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How does $\sqrt{n+1}-\sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1}+\sqrt{n}}$ ?

What are the exact steps to get to the right side from the left?

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    $\begingroup$ You can't because they are not equal. Check your signs. It should be $+$ somewhere, perhaps in the denominator. $\endgroup$ – user147263 Feb 17 '15 at 5:01
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    $\begingroup$ There should be a $+$ in the denominator $\endgroup$ – AvZ Feb 17 '15 at 5:03
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It's the hint with almost solution and quite good technic for such problems.

$$ \bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot 1 = \bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$$

Useful formula for this task:

$$ \bigg( \sqrt{a} - \sqrt{b} \bigg) \cdot \bigg( \sqrt{a} + \sqrt{b} \bigg) = a - b $$

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hint: $ 1 = (n+1) - n = \left(\sqrt{n+1}\right)^2 - \left(\sqrt{n}\right)^2 = \left(\sqrt{n+1} + \sqrt{n}\right)(?-?)$

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Hint: Evaluate $$\sqrt{n+1}-\sqrt{n} = (\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$$

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