How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
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4$\begingroup$ What if you multiply by $1 = \dfrac{\sqrt{2}}{\sqrt{2}}$? $\endgroup$– AmzotiDec 27, 2014 at 18:32
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1$\begingroup$ Commander, you have to learn basic maths elsewhere you will be stuck with this questions for years. $\endgroup$– KarlDec 27, 2014 at 18:35
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4$\begingroup$ @Karl: There is absolutely no requirement on this site that questions must not be about "basic maths". We welcome questions at all levels! $\endgroup$– hmakholm left over MonicaDec 27, 2014 at 18:37
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2$\begingroup$ @Karl: And you're wrong about that. It is completely fine for him to ask the question here. It is simply not correct that he has to go "elsewhere" to learn basic before he can ask here! $\endgroup$– hmakholm left over MonicaDec 27, 2014 at 18:44
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3$\begingroup$ @Karl: Your spat with Henning is simply a misunderstanding! "Elsewhere" certainly does not mean the same as "otherwise": it means, unambiguously, "in another place", i.e. not here on math.stackexchange. Probably you were thinking of "else"; but "otherwise" is more correct here. $\endgroup$– TonyKDec 27, 2014 at 19:05
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6 Answers
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Since $2=\sqrt{2}\cdot\sqrt{2}$ you have that $$\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{1}{\sqrt{2}}$$
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One thing we tend to like to do when dealing with this stuff is have the denominator of fractionish things as a plain number, as much as possible. To do this, we'll multiply both top and bottom by something that will cancel out any radicals in the bottom. $$\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$ This even works for more complicated stuff. $$\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{\sqrt{5}+\sqrt{3}}\cdot\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}=\frac{\sqrt{5}-\sqrt{3}}{2}$$
answered Dec 27, 2014 at 18:38
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$\begingroup$ I would argue that $\frac{1}{\sqrt{2}}$ is no less and probably more simple than $\frac{\sqrt{2}}{2}$, but most middle- and high-school math teachers inexplicably disagree. $\endgroup$– anomalyJul 16, 2016 at 18:58
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$\begingroup$ Of the three common formats, I'd pick $\sqrt{\frac{1}{2}}$ over $\frac{\sqrt{2}}{2}$ over $\frac{1}{\sqrt{2}}$. Halving is a lot easier to think at than dividing by an irrational number. $\endgroup$ Jul 16, 2016 at 19:03
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$\begingroup$ I don't know what you think is complicated about dividing by an irrational number, but it ultimately doesn't matter. $\endgroup$– anomalyJul 17, 2016 at 22:46
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Let's square both of them: \begin{align} \sqrt{\frac{1}{2}}^2 &= \frac{1}{2}, \text{ while} \\ \left(\frac{\sqrt{2}}{2}\right)^2 &=\frac{\sqrt{2}^2}{2^2} \\ &=\frac{2}{4} \\ &= \frac{1}{2} \end{align}
So they're both positive numbers, and their squares are the same, so they must be the same.
answered Dec 27, 2014 at 18:35
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The number $\frac{1}{\sqrt{2}}$ is defined to be the number such that, when you multiply it by $\sqrt{2}$, you get $1$. Symbolically, it is the solution to $x\cdot \sqrt{2}=1$. That's how division works - it's the inverse of multiplication. What is $\frac{\sqrt{2}}2\cdot \sqrt{2}$? Why, it's $\frac{2}2=1$ - so it must be $\frac{1}{\sqrt{2}}$, as it satisfies the definition of division for $\frac{1}{\sqrt{2}}$.
answered Dec 27, 2014 at 18:35
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$\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}}$
$\frac{\sqrt{2}}{2}=\frac{2^{\frac{1}{2}}}{2^1}=2^{\frac{1}{2}-1}=2^{-\frac{1}{2}}$
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The sqrt(2)/2 turns into sqrt(2/4) when the two is squared and put under the square root which simplified is sqrt(1/2).
