How do you rationalize the denominator $sqrt2/(2-sqrt2)$ ?

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Answer 1 · 2015-04-01T12:02:46

Multiply by $1$ in the form: $(2+sqrt2)/(2+sqrt2)$ .
(This doesn't change the value of the number, it only changes how the number is written.)

$sqrt2/(2-sqrt2)=sqrt2/(2-sqrt2) * (2+sqrt2)/(2+sqrt2)=(sqrt2(2+sqrt2))/((2-sqrt2)(2+sqrt2))$ . So,

$sqrt2/(2-sqrt2) = (2sqrt2 + (sqrt2)^2)/(4-(sqrt2)^2) = (2sqrt2 + 2)/(4-2)$

$sqrt2/(2-sqrt2) = (2sqrt2 + 2)/2 = (2(1+sqrt2))/2 = 1+ sqrt2$

Why does it work?

We can verify that $(a-b)(a+b)=a^2-b^2$ for any numbers, $a$ and $b$ .

So we can also verify that $(a-sqrtc)(a+sqrtc)=a^2 -c$ . (no square roots).

By the way, we can also verify that $(sqrta-sqrtb)(sqrta+sqrtb)$ has no square roots.

How do you rationalize the denominator sqrt2/(2-sqrt2)sqrt2/(2-sqrt2)?