I am trying to do the first exercise set from Abstract Algebra by Dummit and Foote.

The question defines $$ G=\{x\in\mathbb{R}\mid0\leq x<1\} $$

and asks to prove that it is a group with the multiplication: $$ x\star y=x+y-\lfloor x+y\rfloor $$

I have first tried to find the identity: $$ x=x\star e=x+e-\lfloor x+e\rfloor $$

subtracting $x$ from both sides gives $$ e=\lfloor x+e\rfloor $$

where $e,x\in[0,1)$.

How do I solve such an equation ?

I tried guessing for trivial solutions like $0,x,-x$ but that didn't work. I also thought about proving the existence of such $e$ with something like the intermediate value theorem (IVT), but if I recall correctly $\lfloor\rfloor$ is not a continuous function, and being an Abstract Algebra textbook I suspect that if the solution involves analysis then it would of been hinted.

Can someone please help me to proceed ? (I prefer a \textbf{hint} then a solution)

  • 4
    $\begingroup$ You rejected a simple answer for no good reason. $\endgroup$ Commented Oct 19, 2013 at 8:00
  • $\begingroup$ @AndréNicolas - got it, thanks! $\endgroup$
    – Belgi
    Commented Oct 19, 2013 at 8:04

2 Answers 2


Hint: for all $x \in [0,1): \lfloor x \rfloor = 0$


Hint: Trivial solutions do work. Note that $(\mathbb R,0)\to (G,\star)$, $x\mapsto x-\lfloor x\rfloor$ is a group homomorphism.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .