Stability of the unit interval under the binary operation $x*y=x+y-xy$

Question

Let $I=[0,1]$ the unit interval. I want to show that If $x\in [0,1]$ and $y\in [0,1]$ then $x+y-xy\in [0,1]$

My first attempt is to write $x+y-xy= (1-y)x+y.1$ which is a point in the segment $[x,1]$ since $[x,1]$ is convex. Hence $x+y-xy\in I$.

My second attempt is to use a direct method without mentioning the notion of convexity: we have that $x-xy=x(1-y)\geq 0$ so $x\geq xy$ and similarly $y\geq xy$ hence $x+y\geq xy$ so $x+y-xy\geq 0$. But I don't see how to show that $xy-xy\leq 1$.

Answer 1

Write $x+y-xy$ as $1-(1-x)(1-y)$. Now note that $1-x,1-y \in [0,1]$ and conclude what you want.

Answer 2

You can use geometry too. There's a unit square. Assume point D has coordinates $(x,y)$, so D locates inside (or on the borders of) this unit square.

Notice that $x=area(OGEF)$, $y=area(OABC)$, $xy=area(OGDC)$, and the unit square's area is 1, we get $$ 0 \leq x+y-xy \leq 1 $$