Reformulate $x/(x+y)$ using a single $x$ and $y$

Is it possible to reformulate this expression to only list $$x$$ and $$y$$ once using common math functions? The ranges of $$x$$ and $$y$$ are both $$0$$ to $$1$$ inclusive if that helps.

Common being those typically found on a scientific calculator.

• Use the reciprocal $\frac{x+y}{x}=1+\frac yx$ Oct 13, 2021 at 23:29

This is slightly more complicated, but $$\frac{1}{1+\frac{y}{x}}$$ is one possibility.

• Not to be a pedant, but I would argue that this is not technically a reformulation, as it doesn't include the possibility of $x = 0$ (which is explicitly included in the question). Oct 13, 2021 at 23:38
• if x = 0 then there is no need for reformulaton. Oct 13, 2021 at 23:40

First, if $$x=0$$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $$y\neq 0$$. As $$x,y\in[0,1]$$, it follows that we exclude the case $$x=y=0$$. More specially, if $$x=0$$ then $$y\neq 0$$ and if $$y=0$$ then $$x\neq 0$$. Next, if $$x+y\neq 0$$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $$A$$ gives

$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$

Then since the reciprocal of the reciprocal is our original function we have

$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $$x\neq 0$$ and $$x \neq -y$$. Since $$x,y\in[0,1]$$, the second condition is only valid when $$x=y=0$$ which we excluded.

• Indeed it's a simple check for near zero within a tolerance to avoid floating point underflow. :) Oct 31, 2021 at 14:29