How to prove that $x^2 - \lfloor x\rfloor^2$ is onto?

Question

The function's domain and codomain are nonnegative real numbers. I've given this countless attempts and I'm out of ideas on how to prove this. What is proving to be tricky to me is writing $x$ in terms of $y$ due to the floor function.

Answer 1

Hint: the function takes the interval $[n, n+1)$ to $[0, 2n+1)$.

Answer 2

We want to show that, for any $r > 0$, there is an $x$ such that $f(x) =x^2 - \lfloor x\rfloor^2 =r $.

If $\lfloor x \rfloor =n $ and $x-n = c$ $(0 \le c \lt 1$) then $f(x) =x^2 - \lfloor x\rfloor^2 =(n+c)^2-n^2 =2nc+c^2 $.

From this, $0 \le f(x) \lt n$ (by choosing $c = \frac12$).

Therefore, if $n > r$ then $f(n) < r$ and $f(n+\frac12) =n+\frac14 \gt r$.

Since $f(x)$ is continuous for $n \le x \le n+1$, there is an $x$ such that $f(x) = r$.

Answer 3

I find it helps sometimes to write $x = [x] + \{x\}$ so we wish to prove that for any $y \in \mathbb R$ there is an ineger $n =[x]$ and a real number $r=\{x\}; 0\le r < 1$ where

$x^2 - [x]^2 = (n+r)^2 -n^2 = 2rn + r^2 = y$.

Well.... if we use quadratic formula to solve for $r$ we have

$r^2 - 2nr - y = 0$ so $r= n \pm \sqrt{n^2 + y}$

So we need 1) $n^2 +y \ge 0$ and 2) then $0\le n\pm \sqrt{n^2 + y} < 1$

If $n = 0$ this will be a solution if $0\le y < 1$ and $r = \sqrt y$.

If $n=-1$ this well be a solution if $1 \le \sqrt{1 + y}< 2$ or if $1 \le 1+y < 4$ or if $0 < y < 3$ and $r = -1+\sqrt{1+y}$.

In general for a positive $n$ this will be a solution if $-n \le - \sqrt{n^2+y} < 1-n$ or if $n-1 < \sqrt{n^2 + y} \le n$ or if $n^2 - 2n+1 < n^2 + y \le n^2$ or if $-2n + 1< y < 0$ and $r = n-\sqrt{n^2 +y}$. Now for all negative $y$ we can find such an integer.

And for any negative $n$ this will be a solution if $-n=|n| \le \sqrt{n^2+y} < |n| + 1$ or $n^2 \le n^2 + y < n^2 + 2|n| + 1$ or $0 \le y < 2|n| + 1$ so the is possible for all positive $y$.

Note: it's onto, but it sure as heck ain't one-to-one.

Answer 4

HINT.-Let the real $x=\lfloor x\rfloor+\{x\}$ so we have $$f(x)=2\lfloor x\rfloor\{x\}+(\{x\})^2$$ Now for arbitrary real $y=\lfloor y\rfloor+\{y\}$ the equation $$2\lfloor x\rfloor\{x\}+(\{x\})^2=\lfloor y\rfloor+\{y\}$$ has infinitely many solutions (Why?)