Question

write the expression in simplest form. $\frac{1}{1+sqrt{3}} = \frac{square}{square}$

Explanation:

Step1: Rationalize the denominator

Multiply numerator and denominator by the conjugate $1-\sqrt{3}$:
$\frac{1}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{1-\sqrt{3}}{(1+\sqrt{3})(1-\sqrt{3})}$

Step2: Expand the denominator

Use difference of squares $(a+b)(a-b)=a^2-b^2$:
$\frac{1-\sqrt{3}}{1^2 - (\sqrt{3})^2} = \frac{1-\sqrt{3}}{1 - 3}$

Step3: Simplify the denominator

Calculate the denominator value:
$\frac{1-\sqrt{3}}{-2}$

Step4: Rewrite the expression

Rearrange to standard form:
$\frac{\sqrt{3}-1}{2}$

Answer:

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