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Why $\sqrt {12} = 2 \sqrt 3$? It is obvious? If we considered the function $f(s) = s^2 $ it is injective on positive numbers so we obtain the conclusion. But in the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots.

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    $\begingroup$ Well, $\sqrt{\frac{1}{2}}=\frac{\sqrt2}{2}$. Irrational numbers can still be equal to one another. If this were a case of irrational$=$rational, I can see how it might be a problem, but that's not what's happening here. $\endgroup$
    – πr8
    Jan 11, 2016 at 7:33
  • $\begingroup$ $\sqrt {ab}=\sqrt a\cdot \sqrt b$ and $\sqrt {c^2}=\pm c$. You must choose here the positive sign. $\endgroup$
    – Piquito
    Jan 12, 2016 at 11:47
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    $\begingroup$ Put another way, $\sqrt{12} = \sqrt{2^2 \times 3}$. $\endgroup$ Jan 13, 2016 at 3:18
  • $\begingroup$ This derives from $\sqrt{ab}=\sqrt a\sqrt b$, itself from $(cd)^2=c^2d^2$. $\endgroup$
    – user65203
    Jan 13, 2016 at 15:22

4 Answers 4

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Let's assume (unless that's the part you want proven) that there is only 1 positive number $x$ with the property $x^2=12$. We'll denote that number by $\sqrt{12}$. Likewise let $\sqrt{3}$ be the positive $y$ such that $y^2=3$.

We know that $\Bbb R$ is a field and thus multiplication is commutative. We also know that because $2\gt 0$ and $\sqrt{3}\gt 0$ that $2\sqrt{3}\gt 0$. Then $$(2\sqrt{3})^2 = (2\sqrt{3})(2\sqrt{3}) = (2^2)(\sqrt{3})^2 = 4\cdot 3=12=\sqrt{12}^2 \\ \implies 2\sqrt{3}=\sqrt{12}$$

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  • $\begingroup$ It is just a euristic question.Let's suppose that we know just to compute the square roots. $\endgroup$
    – all
    Jan 11, 2016 at 7:41
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    $\begingroup$ I don't know what you mean by "computing" the square roots. If you mean using one of the myriad methods of finding a rational number that approximates the square root to whatever precision we like, then we still need a definition of the number to know if our approximation is at all good. That definition is $\sqrt{a}$ is the nonnegative number $x$ such that $x^2=a$ where $a\ge 0$. $\endgroup$
    – user137731
    Jan 11, 2016 at 7:50
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By definition, $\sqrt3$ is the positive number $s$ such that $s^2=3.$ Likewise, $\sqrt{12}$ is the positive number $t$ such that $t^2=12.$ The claim, then, is that $t=2s,$ but is it true? Well, $2s$ is a positive number, since it is a product of two positive numbers. Moreover, $$(2s)^2=2^2s^2=4s^2=4\cdot 3=12.$$ Hence, $2s=t,$ as desired.

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  • $\begingroup$ I am agree , but my question is like that:2/4=1/2 because they are 0,5.They have the same digits $\endgroup$
    – all
    Jan 11, 2016 at 7:38
  • $\begingroup$ $\sqrt{12}$ and $2\sqrt{3}$ also have all the same digits. They just happen to have infinitely-many digits, and those digits don't eventually cycle. $\endgroup$ Jan 11, 2016 at 7:40
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    $\begingroup$ It is a truly terrible idea to try checking the decimal expansions to see if two irrational numbers are equal. Who can ever check infinitely-many digits? Likewise, it isn't necessary to check digits in the rational case! By definition, $\frac12$ is the number $s$ such that $2s=1;$ $\frac24,$ the number $t$ such that $4t=2$. Since $$4s=2\cdot2s=2\cdot1=2,$$ then $s=t,$ meaning that $\frac12=\frac24.$ $\endgroup$ Jan 11, 2016 at 7:58
  • $\begingroup$ More briefly, $\frac12=\frac24$ because $1\cdot4=2\cdot2.$ Likewise, $\frac24=\frac5{10}$ because $2\cdot10=4\cdot5.$ By transitivity of equality, $\frac12=\frac5{10}.$ Recalling that $0.5$ is simply another way of writing $\frac5{10},$ we're done. No calculator or computer required. $\endgroup$ Jan 11, 2016 at 8:07
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Since,$$12 = 4\times 3 = 2^2 \times 3$$ $$\sqrt{12} = \sqrt{2^2\times 3} = \sqrt{2^2}\times\sqrt{3}=2\sqrt{3}$$

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  • $\begingroup$ but he asks why $\sqrt{4}=2$ not $\sqrt{4}=-2.$ $\endgroup$
    – Leox
    Jan 11, 2016 at 7:29
  • $\begingroup$ No.My question is:Why they are equal.They are irațional numbers , if we compute the square roots. $\endgroup$
    – all
    Jan 11, 2016 at 7:30
  • $\begingroup$ @Leox: I don't think the OP is asking about this. $\endgroup$ Jan 11, 2016 at 7:30
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Let's consider

\begin{aligned} f : \mathbf{R_+} &\to \mathbf{R_+}\\ x &\to x^2 \end{aligned}

then $f$ is bijection, for it is strictly increasing.

$f(\sqrt{12}) = f(2\sqrt{3}) = 12$, so $\sqrt{12} = 2\sqrt{3}$.

Note that $\sqrt{12}$ is just a notation, there is no problem to denote one number with different notations. For example, we can also use $[3, \overline{[2,6]}]$ which means to be a continued fraction to denote $\sqrt{12}$.

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  • $\begingroup$ @Bye_World ah, It is surely a mistake. I have corrected it. Thanks! $\endgroup$
    – wangjiezhe
    Jan 13, 2016 at 1:23

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