7

I definitely feel like an idiot because of the fact that I have to ask you the following question, but anyway I just don't know any other option.
If just proven two things which are the opposites of each other, so at least one of my proofs have to be wrong.
I tried to solve the following problem:

Definition: $a \boxplus b = 2a+2b$. $\boxplus$ is commutative.
Is $\boxplus$ assosiativ?

So I found two ways to solve the problem:
1. per definition
We want to show $(a \boxplus b) \boxplus c = a \boxplus (b \boxplus c)$.
$(a \boxplus b) \boxplus c = (2a+2b)*2+2c = 4a+4b+2c.$
And on the other side $a \boxplus (b \boxplus c) = 2a+2(2b+2c)= 2a+4b+4c$
$4a+4b+2c \neq 2a+4b+4c$. So $\boxplus$ is not assosiative.

2. Show that commutativity implicates associativity
We want to proof $ (a \circ b) \circ c = a \circ (b \circ c)$ [$\circ$ shall be commutative]
$ (a \circ b) \circ c = a \circ b \circ c = b \circ a \circ c = b \circ c \circ a = (b \circ c) \circ a. $

I know that my mistake have to look stupid and obvious, but I just can't see where I'm wrong.
Thank you all for helping me!

Dominik
  • 460
  • 6
    If a binary operation is not associative it does not make sense to talk about $a\circ b \circ c$ because there is ambguity about which operation to do first. – daruma Jun 04 '18 at 11:41
  • 3
    @ArnaudD. This is not a duplicate, since what the OP wants is that someone checks his or her proof. – José Carlos Santos Jun 04 '18 at 11:44
  • 3
    It's worth pointing out that certain non-associative operations have established conventions to evaluate $a \circ b \circ c$. Most notably, if $a \circ b$ is interpreted to be exponentiation, i.e. $a^b$, then $a \circ b \circ c$ is defined, by convention to be right associative, that is, $$a \circ b \circ c := a \circ (b \circ c).$$(It's not too hard to see why that's the convention, as opposed to left associativity, if you think about it!) But, these are specific conventions for specific operations. In general $a \circ b \circ c$ is ambiguous without associativity. – Theo Bendit Jun 04 '18 at 12:00
  • @JoséCarlosSantos It may be possible to justify a question like this as "it is asking about a specific problem with their attempted solution" but "it's an attempted solution" is a nonsense reason. By that logic, we would be forced to tolerate any number of slightly varied proof attempts for a single question. As it happens, we have a system for handling and giving feedback on attempted solutions: posting solutions as solutions. Suggesting that solution attempts are immune to the duplicate button is a definite way to hinder searches for answers to the original question. – rschwieb Jun 04 '18 at 13:43
  • Dear dominik, here is some friendly advice: Next time you have a question of the form "I have a solution to a question but I don't think it is right" please do a search for the question first. If you find the same question (as you almost certainly would have found https://math.stackexchange.com/q/160945/29335) you should post it as an answer. No doubt someone would quickly point out the mistakes, since people are keen on critiquing solutions, especially latecoming ones. This will help shield you from duplication closure. – rschwieb Jun 04 '18 at 13:48
  • (cont'd) On the other hand, if your question is nowhere to be found, consider posting it with the note that you will momentarily self-answer, and then self-answer it with your attempt. No doubt if you are incorrect, people will quickly point it out and/or provide their own solutions. Organizing questions and solutions like this helps keep the site searchable and probably gets feedback faster. – rschwieb Jun 04 '18 at 13:50
  • I don't feel strongly either way about closing the question since "the horses are already out of the barn" already, and at least the answer was not just "everything is correct." It's not optimal, but whatever. I'm mainly concerned with encouraging better overall posting policy where possible. – rschwieb Jun 04 '18 at 13:56

3 Answers3

15

In your second way to solve the problem, you write the expression $a\circ b\circ c$ which is not defined.

We usually only define $a\circ b\circ c$ if we already know that $\circ$ is associative, in which case we can shorten $(a\circ b)\circ c$ and $a\circ(b\circ c)$ to simply $a\circ b\circ c$ because we already know that both of those expressions are equal. If we do not know that, then the expression $a\circ b\circ c$ is not yet defined, because it could mean either one or the other, and there is no guarantee that they are the same.

Certainly, we can define $a\circ b\circ c$ to mean $a\circ(b\circ c)$, but if you define it like that, then your claim that

$$a\circ b\circ c = b\circ a \circ c$$

becomes (using the definition) the claim that

$$a\circ(b\circ c) = b\circ(a\circ c)$$

which is not a claim that can be proven by simply claiming commutativity.

Similarly, if you define $a\circ b\circ c=(a\circ b)\circ c$, you hit a problem because then you actually can claim that $$a\circ b\circ c= (a\circ b)\circ c=(b\circ a)\circ c=b\circ a\circ c$$ but then the claim

$$b\circ a\circ c=b\circ c\circ a$$

is equivalent to

$$(b\circ a)\circ c = (b\circ c)\circ a$$

which can again not be proven using commutativity alone.

5xum
  • 126,227
  • 6
  • 135
  • 211
  • 1
    This is rather pedantically phrased and in fact just not true, unless you value Principia Mathematica or Lisp-like verbosity. As Theo Bendit commented, it is possible to define that $a \circ b \circ c$ should mean e.g. $a \circ (b \circ c)$, for a nonassociative operation. Exponentiation is the typically given example with $a^{b^c} \equiv a^{(b^c)}$ by convention, but it's even needed for ordinary subtraction: $a-b-c \equiv (a-b)-c \neq a-(b-c)$. The point is just, it needs to be defined somewhere. It can be done for sure. – leftaroundabout Jun 04 '18 at 12:11
  • @leftaroundabout Agreed. I added a lot to my answer to make it more technically correct. – 5xum Jun 04 '18 at 12:17
6

Your first proof is correct, other than the fact that you write (twice) “$=2c$” instead of “$+2c$”.

The other proof is wrong from the start. The expression $a\circ b\circ c$ is ambiguous if you don't assume associativity. A commutative operation doesn't have to be associative.

José Carlos Santos
  • 440,053
0

You are using associative property in your proof. $$(a \circ b) \circ c = a \circ b \circ c = b \circ a \circ c = b \circ c \circ a = (b \circ c) \circ a.$$

Without associativity the following $$a \circ b \circ c $$ simply does not make sense.

Mohammad Riazi-Kermani
  • 69,891