# why does $\sqrt2 = \frac{2}{\sqrt2}$?

I noticed just now that $\sqrt2 = \frac{2}{\sqrt2}$

I'm suprised because isn't this like saying $x = \frac{2}{x}$?

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Where is the surprise? If you noticed $\sqrt{4}=\frac{4}{\sqrt{4}}$, i.e. $2=\frac{4}{2}$, would you be surprised? –  Henry Apr 12 '11 at 8:28
@Henry I think the source of my confusion was that I hadn't noticed that $x= \frac{2}{x}$ iff $x = \sqrt2$ whereas for any other $x$ it would not be true. –  Danny King Apr 12 '11 at 9:05
@Danny $x=\frac{n}{x}$ if $x=\sqrt{n}$.. did this rid it of the mystique? –  Please Delete Account Apr 12 '11 at 19:07
Perhaps someone should correct the title? –  wildildildlife Apr 12 '11 at 20:44
Oops sorry, didn't see that! Corrected. –  Danny King Apr 13 '11 at 8:02

Yes, just make sure you make sure you understand the distinctions below:

$\frac{2}{\sqrt2}$ is obtained by multiplying the $\sqrt2$ by 1 or $\frac{\sqrt2}{\sqrt2}$.

$$\left(\frac{\sqrt2}{1}\right)\left(\frac{\sqrt2}{\sqrt2}\right) = \frac{2}{\sqrt2}$$

This happens clearly because the two square roots cancel each other out due to the fact that the squaring operator is the inverse of the square-root operator, and vice versa.

Now let's perform the same method on x.

$$\left(\frac{x}{1}\right)\left(\frac{x}{x}\right) = \frac{x^2}{x}$$

$x \neq \frac{2}{x}$ unless we first explicitly state that x = $\sqrt2$.

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Mr_CryptoPrime's answer is the right way to think about this, but here is another way to see it. Since $\sqrt{2}$ can be defined as a positive real $x$ number such that $x^2=2$, we can ask ourselves whether $2/\sqrt{2}$ satisfies these two properties:

• $2/\sqrt{2}$ is a positive real number (because it is a quotient of two positive real numbers), and
• $\displaystyle \left(\frac{2}{\sqrt{2}}\right)^2 = \frac{2^2}{(\sqrt{2})^2} = \frac{4}{2} = 2.$

Thus, $2/\sqrt{2}$ satisfies the properties that define $\sqrt{2}$, and the two numbers must be equal.

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

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I think I have simply stated the definition of the square root, since multiplying $x = \frac{2}{x}$ by $x$ gives $x^2 = 2$, so $x = \sqrt2$

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Two things: First, you need to be careful: you omitted the solution $x = -\sqrt{2}$. Second: If you have a square root in the denominator it is good practice to eliminate it by expanding the fraction as described in Mr_CryptoPrime's answer. –  t.b. Apr 12 '11 at 8:08