# One particular equation involving floor function

I've got this equation to solve

$\left\lfloor{\frac {x + 1}{x+7}}\right\rfloor = \left\lfloor{\frac {x }{x+3}}\right\rfloor$

It's not very hard to solve it if using properties of rational functions but I wonder if it can be solved using only arithmetics. Anyone got idea?

• It's not clear what you mean by "using properties of rational functions" here. – Ben Grossmann Apr 2 '17 at 15:16
• or "only arithmetics" – mdave16 Apr 2 '17 at 15:17
• Properties of such rational functions (quotient of two linear functions) would be injectivity, existence of horizontal asymptote at y=1 and, of course, continuity – Martin Apr 2 '17 at 15:25
• Your explanation of what you meant by rational functions is harder to understand than the original statement. – fleablood Apr 2 '17 at 15:49
• en.wikipedia.org/wiki/Rational_function – Martin Apr 2 '17 at 16:07

If I were trying to find the most elementary way, I think first I'd rewrite as

$$\left\lfloor 1+\frac{-6}{x+7}\right\rfloor = \left\lfloor 1+\frac{-3}{x+3} \right\rfloor.$$

Then the $1$'s come outside and cancel, leaving

$$\left\lfloor \frac{-6}{x+7}\right\rfloor = \left\lfloor \frac{-3}{x+3} \right\rfloor.$$

Then I'd ask myself, "When are both sides equal to $0$?" which gives two inequalities:

$$0\leq \frac{-6}{x+7} <1 \mbox{ and } 0\leq \frac{-3}{x+3} <1$$

Then I'd solve these inequalities and see where the solution sets overlap.

Then I'd repeat by answering "When are both sides equal to $1$?" And "When are both sides equal to $-1$?" That should turn out to be most of the solution set.

There might be some other pieces, but work out "When are both sides equal to $k$?" and you should get a sequence of little intervals.

If $x \ge 0$ then $0 < \frac {x+1}{x+7} < 1$ and $0 \le \frac x{x +3} < 1$ so $[\frac{x+1}{x+1}] = [\frac x{x+3}] = 0$. So all $x \ge 0$ are solutions.

If $x < 0$ is a matter of solving $[\frac {y - 1}{y-7}] = [\frac y{y- 3}]$ for $y = - x > 0$.

If $y = 1$ then $[\frac {y - 1}{y-7}] = 0 \ne [\frac y{y- 3}] = [- 1/2] = -1$.

Otherwise $\frac {y-1}{y-7} = 1 + \frac 6{y-7}$ and $\frac y{y-3} = 1 + \frac 3{y-3}$

so $[\frac {y - 1}{y-7}]=[\frac y{y- 3}]\iff [\frac 6{y-7}] =[\frac 3{y-3}]$

$[\frac 6{y-7}] = k \iff k \le \frac 6{y-7}< k + 1 \iff \frac 6{k}+7\ge y >\frac6{k+1} +7$

Likewise $[\frac 3{y-3}] = k \iff \frac 3k + 3 \ge y > \frac 3{k+1} +3$

So $y >0$ $[\frac 3{y-3}]= [\frac 6{y-7}] \iff \min(\frac 6k + 7,\frac 3k + 3) \ge y > \max (\frac 6{k+1} + 7, \frac 3{k+1} + 3)$.

If $k = 0$ then this is true if $y > 13$ and so $x < -13 \implies [\frac {x+1}{x+7} ]=[\frac {x}{x+3}] = 1$

If $k \ge 1$ then this is true if $\frac 3k + 3 \ge y > \frac 6{k+1} + 7 \ge \frac 6{2k} + 7 = \frac 3k +7$ which is impossible.

If $k = -1$ then to be true, $0 \ge y$ which is impossible.

If $k = -2$ then $\frac 32 \ge y > 1$ allows $[\frac{x+1}{x+7}] =[\frac{x}{x+3}] = -1$.

if $k = -3$ then $\frac 3{k} + 3 \ge y > \frac 6{k + 1} + 7 > \frac 6{2k} + 7 = \frac 3k + 7$ which is impossible.

So solution is $x < -13; -\frac 32 \le x < -1$; $x \ge 0$