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

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

  • 10
    Where is the surprise? If you noticed $\sqrt{4}=\frac{4}{\sqrt{4}}$, i.e. $2=\frac{4}{2}$, would you be surprised?2011-04-12
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    @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.2011-04-12
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    @Danny $x=\frac{n}{x}$ if $x=\sqrt{n}$.. did this rid it of the mystique?2011-04-12
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    Perhaps someone should correct the title?2011-04-12
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    Oops sorry, didn't see that! Corrected.2011-04-13
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    In trying to answer the question you added I ask (**Added**: I didn't see that you just fixed the title before posting this comment): Yes, sure, but what's the definition of $\sqrt{2}$? It's one of the two solutions of $x^2 = 2$. Since $x$ can't be zero, we may divide this equation by $x$ and get for a solution that $x = \frac{2}{x}$. As I pointed out in another comment, there are two solutions to this equation, namely $\sqrt{2}$ and $-\sqrt{2}$.2011-04-13
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    You could have taken about ten seconds to try to solve the simple equation you wrote in the second line...2013-06-13

4 Answers 4

12

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 = $\sqrt2$.

3

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 $x = \sqrt2$

  • 4
    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.2011-04-12
3

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

2

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.