Proof that if $\sqrt {ab}$ is rational then $\sqrt {\frac{a}{b}}$ is also rational.

Question

I can reduce it using the fact that square root of a number is rational if it is a square number. Now it would be something like : If $ab=k^2$ for some $k$ in $N$ then show that $\frac{a}{b}=m^2$ for some $m$ in $N$

Where $a$,$b$ belongs to $N$

$N$ is the set of natural numbers.

I have tried prime factorisation also But do not know how to proceed.

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CLAIM

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If $a$,$b$ belongs to $N$ and if $\sqrt ab$ is integer then prove that $\sqrt \frac{max(a,b)}{min(a,b)}$ is also an integer. Where $N$ is the set of natural numbers.

Answer 1

Hint $ $ Employ $\,b\sqrt{a/b} = \sqrt{ab}.\,$ This is a special case of a general method. Let's view it that way.

Observe that $\, \overbrace{bx^2 = a}^{\large\ \ x\ =\ \sqrt{a/b}}\!\!\!\!\overset{\large \times\ b}\iff\! \overbrace{(bx)^2 = ab}^{\large\,\ bx\ \ =\ \ \sqrt{ab}}\ $ for $\,b\neq 0$

We scaled the LHS by $b$ to get $\,X^2 = ab,\ X := bx\,$ which is now monic (lead coeff $=1).\,$ This is a special case of the monicization method used in the AC method for factoring non-monic quadratics.

The same method also works for higher-degree polynomials (as I explain in the linked post), which leads to higher-degree generalizations of the above identity for square-roots.

Generally the method shows that any algebraic number can be written as $\,\alpha = \beta/n\,$ where $\,n\in\Bbb Z\,$ and $\,\beta\,$ is an algebraic integer (i.e. a root of a monic $f(x)\in\Bbb Z[x]),$ e.g. above $\,\sqrt{a/b} = \sqrt{ab}/b,\,$ where $\,\sqrt{ab}\,$ is a root of the monic $\,x^2\!-ab\,$ so is an algebraic integer.

In higher-degree cases such equations are generally less obvious, so it is worth being aware of the general viewpoint (even though it is a bit overkill for this particular case).

Answer 2

This hint will help you not only prove something you want to, but also state more clearly what you meant.

If $a,\,b$ are rational with $b\ne 0$ then $\sqrt{\frac{a}{b}}=\frac{\sqrt{ab}}{b}$ is the ratio of two rationals, the denominator non-zero. Any such ratio is rational (proof is an exercise).

Answer 3

An "easy" algebraic proof:

if (assuming that $a$ and $b$ are rational numbers) $\sqrt {ab}={\frac{x}{y}}$, then $ab$ is a perfect "square fraction" i.e. $ab=\frac{x^2}{y^2}$. then $a=\frac{x^2}{by^2}$, and $\sqrt \frac{a}{b} = \sqrt{\frac{x^2}{b^2y^2}}=\frac{x}{by}=\frac{n}{m}$.

Answer 4

First, point out that your statement doesn’t hold for $a,b,k,m \in \mathbb{N}$.

Take for example $(a,b) = (2,18)$. Then $k = \sqrt{ab} = 6 \in \mathbb{N}$, but $m = \sqrt{\frac{a}{b}} = \frac{1}{3} \notin \mathbb{N}$.

But the following statement does hold:

If $ab = k^2$ with $a,b,k \in \mathbb{Q}$, then $\frac{a}{b} = m^2$ for some $m \in \mathbb{Q}$, where $\mathbb{Q}$ is the set of rational numbers.

Proof: If $ab = k^2$, then $\frac{a}{b} = \frac{ab}{b^2} = \frac{k^2}{b^2} = (\frac{k}{b})^2$, so $m = \frac{k}{b}$ is rational, as desired.

Answer 5

Your claim works if and only if $b$ is rational.

Note that

$\frac{\sqrt{ab}}{\sqrt{\dfrac{a}{b}}}=b$

If numerator is rational then denominator is rational too.

Counterexample for the case when $b$ is irrational take $a=2\sqrt{2}$ and $b=\sqrt{2}$

Here $\sqrt{ab}=2$ however $\sqrt{\dfrac{a}{b}}=\sqrt{2}$