# Sums and products

On reddit, somebody asked: why does $\frac{7}{3} + \frac{7}{4} = \frac{7}{3}\times\frac{7}{4}$?

My first thought: that’s a neat variation of $2+2=2\times 2$. My second: when *does* that hold true?

It’s not too tricky to solve algebraically:

- $x + y = xy$
- $x = xy - y$
- $x = y(x-1)$
- $y =\frac{x}{x-1}$
- $y = 1 + \frac{1}{x-1}$.

That’s clearly true for $x=y=2$, and less clearly true (but still true) for $x= \frac{7}{4}$. So that’s neat: we can construct pairs of numbers that work for any $x$ except for 1.

I also wondered whether it turned out more neatly if, instead of starting from 0, we started from 2.

Suppose instead of calling our numbers $x$ and $y$, we call them $(2+p)$ and $(2+q)$.

Then we need to make $(2+p)(2+q) = 4+p+q$.

Multiplying out gives $4 + 2(p+q) + pq = 4 + p+q$

Then we get $p+q + pq = 0$, or $p + q(1+p) = 0$, so $q = -\frac{p}{1+p}$.

We could also make that $q = \frac{1}{1+p} - 1$. That’s not, honestly, much better. What about around 1?

If our numbers are $(1+a)$ and $(1+b)$, then the sum is $1 + a + b + ab = 2 + a + b$, so we get $ab = 1$. That’s significantly nicer! If we take two reciprocals and add them to 1, we get a solution.

The (2,2) solution comes from $a=b=1$; the $\left(\frac{7}{3}, \frac{7}{4}\right)$ one comes from $a = \frac{4}{3}$ and $b =\frac{3}{4}$.