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# surds

+2
193
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show that $$\frac{1}{1+\frac{1}{\sqrt{}2}}$$ can be written as $$2-\sqrt{2}$$

Guest Nov 27, 2017

#1
+7266
+3

$$\frac{1}{1+\frac{1}{\sqrt2}}$$

Rewrite the  1  in the denominator as  $$\frac{\sqrt2}{\sqrt2}$$ .

$$\frac{1}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt2}}$$

Add the fractions in the denominator together.

$$\frac{1}{\frac{\sqrt2+1}{\sqrt2}}$$

That is the same as...

$$1\div\frac{\sqrt2+1}{\sqrt2}$$

Invert the second fraction and multiply.

$$1\,\cdot\,\frac{\sqrt2}{\sqrt2+1}$$

$$\frac{\sqrt2}{\sqrt2+1}$$

Multiply the numerator and denominator by  $$\sqrt2-1$$ .

$$\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}$$

$$\frac{2-\sqrt2}{2-1}$$

$$\frac{2-\sqrt2}{1}$$

$$2-\sqrt2$$

hectictar  Nov 27, 2017
#1
+7266
+3

$$\frac{1}{1+\frac{1}{\sqrt2}}$$

Rewrite the  1  in the denominator as  $$\frac{\sqrt2}{\sqrt2}$$ .

$$\frac{1}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt2}}$$

Add the fractions in the denominator together.

$$\frac{1}{\frac{\sqrt2+1}{\sqrt2}}$$

That is the same as...

$$1\div\frac{\sqrt2+1}{\sqrt2}$$

Invert the second fraction and multiply.

$$1\,\cdot\,\frac{\sqrt2}{\sqrt2+1}$$

$$\frac{\sqrt2}{\sqrt2+1}$$

Multiply the numerator and denominator by  $$\sqrt2-1$$ .

$$\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}$$

$$\frac{2-\sqrt2}{2-1}$$

$$\frac{2-\sqrt2}{1}$$

$$2-\sqrt2$$

hectictar  Nov 27, 2017
#2
+2

Great job, hectictar !. Thank you.

Guest Nov 27, 2017
edited by Guest  Nov 27, 2017