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Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length elsewhere, beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.

Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

This idea immediately generalizes to a proof by induction on degree that $\Bbb Z$ is integrally closed (i.e. the monic case of the rational root test).

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length elsewhere, beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.

Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length [elsewhere][1], beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs. [1]:https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313

Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length elsewhere, beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.

Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length [elsewhere][1], beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs. [1]:http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313

Below is a simple direct proof that I found as a teenager:

THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.

Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout

so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$

Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length [elsewhere][1], beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs. [1]:https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313