Estonian Mathematical Olympiad

Any number that can be obtained after a finite number of steps is of the form $r\sqrt{2}+s$, where $r$ is positive and $s$ is a nonnegative rational number. Indeed, the initial number $\sqrt{2}$ is of the form $1\sqrt{2}+0$, and if we multiply a number of the form $r\sqrt{2}+s$ by a positive rational number $t$ or add a natural number $k$ to it, the result is $rt\sqrt{2}+st$ or $r\sqrt{2}+(s+k)$ correspondingly, which is of the same form.

Show that any number of the form $r\sqrt{2}+s$, where $r$ is positive and $s$ is a nonnegative rational number, can be obtained by at most 3 steps. If $s>0$ then multiply $\sqrt{2}$ by a positive rational number $\frac{r}{s}$, add 1 to the result, and then multiply the result by the positive rational number $s$. If $s=0$ then it is enough to multiply $\sqrt{2}$ by the positive rational number $r$.

Finally, we show that 2 steps are not always enough. First, make sure that in the representation $r\sqrt{2}+s$ of a number, $r$ and $s$ are uniquely determined. Indeed, if $r\sqrt{2}+s=r^{\prime }\sqrt{2}+s^{\prime }$ then $(r-r^{\prime })\sqrt{2}=s^{\prime }-s$. If $r-r^{\prime }\ne 0$ then $\sqrt{2}=\frac{s^{\prime }-s}{r-r^{\prime }}$, which is not possible, because $\frac{s^{\prime }-s}{r-r^{\prime }}$ is a rational number, but $\sqrt{2}$ is an irrational number. Hence $r-r^{\prime }=0$ or $r^{\prime }=r$; from $(r-r^{\prime })\sqrt{2}=s^{\prime }-s$ also $s^{\prime }=s$.

Consider the number $r\sqrt{2}+s$ where $r=1$ and $s$ is an arbitrary fractional number. We cannot get this number by multiplications only, because then $s$ remains zero, or with additions only, because then $s$ would be an integer. If we multiply by a rational number $t$ and add a natural number $k$ in either order then the result is either $t\sqrt{2}+k$ or $t(\sqrt{2}+k)$. In both cases we must have $t=r=1$. But multiplication by 1 does not change the result, and we already saw that by addition only it is not possible to get the result.