Condition for Rational Number to be Square of Rational Number
Theorem
Let $m$ and $n$ be (strictly) positive integers which are coprime.
Then:
- $\dfrac m n$ is the square of a rational number
- both $m$ and $n$ are square numbers.
Proof
Let $m$ and $n$ be (strictly) positive integers which are coprime.
Sufficient Condition
Suppose that $\dfrac m n$ is the square of a rational number.
Then there exists a rational number $r$ such that:
- $\dfrac m n = r^2$
By Existence of Canonical Form of Rational Number, we can find $p \in \mathbb Z, q \in \mathbb Z_{>0}, p \perp q$ such that:
- $r = \dfrac p q$
Now we have:
- $\dfrac m n = \paren {\dfrac p q}^2 = \dfrac {p^2} {q^2}$
By Powers of Coprime Numbers are Coprime, $p^2$ and $q^2$ are coprime.
Hence both $\dfrac m n$ and $\dfrac {p^2} {q^2}$ are in canonical form.
By Canonical Form of Rational Number is Unique, we must have:
- $m = p^2, n = q^2$
which shows that both $m, n$ are square numbers.
$\Box$
Necessary Condition
Suppose that both $m$ and $n$ are square numbers.
Then there exists integers $x, y$ such that:
- $m = x^2, n = y^2$
Since $n > 0$, we must have $y \ne 0$.
Therefore we can write:
- $\dfrac m n = \dfrac {x^2} {y^2} = \paren {\dfrac x y}^2$
By definition, $\dfrac x y$ is a rational number.
Hence we have shown that $\dfrac m n$ is the square of a rational number.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 5$