Simplify and rationalize the denominator: $$\frac{1}{1+ \frac{1}{\sqrt{3}+1}}.$$
To begin, we first consider the $\frac{1}{\sqrt{3} + 1}$ term. We can multiply both the numerator and the denominator by the conjugate of the denominator to get $$\frac{1}{\sqrt{3} + 1} = \frac{1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{\sqrt{3}-1}{3-1} = \frac{\sqrt{3}-1}{2}.$$We can then substitute this back into our original expression and multiply both the numerator and denominator by $2$ to get  \begin{align*}
\frac{1}{1+ \frac{1}{\sqrt{3}+1}} & = \frac{1}{1 + \frac{\sqrt{3} - 1}{2}} \\
& = \frac{2}{2 + \sqrt{3} - 1} \\
& = \frac{2}{\sqrt{3} + 1}.
\end{align*}If we multiply both the numerator and denominator of this expression by $\sqrt{3}-1$ and simplify, we end up with  \begin{align*}\frac{2}{\sqrt{3} + 1} &= \frac{2}{\sqrt{3} + 1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\&= \frac{2(\sqrt{3}-1)}{3 - 1} = \frac{2(\sqrt{3}-1)}{2} = \boxed{\sqrt{3}-1}.\end{align*}
\sqrt{3}-1
Level 4
Algebra
Problem: Simplify and rationalize the denominator: $$\frac{1}{1+ \frac{1}{\sqrt{3}+1}}.$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first, we need to address the inner fraction. The term $\frac{1}{\sqrt{3}+1}$ can be simplified by rationalizing the denominator. We multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3}-1$. This gives us:

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

Next, we calculate the denominator:

$$(\sqrt{3}+1)(\sqrt{3}-1) = \sqrt{3}^2 - 1^2 = 3 - 1 = 2.$$

Thus, we have:

$$\frac{1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{2}.$$

Now, substituting this back into the original expression, we get:

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

To combine the terms in the denominator, we rewrite 1 as $\frac{2}{2}$:

$$1 + \frac{\sqrt{3}-1}{2} = \frac{2}{2} + \frac{\sqrt{3}-1}{2} = \frac{2 + \sqrt{3} - 1}{2} = \frac{1 + \sqrt{3}}{2}.$$

Now, we can rewrite the entire expression:

$$\frac{1}{\frac{1+\sqrt{3}}{2}} = \frac{2}{1+\sqrt{3}}.$$

Next, we rationalize the denominator again by multiplying by the conjugate of $1+\sqrt{3}$, which is $1-\sqrt{3}$:

$$\frac{2}{1+\sqrt{3}} \cdot \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{2(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}.$$

Calculating the denominator:

$$(1+\sqrt{3})(1-\sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2.$$

Thus, we have:

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

Therefore, the final answer is $\sqrt{3}-1$.<|END_OF_LATENT|> The final answer is $\sqrt{3}-1$.