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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 $\sqrt{2}$.

Thanks in advance

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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
Related. – Cameron Buie Oct 22 '13 at 13:59
up vote 44 down vote accepted

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


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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 $$\left({\frac{\sqrt 8}2}\right)^2 = \frac{{\left(\sqrt 8\right)}^2}{2^2} = \frac84 = 2. $$

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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:

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


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

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

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+1 Great transformation ! – crypticous Feb 10 '13 at 8:59

Geometric approach:

enter image description here

Now what is $\displaystyle\text{AB}/2$ ?

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I've loved your aproach. Thanks! – Julio Nobre Dec 1 '13 at 22:00
You're welcome @JulioNobre ! :-) – user93957 Dec 20 '13 at 11:38
I really love the post full of graphs. +1 – Babak S. Dec 27 '13 at 12:36
Thank you @B.S. :^) – user93957 Dec 27 '13 at 12:42

Other techniques to prove that result:

Technique $\mathbf 1$:

We may also prove that by solving this equation for $x$:

$$\begin{align}\frac{\sqrt x}{2}&=\sqrt2\\\sqrt x&=2\sqrt 2\\\left(\sqrt x\right)^2&=\left(2\sqrt 2\right)^2\\x&=4\cdot2=\color{green}{\boxed{\color{black}{8}}}\end{align}$$

Therefore: $$\frac{\sqrt8}{2}=\sqrt 2$$ I know, this is a terrible way. But hey, it works! ;-)

Technique $\mathbf 2$:

Just like in Technique $\mathbf 1$: We may also prove that result by solving this equation for $x$:

$$\begin{align}\frac{\sqrt 8}{2}&=\sqrt x\\ \left(\frac{\sqrt 8}{2}\right)^2&=\sqrt x^2\\ \frac84&=x \implies x=\color{green}{\boxed{\color{black}2}} \end{align}$$

Therefore: $$\frac{\sqrt8}{2}=\sqrt 2$$

Technique $\mathbf 3$:

If $a=b$ then this means $a-b=0$, and the opposite is true. So let's check for $\sqrt 2$ and $\sqrt 8/2$:$$\begin{align}\dfrac{\sqrt 8}{2}-\sqrt{2}&=\dfrac{\sqrt 8}{2}-\left(\sqrt{2} \cdot \dfrac{\sqrt{4}}{2}\right)\quad\to\quad\text{since $\sqrt 4=2$}\\ &=\dfrac{\sqrt 8}{2}-\dfrac{\sqrt 2\cdot\sqrt 4}{2}\\ &=\dfrac{\sqrt{8}-\sqrt{2}\cdot\sqrt{4}}{2}\\ &=\dfrac{\sqrt{8}-\sqrt{2\cdot4}}{2}\\ &=\dfrac{\sqrt 8-\sqrt 8}{2}=\dfrac{0}{2}=0 \end{align}$$ Therefore, we can easily conclude that: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$

Technique $\mathbf 4$:

If $a=b$ then it follows that $a/b=1$, and the opposite is true. So let's check for $\sqrt 2$ and $\sqrt 8/2$: $$\begin{align}\dfrac{\sqrt2}{\dfrac{\sqrt8}{2}}&=\sqrt{2}\cdot\dfrac{2}{\sqrt{8}}=\dfrac{\sqrt2\cdot\sqrt 4}{\sqrt8}=\dfrac{\sqrt{2\cdot 4}}{\sqrt 8}=\dfrac{\sqrt8}{\sqrt8}=1 \end{align}$$ Therefore: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$

Technique $\mathbf 5$:

We will do a proof by contradiction: We suppose that $\sqrt 8/2\ne\sqrt 2$, therefore there must be a difference between them which is not zero $\sqrt 8/2-\sqrt 2\ne0$, that we will call $x$. So let's check if $x\ne0$: $$\begin{align}x=\dfrac{\sqrt 8}{2}-\sqrt{2}&=\dfrac{\sqrt 8}{2}-\left(\sqrt{2} \cdot \dfrac{\sqrt{4}}{2}\right)\\ &=\dfrac{\sqrt 8}{2}-\dfrac{\sqrt 2\cdot\sqrt 4}{2}\\ &=\dfrac{\sqrt{8}-\sqrt{2}\cdot\sqrt{4}}{2}\\ &=\dfrac{\sqrt{8}-\sqrt{2\cdot4}}{2}\\ &=\dfrac{\sqrt 8-\sqrt 8}{2}=\dfrac{0}{2}=0 \end{align}$$But this is a contradiction, since $x$ must be non-zero but it turns out that it is equal to $0$. Therefore: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$

Generalization: Let $x$ be a strictly positive number in $\Bbb R$, then: $$\dfrac{\sqrt{x^3}}{x}=\sqrt{x}$$

Proof: $$\dfrac{\sqrt{x^3}}{x}=\dfrac{\sqrt{x^3}}{\sqrt{x^2}}=\sqrt{\dfrac{x^3}{x^2}}=\sqrt x\qquad\blacksquare\\\,\\\textit{inspired by Paul Dirac's post :)}$$

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protected by Asaf Karagila Dec 6 '13 at 19:21

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