WakeUp wrote: Prove that for any integer $n$ one can find integers $a$ and $b$ such that \[n=\left[ a\sqrt{2}\right]+\left[ b\sqrt{3}\right] \] Just consider the function $f(x)$ from $\mathbb Z\to\mathbb Z$ defined as $f(x)=\left\lfloor x\sqrt 3\right\rfloor$ + $\left\lfloor -x\sqrt 2\right\rfloor$ It's rather easy to show that $f(x+1)-f(x)< 2$ and $\lim_{x\to+\infty}f(x)=+\infty$ and $\lim_{x\to-\infty}f(x)=-\infty$ So $f(x)$ is a surjection. Hence the result.

Am I right in thinking this proof would work just as well if, the $\sqrt 2$and $\sqrt 3$ were replaced with, say, $1.4$ and $1.7$ ? Or does it depend on these numbers being irrational? Merlin

Merlinaeus wrote: Am I right in thinking this proof would work just as well if, the $\sqrt 2$and $\sqrt 3$ were replaced with, say, $1.4$ and $1.7$ ? Or does it depend on these numbers being irrational? Merlin Yes, you are perfectly right. This proof works for example with any $u>v>0$ such that $[u]=[v]$

is there any easier way or at least without using those terms?

/Bump , is there any 'elementary solution' except limits and stuff ? edit - btw can we try induction ? edit - also i couldn't understand how did he conclude surjection with that

MrOreoJuice wrote: /Bump , is there any 'elementary solution' except limits and stuff ? edit - btw can we try induction ? What pco said is very elementary. He just used a limit to show that the function can get arbitrarily big or small. No need to use the $\lim$ sign. Just say. "$f$ can get arbitrarily big or small"

pco wrote: WakeUp wrote: Prove that for any integer $n$ one can find integers $a$ and $b$ such that \[n=\left[ a\sqrt{2}\right]+\left[ b\sqrt{3}\right] \] Just consider the function $f(x)$ from $\mathbb Z\to\mathbb Z$ defined as $f(x)=\left\lfloor x\sqrt 3\right\rfloor$ + $\left\lfloor -x\sqrt 2\right\rfloor$ It's rather easy to show that $f(x+1)-f(x)< 2$ and $\lim_{x\to+\infty}f(x)=+\infty$ and $\lim_{x\to-\infty}f(x)=-\infty$ So $f(x)$ is a surjection. Hence the result. Could you explain how the difference $f(x+1)-f(x)$ becomes less than $2$ in a more detailed way? Moreover, why does showing that f goes to infinity whenever $x$ goes to infinity imply that f is surjective?

Arslan wrote: Could you explain how the difference f(x+1)-f(x) becomes less than 2 in a more detailed way? Moreover, why does showing that f goes to infinity whenever x goes to infinity imply that f is surjective? $\lfloor x\sqrt 3+\sqrt3\rfloor-\lfloor x\sqrt 3\rfloor \in\{1,2\}$ $\lfloor -x\sqrt 2-\sqrt2\rfloor-\lfloor -x\sqrt 2\rfloor \in\{-2,-1\}$ So (adding) $f(x+1)-f(x)\in\{-1,0,1\}$ So $f(x+1)-f(x)$ can, at each step, move up or down of max one step. Since it may be as great as we want, it must, starting from $f(1)-f(0)=-1$, reach at least once all the nonnegative integers Since it may be as small as we want, it must, starting from $f(1)-f(0)=-1$, reach at least once all the negative integers