Solve 1/1+frac{1{sqrt{2}}}

$$\frac { 1 } { 1 + \frac { 1 } { \sqrt { 2 } } }$$

Evaluate

$$2-\sqrt{2}\approx 0.585786438$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{1}{1+\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}}$$

The square of \sqrt{2} is 2.

$$\frac{1}{1+\frac{\sqrt{2}}{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

$$\frac{1}{\frac{2}{2}+\frac{\sqrt{2}}{2}}$$

Since \frac{2}{2} and \frac{\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{2+\sqrt{2}}{2}}$$

Divide 1 by \frac{2+\sqrt{2}}{2} by multiplying 1 by the reciprocal of \frac{2+\sqrt{2}}{2}.

$$\frac{2}{2+\sqrt{2}}$$

Rationalize the denominator of \frac{2}{2+\sqrt{2}} by multiplying numerator and denominator by 2-\sqrt{2}.

$$\frac{2\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}$$

Consider \left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{2\left(2-\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 2. Square \sqrt{2}.

$$\frac{2\left(2-\sqrt{2}\right)}{4-2}$$

Subtract 2 from 4 to get 2.

$$\frac{2\left(2-\sqrt{2}\right)}{2}$$

Cancel out 2 and 2.

$$2-\sqrt{2}$$