Solve sqrt{2^-1}=frac{sqrt{1}}{sqrt{2}}

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

Verify

Solution Steps

Calculate 2 to the power of -1 and get \frac{1}{2}.

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

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{2} is 2.

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$$

Subtract \frac{\sqrt{2}}{2} from both sides.

$$\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=0$$

Combine \frac{\sqrt{2}}{2} and -\frac{\sqrt{2}}{2} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$