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$$ \left\lfloor \frac{a_2-a_1}{3}\right\rfloor = z $$

Is it valid to multiply the left side by $3$ to get rid of the rational number and negate the need for the floor function? I am dealing with a function of the form $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ with $(a_1,a_2)\in \mathbb{Z}\times\mathbb{Z} $ and $z\in\mathbb{Z}$. so my plan was to destroy the fraction and solve for the $z$ in terms of $a_1$ and $a_2$.

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    $\begingroup$ Try with a few numbers. $\endgroup$
    – user65203
    Oct 23 '20 at 19:04
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Suppose that you could. Then your work might look like

$$ \left\lfloor \frac{a_2-a_1}{3}\right\rfloor = z $$

$$ 3\left\lfloor \frac{a_2-a_1}{3}\right\rfloor = 3z $$

$$\lfloor a_2-a_1\rfloor=3z$$

$$a_2-a_1=3z$$

$$z=\frac{a_2-a_1}{3}$$

However, this isn't always an integer, so your method of 'canceling' the $3$ is invalid. In fact, you basically have the simplest form already (although that is a subjective term so I use it loosely).

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  • $\begingroup$ Thank you I appreciate it. $\endgroup$
    – GildenNate
    Oct 23 '20 at 20:19
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The equation

$$ \left\lfloor \frac{a_2-a_1}{3}\right\rfloor = z $$

is equivalent to

$$ z \le \frac{a_2-a_1}{3} \lt z + 1 $$

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