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