Find the Reciprocal - square root of 3

$- \sqrt{3}$

Step 1

To find the reciprocal, divide $1$ by the number given.

$\frac{1}{- \sqrt{3}}$

Step 2

Simplify.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- 1 (- 1)}{- \sqrt{3}}$

Move the negative in front of the fraction.

$- \frac{1}{\sqrt{3}}$

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$- (\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$- \frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$- \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$- \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$- \frac{\sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$- \frac{\sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$- \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$- \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$- \frac{\sqrt{3}}{3^{1}}$

Evaluate the exponent.

$- \frac{\sqrt{3}}{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$- \frac{\sqrt{3}}{3}$

Decimal Form:

$- 0.57735026 \dots$