Divide a rectangle $a \times b$ in half and solve for the ratio $x=\frac{b}{a}$ given the ratio remains the same. I'm doing some Calculus. The instructor gave this example: 
$$x = \frac{b}{a} = \frac{a}{b/2} = \frac{2a}{b} = \frac{2}{b/a} = \frac{2}{x}$$
I have this problem right here:
$$x = \frac{2a}{b} = \frac{b}{a/2}$$
and another problem just like it:
$$x = \frac{a}{b} = \frac{b}{a/3}$$
I'm taking an introduction to Calculus on Coursera and although I think it's a great resource they kind of go through topics very quickly and don't explain much. So I am very confused as to how to solve this equation.
Any help would be much appreciated. Thanks.
EDIT: I think this image will help more.

I can't embed images yet.
 A: An example of your setting is the A series of ISO paper format.
For instance, if you take a sheet of A$4$ paper and fold it along the line cutting it in halves along the line connecting the middle points of the longer sides, you end up with two sheets of A$5$ paper.
The property is that the ratio “long side/short side” remains constant under the operation of halving the sheet in the way described above.
Thus, if $x$ is this ratio and $a$ and $b$ are the lengths of the sides, with $a<b$, when you halve the sheet, the sides become $a$ and $b/2$ (with $b/2<a$).
The required property is that the ratio remains invariant: then
$$
x=\frac{b}{a}\qquad x=\frac{a}{b/2}
$$
Now
$$
\frac{a}{b/2}=\frac{2a}{b}=\frac{2}{b/a}=\frac{2}{x}
$$
Hence
$$
x=\frac{2}{x}
$$
and therefore $x^2=2$, so $x=\sqrt{2}$.
Indeed the A series is exactly defined this way. An A$0$ sheet has the sides $a$ and $b$ so that $b/a=\sqrt{2}$ and the area is $1\,\mathrm{m}^2$. Therefore $ab=1\,\mathrm{m}^2$ and, since $b=a\sqrt{2}$, we get that 
$$
a=\frac{1}{\sqrt[4]{2}}\,\mathrm{m}\approx 841\,\mathrm{mm}
\qquad
b=\sqrt[4]{2}\,\mathrm{m}\approx 1189\,\mathrm{mm}
$$
The physical lengths are rounded at the millimeter.
Dividing each time by $2$ and switching the sides:
\begin{align}
\text{A}0&& a&=841\,\mathrm{mm} & b&=1189\,\mathrm{mm}\\
\text{A}1&& a&=594\,\mathrm{mm} & b&=841\,\mathrm{mm}\\
\text{A}2&& a&=420\,\mathrm{mm} & b&=594\,\mathrm{mm}\\
\text{A}3&& a&=297\,\mathrm{mm} & b&=420\,\mathrm{mm}\\
\text{A}4&& a&=210\,\mathrm{mm} & b&=297\,\mathrm{mm}\\
\text{A}5&& a&=148\,\mathrm{mm} & b&=210\,\mathrm{mm}
\end{align}
which agrees with ISO 216 on Wikipedia
A: I'm guessing about the meaning of this: are you supposed to rewrite the third term so that the second term (and hence $x$) reappears?
$$
x = \frac{2a}{b} = \frac{b}{\frac{a}{2}} = \frac{2}{\frac{a}{b}} = \frac{4}{\frac{2a}{b}}=\frac{4}{x}
$$
A: This is how the $A_0,A_1,A_2,\cdots$ sheet formats are defined.
Let the ratio of the long size over the short one be $$x=\dfrac ba.$$ Then the half sheet has the same ratio and
$$x=\dfrac a{\frac b2}=\frac{2a}b.$$
Mutiplying the two equations, we obtain
$$x^2=2,$$ i.e. $$x=\sqrt 2.$$

Knowing that the $A_0$ is $1\,m^2$,
$$1=ab=\frac{b^2}x$$ and
$$b^2=\sqrt2,$$ $$b=\sqrt[4]2\,m,\\a=\frac1{\sqrt[4]2}\,m.$$
