Below is a conceptual proof of the irrationality of square-roots. It shows that this result follows immediately from unique fractionization -- the uniqueness of the denominator of any reduced fraction -- i.e. the least denominator divides every denominator. This in turn follows from the key fact that the set of all possible denominators of a fraction is closed under subtraction so comprises an ideal of $\,\mathbb Z,\,$ necessarily principal, since $\,\mathbb Z\,$ is a $\rm PID$. But we can eliminate this highbrow language to obtain the following conceptual high-school level proof:
Theorem $\ $ Let $\;\rm n\in\mathbb N.\;$ Then $\;\rm r = \sqrt{n}\;$ is integral if rational.
Proof $\ $ Consider the set $\rm D$ of all possible denominators $\,\rm d\,$ for $\,\rm r, \,$ i.e. $\,\rm D = \{ d\in\mathbb Z \,:\: dr \in \mathbb Z\}$. Notice $\,\rm D\,$ is closed under subtraction: $\rm\, d,e \in D\, \Rightarrow\, dr,\,er\in\mathbb Z \,\Rightarrow\, (d-e)\,r = dr - er \in\mathbb Z.\,$
Further $\,\rm d\in D \,\Rightarrow\, dr\in D\ $ by $\rm\ (dr)r = dn\in\mathbb Z, \,$ by $\,\rm r^2 = n\in\mathbb Z.\,$ Thus by the Lemma below,
with $\,\rm d =$ least positive element in $\rm D,\,$ we infer that $\ \rm d\mid dr, \ $ i.e. $\rm\ r = (dr)/d \in\mathbb Z.\ \ $ QED
Lemma $\ $ Suppose $\,\rm D\subset\mathbb Z \,$ is closed under subtraction and that $\rm D$ contains a nonzero element.
Then $\rm D \:$ has a positive element and the least positive element of $\,\rm D\,$ divides every element of $\,\rm D$.
Proof $\rm\,\ \ 0 \ne d\in D \,\Rightarrow\, d-d = 0\in D\,\Rightarrow\, 0-d = -d\in D.\, $ Hence $\rm D$ contains a positive element. Let $\,\rm d\,$ be the least positive element in $\,\rm D.\,$ Since $\rm\: d\,|\,n \!\iff\! d\,|\,{-}n,\,$ if $\rm\ c\in D\,$ is not divisible by $\,\rm d\,$ then we
may assume that $\,\rm c\,$ is positive, and the least such element. But $\rm\, c-d\,$ is a positive element of $\,\rm D\,$ not divisible by $\,\rm d\,$
and smaller than $\,\rm c,\,$ contra leastness of $\,\rm c.\,$ So $\,\rm d\,$ divides every element of $\,\rm D.\ $ QED
The proof of the theorem exploits the fact that the denominator ideal $\,\rm D\,$ has the special property that it is closed under multiplication by $\rm\, r.\: $ The fundamental role that this property plays becomes clearer when one learns about Dedekind's notion of a conductor ideal. Employing such yields a trivial one-line proof of the generalization that a Dedekind domain is integrally closed since conductor ideals are invertible so cancellable. This viewpoint serves to generalize and unify all of the ad-hoc proofs of this class of results - esp. those proofs that proceed essentially by descent on denominators. This conductor-based structural viewpoint is not as well known as it should be - e.g. even some famous number theorists have overlooked this. See my post here for further details.