Square root property proof

Question

Can anyone provide a link to a proof of the following square root property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Could not find it online anywhere.

Answer 1

In the following proof, we will use the following definition of the (principal) square root of a non-negative number:

• For a given real number $p\ge 0$, we say that a real number $q$ is its (principal) square root ($q=\sqrt{p}$), iff $q\ge 0$ and $q^2=p$.

It is possible to prove that such a number $q$ exists (not easily - the proof uses some fundamental properties of real numbers), and is unique (fairly easy), so the above definition allows us to view the square root as a function of non-negative real numbers.

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Now, to the proof. Let $a,b\ge 0, b\ne 0$ - real numbers, and let $x=\sqrt{a}, y=\sqrt{b}$. This means that $x,y\ge 0$ and $x^2=a, y^2=b$. Now this implies that $y\ne 0, \frac{x}{y}\ge 0$, and also:

$$\left(\frac{x}{y}\right)^2=\frac{x^2}{y^2}=\frac{a}{b}$$

Therefore, $\frac{x}{y}=\sqrt\frac{a}{b}$. Remembering what $x$ and $y$ were in the first place, we conclude:

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt\frac{a}{b}$$.

Answer 2

Say $a,b\geq 0$. If $x=\sqrt{a}$ and $y= \sqrt{b}$ then $x^2=a$ and $y^2= b$ so $$\Big({x\over y}\Big)^2={x^2\over y^2} = {a\over b}\implies {x\over y} = \sqrt{a\over b}$$

and we are done.

Answer 3

By definition of $\sqrt{\phantom 3}$, and assuming $a\geq0, b>0$, we have that $ \sqrt{\frac ab} $ is the unique non-negative number such that $$ \left(\sqrt{\frac ab}\right)^2=\frac ab $$ (One must, of course, be convinced that this number is indeed unique.) Now note that $$ \left(\frac{\sqrt a}{\sqrt b}\right)^2=\frac{\sqrt a}{\sqrt b}\cdot \frac{\sqrt a}{\sqrt b}=\frac{(\sqrt a)^2}{(\sqrt b)^2}=\frac ab $$ so $\frac{\sqrt a}{\sqrt b}$ fulfills the defining property of $\sqrt{\frac ab}$. They must therefore be equal.

Answer 4

Assuming $a,b\in\mathbb R$, $a\geqslant 0$, and $b>0$, we have \begin{align} \frac{\sqrt a}{\sqrt b} &= \frac{a^{1/2}}{b^{1/2}}\\ &=\frac{e^{\log(a^{1/2})}}{e^{\log(b^{1/2})}}\\ &=\frac{e^{\frac 12\log a}}{e^{\frac12\log b}}\\ &=e^{\frac12\log a - \frac12\log b}\\ &= e^{\frac12 \log\left(\frac ab\right)}\\ &=\left(\frac ab\right)^{1/2}\\ &=\sqrt{\frac ab}. \end{align}

Answer 5

By the definition of square-root, $x=\sqrt{y}\hspace{.15cm}$ means that $x^2=y$.

Assuming that $a,b\in\mathbb{R}$ with $a\geq 0$ and $b>0$, then we can simply compute: \begin{align*} \left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2&=\frac{\sqrt{a}}{\sqrt{b}}\cdot \frac{\sqrt{a}}{\sqrt{b}} =\frac{\sqrt{a}\cdot \sqrt{a}}{\sqrt{b}\cdot\sqrt{b}}=\frac{a}{b} \end{align*} Therefore, $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\hspace{.25cm}$

Note that I used the fact that $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$

Answer 6

I think that $$\sqrt\frac{a}{b}=\frac{\sqrt |a|}{\sqrt |b|} $$ Because $$\frac{a}{b}\ge0 $$ But $a\ge0, b\ge0 $ This is only a special case, because we may lose the solution $\frac{a}{b} $ then $ a<0,b<0 $ Therefore, you need to use the module ;)