Recently I have been struggling with a problem involving the floor function. The problem is: $$ \lfloor x+5 \rfloor = 3\lfloor x\rfloor-1 $$

I have had a similar question to this however it only involved the floor function on one side of the equation and I was able to set up a pair of inequalities and solve them however I have not been able to do the same for this question.

What I have done so far is attempt the same procedure by solving the equation and then setting up a set of inequalities but have been unable to come up with a reasonable answer. I have also graphed this for clarity but am still unsure what is really going on or how to approach this question.

Thank you.

  • $\begingroup$ It's harder when the multiplication is inside the floor. $\endgroup$ – marty cohen Aug 25 '17 at 3:55

This is equivalent to $\lfloor x\rfloor +5=3\lfloor x\rfloor-1$, i.e. $$ \lfloor x\rfloor=3 \implies x \in [3,4[. $$

  • 3
    $\begingroup$ Ok I had no idea it was that straightforward thank you. I am not sure what I was trying to do! Thank you very much :) $\endgroup$ – Sean. M Aug 16 '15 at 23:57
  • equation: $$\lfloor x+5 \rfloor=3\lfloor x \rfloor−1$$
  • same as: $$\lfloor x\rfloor+5 = 3\lfloor x \rfloor-1 $$
  • add $1$: $$\lfloor x \rfloor+6 = 3\lfloor x \rfloor $$
  • subtract $\lfloor x \rfloor$: $$6 = 2\lfloor x \rfloor$$
  • divide by $2$: $$3 = \lfloor x \rfloor$$
  • do the floor thing: $$3 \le x \lt 4$$
  • $\begingroup$ thank you @dxiv $\endgroup$ – John Porter Aug 27 '17 at 1:31

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