5
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ It's harder when the multiplication is inside the floor. $\endgroup$ – marty cohen Aug 25 '17 at 3:55
11
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • 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
3
$\begingroup$
  • 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: $$3 \le x \lt 4$$
|cite|improve this answer
$\endgroup$
  • $\begingroup$ thank you @dxiv $\endgroup$ – John Porter Aug 27 '17 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.