Question

what is the following quotient?
$\frac{1}{1+sqrt{3}}$
options:
$\frac{1-sqrt{3}}{4}$
$\frac{1+sqrt{3}}{4}$
$\frac{sqrt{3}}{4}$
$\frac{-1+sqrt{3}}{2}$

Explanation:

Step1: Rationalize the denominator

To simplify $\frac{1}{1 + \sqrt{3}}$, we multiply the numerator and denominator by the conjugate of the denominator, which is $1 - \sqrt{3}$.
$$\frac{1}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$$

Step2: Multiply the numerators and denominators

The numerator is $1\times(1 - \sqrt{3}) = 1 - \sqrt{3}$.
The denominator is $(1 + \sqrt{3})(1 - \sqrt{3})$, which is a difference of squares: $1^2 - (\sqrt{3})^2 = 1 - 3 = -2$.
So we have $\frac{1 - \sqrt{3}}{-2}$.

Step3: Simplify the fraction

We can rewrite $\frac{1 - \sqrt{3}}{-2}$ as $\frac{-1 + \sqrt{3}}{2}$ or $\frac{\sqrt{3}-1}{2}$, but let's check the options. Wait, maybe I made a mistake in sign. Wait, actually, let's re - do the sign:
$\frac{1 - \sqrt{3}}{-2}=\frac{-( \sqrt{3}-1)}{-2}=\frac{\sqrt{3}-1}{2}$? No, wait, let's multiply numerator and denominator by - 1: $\frac{1 - \sqrt{3}}{-2}=\frac{\sqrt{3}-1}{2}$. But the option we have is $\frac{-1+\sqrt{3}}{2}$, which is the same as $\frac{\sqrt{3}-1}{2}$. Wait, but also, let's check another way. Wait, maybe I miscalculated the denominator. Wait, $(a + b)(a - b)=a^{2}-b^{2}$, so $(1+\sqrt{3})(1 - \sqrt{3})=1 - 3=-2$. Then $\frac{1\times(1 - \sqrt{3})}{-2}=\frac{1 - \sqrt{3}}{-2}=\frac{\sqrt{3}-1}{2}=\frac{-1+\sqrt{3}}{2}$. But let's check the options again. Wait, maybe there is a miscalculation. Wait, another approach:
Wait, the options include $\frac{1 - \sqrt{3}}{4}$? No, wait, maybe I made a mistake in the conjugate. Wait, no, the conjugate of $1+\sqrt{3}$ is $1 - \sqrt{3}$. Wait, let's recalculate:

$\frac{1}{1+\sqrt{3}}\times\frac{1 - \sqrt{3}}{1 - \sqrt{3}}=\frac{1-\sqrt{3}}{1 - 3}=\frac{1 - \sqrt{3}}{-2}=\frac{\sqrt{3}-1}{2}=\frac{-1+\sqrt{3}}{2}$. So the answer should be $\frac{-1 + \sqrt{3}}{2}$ (which is the same as $\frac{\sqrt{3}-1}{2}$). But looking at the options, one of the options is $\frac{-1+\sqrt{3}}{2}$.

Wait, maybe I made a mistake in the sign when simplifying. Let's do it again:

$\frac{1}{1+\sqrt{3}}\times\frac{1 - \sqrt{3}}{1 - \sqrt{3}}=\frac{1-\sqrt{3}}{(1)^{2}-(\sqrt{3})^{2}}=\frac{1 - \sqrt{3}}{1 - 3}=\frac{1 - \sqrt{3}}{-2}=\frac{-( \sqrt{3}-1)}{-2}=\frac{\sqrt{3}-1}{2}=\frac{-1+\sqrt{3}}{2}$. So the correct option is the one with $\frac{-1+\sqrt{3}}{2}$.

Answer:

$\frac{-1+\sqrt{3}}{2}$ (the option with $\frac{-1+\sqrt{3}}{2}$)