X^{n}+c_{n-1}X^{n}+\cdots+c_{0}$ in $\mathbf{Z}[X]$ has root $r \in \mathbf{Q}$, then $\mathbf{r} \in \mathbf{Z}$.
Which in turn is a consequence of the **rational root theorem**,
> Rational root theorem.
> Suppose $r$ is a rational root to some polynomial $c_{n} X^{n}+ c_{n-1}X^{n-1}+\cdots + c_{0} \in \mathbf{Z}[X]$, $c_{n}\neq 0$. Then $r$ is of the form $\frac{a}{b}$ where $a$ is a factor of $c_{0}$ and $b$ is a factor of $c_{n}$.
Indeed, assuming $r = \frac{a}{b}$ is a rational root, with $\gcd(a,b) = 1$. Then by substitution we get $$
\begin{align*}
c_{n} \left( \frac{a}{b} \right)^{n}+c_{n-1} \left( \frac{a}{b} \right)^{n-1}+\cdots + c_{0} = 0 \\
c_{n} a^{n} + c_{n-1}a^{n-1}b+\cdots + c_{0}b^{n} = 0
\end{align*}
$$This shows $b \mid c_{n}a^{n}$ and $a\mid c_{0}b^{n}$. But $\gcd(a,b) =1$, so we have $b \mid c_{n}$ and $a\mid c_{0}$ as claimed. $\blacksquare$
Now, with rational root theorem, if a monic integer polynomial has a rational root $r$, it must be of the form $r = \frac{a}{b}$ where $b \mid 1$, hence showing $r$ is an integer.
B / 7 2024