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

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

Evaluate

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

Solution Steps

Calculate the square root of 1 and get 1.

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

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

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

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}.

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

Square 1. Square \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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

Anything divided by -1 gives its opposite. To find the opposite of 1-\sqrt{2}, find the opposite of each term.

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

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

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