Rational Square Root of Integer is Integer
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Theorem
Let $n$ be an integer.
Suppose that $\sqrt n$ is a rational number.
Then $\sqrt n$ is an integer.
Proof
Suppose that $\sqrt n = \dfrac a b$, with $a, b$ coprime integers and $b > 0$.
Then we would have:
- $n = \dfrac {a^2} {b^2}$
That is:
- $n b^2 = a^2$
From Number divides Number iff Square divides Square:
- $b^2 \divides a^2 \implies b \divides a$
However, since $a \perp b$ and $b \divides a$, this means that necessarily $b = 1$.
That is, $\sqrt n = a$, an integer.
Hence the result.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.14$