JBMO2020 wrote: Let $S$ be a set of real numbers such that: i) $1$ is from $S$; ii) for any $a, b$ from $S$ (not necessarily different), we have that $a-b$ is also from $S$; iii) for any $a$ from $S$ ($a$ is different from $0$), we have that $1/a$ is from $S$. Show that for every $a, b$ from $S$, we have that $ab$ is from $S$. $1\in S$ $\implies$ $1-1=0\in S$ $a,0\in S$ $\implies$ $0-a=-a\in S$ $a,b\in S$ $\implies$ $a-b\in S$ $\implies$ $\frac 1{a-b}\in S$ for $a\ne b$ $a,b\in S$ $\implies$ $\frac 1a,\frac 1b\in S$ $\implies$ $\frac 1b-\frac 1a=\frac{a-b}{ab}\in S$ for $a,b\ne 0$ $\implies$ $\frac{ab}{a-b}\in S$ for $a\ne b$ (constraint $a,b\ne 0$ is no more useful $1,a,b\in S$ $\implies$ $a+1,b+1\in S$ $\implies$ (see above) $\frac{(a+1)(b+1)}{a-b}\in S$ for $a\ne b$ So $\frac{(a+1)(b+1)}{a-b}-\frac{ab}{a-b}=\frac{a+b+1}{a-b}\in S$ for $a\ne b$ So $\frac{a+b+1}{a-b}-\frac 1{a-b}=\frac{a+b}{a-b}\in S$ for $a\ne b$ Replacing there $(a,b)$ by $(a+\frac 1b,a-\frac 1b)$, this implies $ab\in S$ for $b\ne 0$, still true when $b=0$ Q.E.D.