# Question about solving equations containing the Floor/Ceiling Function

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

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 $$a_2$$.

• Try with a few numbers.
– user65203
Oct 23 '20 at 19:04

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).

• Thank you I appreciate it. Oct 23 '20 at 20:19

The equation

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

is equivalent to

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