Solve 1-(frac{sqrt{2}}{1+sqrt{2}})=

$$1 - ( \frac { \sqrt { 2 } } { 1 + \sqrt { 2 } } ) =$$

Evaluate

$$\sqrt{2}-1\approx 0.414213562$$

Solution Steps

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

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

Consider \left(1+\sqrt{2}\right)\left(1-\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}.

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

Square 1. Square \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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

Anything divided by -1 gives its opposite.

$$1-\left(-\sqrt{2}\left(1-\sqrt{2}\right)\right)$$

Use the distributive property to multiply \sqrt{2} by 1-\sqrt{2}.

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

The square of \sqrt{2} is 2.

$$1-\left(-\left(\sqrt{2}-2\right)\right)$$

To find the opposite of \sqrt{2}-2, find the opposite of each term.

$$1-\left(-\sqrt{2}-\left(-2\right)\right)$$

The opposite of -2 is 2.

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

To find the opposite of -\sqrt{2}+2, find the opposite of each term.

$$1-\left(-\sqrt{2}\right)-2$$

The opposite of -\sqrt{2} is \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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