# Why is $\sqrt{8}/2$ equal to $\sqrt{2}$?

I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.

-
It isn’t: $\frac{\sqrt8}2=\sqrt2\ne 2=\sqrt4$. – Brian M. Scott Nov 12 '12 at 22:32
@BrianM.Scott: But $\sqrt2=2$! :-) – Asaf Karagila Nov 12 '12 at 22:39
Yes, has you all have noted, I made a mistake. Instead of writing sqrt(2), I wrote sqrt(4), which of course is 2. I have already fixed my question. Reading your answers I have already understood the solution. Thanks a lot for that! – Julio Nobre Nov 12 '12 at 23:26
To clarify what the edit to the question was: the question originally asked, incorrectly, why $\sqrt{8}/2$ is equal to $\sqrt{4}$. Now it correctly asks why $\sqrt{8}/2$ is equal to $\sqrt{2}$. – Rory O'Kane Nov 13 '12 at 7:40

It isn't, as originally written. To see why the fixed version is correct, we have:

$$\frac{\sqrt{8}}2=\frac{\sqrt{4\cdot2}}2=\frac{\sqrt{4}\sqrt{2}}2=\frac{2\sqrt{2}}2=\sqrt{2}.$$

-
 Altough Samuel Reid's was also insighfull, I think your explanation was more acessible to me. That is is why I will mark it as right answer. Thanks for both of you! – Julio Nobre Nov 12 '12 at 23:28

I am a bit surprised that nobody suggested $$\left({\frac{\sqrt 8}2}\right)^2 = \frac{{\left(\sqrt 8\right)}^2}{2^2} = \frac84 = 2.$$

-
This is certainly what I would have said. – Lubin Nov 13 '12 at 3:43
Sorry, but this answser is certainly wrong. – Julio Nobre Nov 13 '12 at 11:03
@JulioNobre: no, it isn't. – Martin Argerami Nov 13 '12 at 11:38
@julio The definition of $\sqrt2$ is that it is the unique positive number whose square is 2. $\frac{\sqrt8}2$ is a positive number, and I've demonstrated that its square is 2. – MJD Nov 13 '12 at 12:50
Ok. Now I got it :D – Julio Nobre Nov 13 '12 at 17:10

It’s simple:

$$\frac{\sqrt{8}}{2} = \frac{\sqrt{2 \cdot 2 \cdot 2}}{\sqrt{2 \cdot 2}} = \sqrt{2}$$

:)

-
 Yes, I loved its simplicity :D – Julio Nobre Nov 13 '12 at 11:04

By noting that $2^3 =8$, you have $$\frac{\sqrt{8}}{2} = \frac{\sqrt{2^3}}{2} = \frac{ 2^{3/2}}{2} = 2^{1/2}=\sqrt{2} \neq \sqrt{4}$$

-

$$\frac{\sqrt8}{2}=\frac{\sqrt8}{\sqrt{2^2}}=\sqrt{\frac{8}{2^2}}=\sqrt{\frac{8}{4}}=\sqrt2.$$

-
 +1 Great transformation ! – DaHaKa Feb 10 at 8:59