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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.

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 $b=asqrt{2}$, 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

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?

$$
x = frac{2a}{b} = frac{b}{frac{a}{2}} = frac{2}{frac{a}{b}} = frac{4}{frac{2a}{b}}=frac{4}{x}
$$

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

$$b^2=sqrt2,$$ $$b=sqrt[4]2,m,\a=frac1{sqrt[4]2},m.$$