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I tried to show that commutativity of addition implies associativity. For this I assumed that there is no associative property and $ a + b + c $ should be interpreted as $(a + b) + c $ .

$$(a + b) + c = a + b + c = b + a + c = c + b + a = (c + b) + a=(b + c) + a$$ $$ =a+(b + c).$$

I suppose this is incorrect but I am not sure of where exactly the mistake is. Does it have to do with wrong or lack of use of parentheses?

Am I required to use parentheses to apply the commutative property because the property is defined for only 2 elements so it should be $ (b + a) + c = c +( b + a ) $ instead of $b + a + c = c + b + a$ ?

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    $\begingroup$ It would certainly be clearer if you always used parentheses. $\endgroup$
    – badjohn
    Commented Feb 19, 2022 at 12:48
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    $\begingroup$ I think that your mistake is at the fourth $=$. Adding the parentheses should make this easier to spot. $\endgroup$
    – badjohn
    Commented Feb 19, 2022 at 12:50
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    $\begingroup$ You used associativity in your fourth equality. Don't remove the parantheses! $\endgroup$ Commented Feb 19, 2022 at 12:52
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    $\begingroup$ The mistake is that you write $b+a+c = c+b+a$, which $\underline{\text{in your notation}}$ translates as $(b+a)+c = (c+b)+a$ which is incorrect. Use parenthesis everywhere if you want to work with properties which are parenthesis-dependent. $\endgroup$
    – Kolja
    Commented Feb 19, 2022 at 12:54

5 Answers 5

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The question has been answered, but just to give a concrete example of a commutative but non-associative operation:

Suppose we play rock-paper-scissors, where the "outcome" of the game is the winner's move (or move of either player in a draw). Then given $A=\{R,P,S\}$, the outcome is a binary operation $\bigoplus:A^2\rightarrow A,$ given by

$$\begin{array}{c | c c c} \bigoplus & R & P & S\\ \hline R & R & P & R\\ P & P & P & S\\ S & R & S & S \end{array}$$

Notice how the outcome operation is commutative but not associative since

$$R\oplus(P\oplus S)=R\oplus S=R,\\ (R\oplus P)\oplus S=P\oplus S=S.$$

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    $\begingroup$ A counterexample is a perfect disproof. And this is a nice simple counterexample. The existence of this counterexample means that the search for the error in the "proof" is a mere academic exercise; we know there is an error somewhere in the proof because the conclusion it is supposed to prove is known to be false. $\endgroup$
    – David K
    Commented Feb 19, 2022 at 21:24
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    $\begingroup$ A counter-example also helps to find the error in the proof, since we can substitute in these R, P and S for the original a, b, c and see which equality fails to hold. $\endgroup$
    – kaya3
    Commented Feb 20, 2022 at 3:44
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The reason we can get away with writing $a + b + c$ without being concerned about exactly what it implies about the order of operations is because of the associative property, which is what you are trying to prove.

So you either have to write parentheses in every one of your expressions adding three quantities, or you have to decide whether $x + y + z = (x + y) + z$ or $x + y + z = x + (y + z),$ and then you have to stick with the same choice for the entire proof.

If you rely on being able to write $(a+b)+c=a+b+c$ then you are saying the leftmost addition always occurs first. In that case, we can always interpret anything of the form $x + y + z$ by explicitly putting the parentheses around the first two terms, $(x + y) + z$, and the first few equations of your "proof" then become:

$$ (a+b)+c = (a+b)+c = (b+a)+c \stackrel?= (c+b)+a. $$

Commutativity would support putting $c + (b + a)$ on the right-hand side of the last equation but in the first equation you ruled out that interpretation of the notation $c + b + a.$ You end up with something that cannot be shown merely by commutativity.

(Alternatively, if you claim that $c + b + a$ means $c + (b + a),$ then neither of the first two equations can be shown by commutativity alone.)

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Take for example the Avg (average) operation ⨁:RXR→R $$a⨁b = avg (a,b)={{(a+b)} \over 2}$$

1.the operation is commutative:

$$a⨁b = avg (a,b)={{(a+b)} \over 2}$$

$$b⨁a = avg (b,a)={{(b+a)} \over 2}$$

=>a⨁b =b⨁a

2.the operation is not associative:

$$(a⨁b)⨁c =avg(avg(a,b), c)= avg({{a+b} \over 2}, c)={{{{a+b} \over 2} + c} \over 2}={{a+b+2c} \over 4}$$

$$a⨁(b⨁c)=avg(a, avg(b,c))={{2a+b+c} \over 4}$$

3.in your demostration the first equal is wrong: (a+b)+c=a+b+c

$$(a⨁b)⨁c =avg(avg(a,b), c)= avg({{a+b} \over 2}, c)={{{{a+b} \over 2} + c} \over 2}={{a+b+2c} \over 4}$$

a⨁b⨁c is not defined, but if we are extending (intuitive) the operation to ⨁:RXRXR→R $$a⨁b⨁c = avg (a,b,c)= {{a+b+c} \over 3} $$ So, the equality is false.

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    $\begingroup$ This would be a great answer if it were rendered properly with Mathjax. $\endgroup$
    – Randall
    Commented Feb 19, 2022 at 22:54
  • $\begingroup$ What your last line shows is that an operation that gives the average of two numbers cannot so easily be employed to produce the average of three numbers. But the key point is at the beginning of that sentence: $a\oplus b\oplus c$ is not defined, which goes directly to the error in the question, namely, the proof is never clear about what $a + b + c$ actually means when $+$ is a binary operation. $\endgroup$
    – David K
    Commented Feb 19, 2022 at 23:12
  • $\begingroup$ Try \oplus for $\oplus$. $\endgroup$
    – David K
    Commented Feb 20, 2022 at 16:15
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You write:

Am I required to use parentheses to apply the commutative property because the property is defined for only 2 elements so it should be (𝑏+𝑎)+𝑐=𝑐+(𝑏+𝑎) instead of 𝑏+𝑎+𝑐=𝑐+𝑏+𝑎 ?

Yes. That is exactly it. I believe you have answered your question at least as well as any of the rest of us.

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Indeed, you've gone off-track by trying to write this progression without parentheses. You've accidentally made a "standard" manipulation in the absence of parentheses which only makes sense if association holds.

As noted by others, the problem is really in the third equality, but it's masked because steps were skipped by the lack of parentheses. As originally written by the OP, that appears as:

$$b + a + c = c + b + a$$

Let's try to be more precise by explicating the ordering with parentheses, and expanding the hidden steps:

$$b + a + c = (b + a) + c = c + (b + a) = (c + b) + a = c + b + a$$

Among the four equalities here, the 1st, 2nd, and 4th are all correct by our assumptions. It's the 3rd that's in error, and by being explicit, we can see that's the association assumption itself. While the OP wanted to get to $c + b + a$, i.e., $(c + b) + a$, in the sequence, there's no way to say that's the same as the prior $c + (b + a)$ without association.

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