The polynomial $x^2-2$ is irreducible over $\mathbb{Q}$

Given: The polynomial $f=x^2-2\in \mathbb{Z}[x]\subseteq \mathbb{Q}[x]$.

A reduced rational root $r\mathrm{/}s$ of an integer polynomial has numerator dividing the constant coefficient and denominator dividing the leading coefficient (Rational root theorem).

A quadratic over a field is irreducible exactly when it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

An integer is prime when it exceeds $1$ and has no positive divisors other than $1$ and itself (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$).

A divisor of a nonzero integer has absolute value at most the absolute value of that integer (If $d\mid a$ and $a\mathrm{\ne }0$ then $d\mathrm{\ne }0$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; hence the set of divisors of a nonzero integer is bounded above by $\mathrm{\mid }a\mathrm{\mid }$).

Integer absolute value is given by the positive and negative cases (The absolute value $\mathrm{\mid }a\mathrm{\mid }$ of an integer).

Integer absolute value is nonnegative and vanishes exactly at zero (Absolute value in $\mathbb{Z}$: $\mathrm{\mid }a\mathrm{\mid }\ge 0$; $\mathrm{\mid }a\mathrm{\mid }=0$ exactly when $a=0$; $\mathrm{\mid }-a\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$; $\mathrm{\mid }ab\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }$; $-\mathrm{\mid }a\mathrm{\mid }\le a\le \mathrm{\mid }a\mathrm{\mid }$; and $\mathrm{\mid }a\mathrm{\mid }\le c$ exactly when $-c\le a\le c$).

The integers form an ordered ring (The integers form a totally ordered ring).

The natural numbers embed in the integers preserving their arithmetic (The naturals embed in the integers).

The natural-number order is discrete (Discreteness: $\sigma (n)$ is the immediate successor).

The rational numbers form a field (The rationals form a field).