2
$\begingroup$

I'm in the process of doing some practice problems to better understand binary operations. I'm currently working on the following binary operation

On the set, $\mathbb{Z}$, the binary operation $a * b = a + 2b$

I've found that the operation is not associative and commutative (I could be wrong) but now when determining if this operation has an identity element, is it enough that the operation is non commutative and non associative to say that it has no identity element?

I guess to further strengthen the claim, I would look for cases where the identity element fails?

All help is greatly appreciated!

Thanks for looking!

| cite | improve this question | |
$\endgroup$
2
$\begingroup$

No, you can't say that the operation has no identity just because it is not associative or commutative. To say that $e$ is an identity for the operation is to say that $e*a=a=a*e$ for all $a$. To prove that there is no identity, you would need to prove that no such $e$ exists; there's no reason that an identity automatically can't exist just because the operation is not commutative or associative.

To figure out whether there is an identity, then, you should look at what the equation $e*a=a=a*e$ says. For your operation, it says that $e$ must satisfy $$e+2a=a=a+2e$$ for all $a\in\mathbb{Z}$. Is there any such integer $e$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Alright thanks for writing it like that! In terms of integers for e, I could be looking at this wrong but what if I were to solve each side to find a solution for e. So isolating e on the LHS would give me e = -a and on the RHS, e = 0. With the LHS, with e = -a, it would result in a, but plugging that into the RHS wouldn't. As with the RHS with e = 0, we would get a but on the LHS, we'd still have 2a. Is this the right kind of idea? $\endgroup$ – Jeremy Cho Jan 16 '17 at 7:56
  • $\begingroup$ Yeah, that's the right kind of idea. So is there a single integer $e$ that makes the equation true for every value of $a$? $\endgroup$ – Eric Wofsey Jan 16 '17 at 7:57
  • $\begingroup$ Outside the method I tried to do, I can't think of any other methods of obtaining an integer where the equation holds true for every value of a. I could be wrong but I am to believe that this operation will not have an identity $\endgroup$ – Jeremy Cho Jan 16 '17 at 16:47
  • $\begingroup$ That's right. What "isolating $e$" does is proves that if the equation holds for some particular value of $a$, you must have $e=0$ and also $e=-a$. But this needs to be true for every value of $a$ at once, and there is no integer $e$ that is equal to $-a$ no matter what $a$ is. $\endgroup$ – Eric Wofsey Jan 16 '17 at 16:52
  • $\begingroup$ Awesome, thanks for all the help Eric! Logic math is definitely my weakest suit so it's great that you explained it clearly $\endgroup$ – Jeremy Cho Jan 16 '17 at 17:09

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.