Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
 A: \begin{align*}\frac{3}{\frac{\sqrt{5}}{5}} &= 3 \cdot \frac{5}{\sqrt 5}\\
&= 3 \cdot \frac{5}{\sqrt 5} \cdot 1\\
&= 3 \cdot \frac{5}{\sqrt 5} \cdot \frac{\sqrt 5}{\sqrt 5}\\
&= 3 \cdot \frac{5\sqrt 5}{5}\\
&= 3\sqrt 5
\end{align*}
A: You multiplied the original fraction by $\dfrac5{\sqrt5}$, which is not $1$, so of course you changed the value. The correct course of action is to multiply by $1$ in the form
$$\frac{5/\sqrt5}{5/\sqrt5}$$
to get
$$\frac3{\frac{\sqrt5}5}=\frac3{\frac{\sqrt5}5}\cdot\frac{\frac5{\sqrt5}}{\frac5{\sqrt5}}=\frac{3\cdot\frac5{\sqrt5}}1=3\cdot\frac5{\sqrt5}=3\sqrt5\;,$$
since $\dfrac5{\sqrt5}=\sqrt5$.
More generally,
$$\frac{a}{b/c}=\frac{a}{\frac{b}c}\cdot\frac{\frac{c}b}{\frac{c}b}=\frac{a\cdot\frac{c}b}1=a\cdot\frac{c}b\;,$$
this is the basis for the invert and multiply rule for dividing by a fraction.
A: This means
$$
3\cdot \frac{5}{\sqrt{5}}=3\cdot\frac{(\sqrt{5})^2}{\sqrt{5}}
=3\sqrt{5}
$$
You're multiplying twice for the reciprocal of the denominator.
Another way to see it is multiplying numerator and denominator by the same number:
$$
\frac{3}{\frac{\sqrt{5}}{5}}=\frac{3\sqrt{5}}{\frac{\sqrt{5}}{5}\cdot\sqrt{5}}
=\frac{3\sqrt{5}}{1}
$$
A: Start with $$\frac{3}{\sqrt{5}/5}=\frac{15}{\sqrt{5}},$$
and then rationalize the denominator (multiply both numerator and denominator by $\sqrt{5}$) to get
$$\frac{15\sqrt{5}}{5}=3\sqrt{5}.$$
A: You have the following fraction to simplify:
$$\begin{align} \frac{3}{\sqrt{5}/5} &=\frac{5\times 3}{\sqrt{5}} \\ &=\frac{15}{\sqrt{5}} \\ &=\sqrt{\bigg(\frac{15}{\sqrt{5}}\bigg)^2} \\ &=\sqrt{\frac{15^2}{\sqrt{5}^2}} \\ &=\sqrt{\frac{225}{5}} \\ &=\sqrt{45} \\ &= \sqrt{9\times 5} \\ &= \sqrt{9}\sqrt{5} \\ &= 3\sqrt{5} \\ \therefore \frac{15}{\sqrt{5}}\times \frac{5}{\sqrt{5}} &=\frac{15}{\sqrt{5}}\times \frac{\big(\frac{15}{\sqrt{5}}\big)}{3} \\ &=\frac{15}{\sqrt{5}}\times \frac{15}{\sqrt{5}}\times \frac{1}{3} \\ &=\bigg(\frac{15}{\sqrt{5}}\bigg)^2 \times \frac{1}{3} \\ &=45\times \frac{1}{3} \\ &=\frac{45}{3} \\ &=15 \end{align}$$
A: One thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.
$$\begin{align}
\frac{3}{\frac{5}{\sqrt{5}}}&=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}\cdot\frac{\frac{\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{1}=\frac{3}{1}\cdot\frac{\sqrt{5}}{5}=\text{etc.}\\
\end{align}$$
A: In the expression $\frac{3}{\frac{\sqrt{5}}{5}}$, to find the reciprocal of that expression's denominator, $\frac{\sqrt{5}}{5}$, simply swap its numerator and denominator. 
Thus the reciprocal of $\frac{\sqrt{5}}{5}$ is $\frac{5}{\sqrt{5}} $, and $$ \frac{3}{\frac{\sqrt{5}}{5}} = 3 \cdot \frac{5}{\sqrt{5}}$$
A: I think the result you have to remember is 
$$\boxed{\dfrac 1{\sqrt{a}}=\dfrac{\sqrt{a}}a}$$
You will see this kind of manipulation very often in your studies, and it allows to get rid of square roots on denominator.
Here your expression is just $\dfrac{3}{\frac{\sqrt{5}}{5}}=\dfrac{3}{\frac 1{\sqrt{5}}}=3\sqrt{5}$
