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I think I can prove that $ab = ba$ by other axioms. Am I wrong? Why?

Edit: proof updated with notes showing where I actually used the commutative property without noticing.

Proof. For any numbers $a$ and $b$, we know this is true: $$ ab + ab = ab + ab $$

By using the distribution axiom: $$\begin{split} a(b + b) &= \underbrace{b(a + a)}_{\text{mistake: implies commutivity}}\\ a(2b) &= b(2a)\\ a2b &= b2a\\ \end{split}$$

By multiplying both sides by $2^{-1}$: $$\begin{split} a2b2^{-1} &= b2a2^{-1}\\ \end{split}$$

By the multiplication associative axiom: $$\begin{split} ab(22^{-1}) &= ba(22^{-1}) \qquad\text{(mistake: requires commutivity)}\\ \end{split}$$

By the multiplication inverse axiom: $$\begin{split} ab(1) &= ba(1)\\ \end{split}$$

By the multiplication identity axiom: $$ab = ba$$

$\blacksquare$

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    $\begingroup$ At your first step, I got $a(b+b)=(a+a)b$, not $b(a+a)$. $\endgroup$ – Lord Shark the Unknown Feb 19 at 17:06
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    $\begingroup$ "By the multiplication associative axiom: " Um ... $a2b2^{-1} = ab(22^{-1})$ doesn't follow by associativity. It require commutivity to make the claim. $\endgroup$ – fleablood Feb 19 at 17:06
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    $\begingroup$ Oh and as Lord Shark points out you can say $ab+ab=(a+a)b$ but you can not say $ab+ab = b(a+a)$ unless you assume commutivity in the first place. $\endgroup$ – fleablood Feb 19 at 17:08
  • $\begingroup$ Lord Shark makes a valid point, though doing that step properly still gets you to $2ab$, and you still need to swap $2$ and $a$ in order to get to what the other side gives which is $a2b$. $\endgroup$ – Ian Feb 19 at 17:22
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Two things. If you don't have commutativity we can claim $b(a+a) = (a+a)b,$ and so distribution would allow only $ab +ab = (a+a)b= 2ab$ and you can't get $ab+ab =b(a+a)= b2a$.

Second you claim you can get $a2b2^{-1}= ab(22^{-1})$ from associativity. I... just can't see any justification as to how associativity implies that, as the $b$ and the $2$ have "switched places"... I can't see any reason that could be without associativity.

....

Interestingly though for all the "integers" -- terms that arise as $1+1 = 2$ and $2+1 = 3$, etc. -- you can claim limited commutativity as

$2b = (1+1)b = b+b = b(1+1) = b2$.

But I don't think you can extend that to elements that are not "rational". (i.e., not a sum of $1$s or a sum of $1$ multiplied by an inverse of another sum of squares). But I think you can prove by induction all "rationals" in a "non-commutative field" commute.

But I don't think a non-commutative field, in and of itself, is internally inconsistent. (I could be wrong.)

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  • $\begingroup$ Did transition from $b+b$ to $b(1+1)$ imply commutivity? E.g. I wonder if non-$1$ coefficients are usually written, but if not written, I wonder if they are implied to be a $1$ on the left side? E.g. $1b+1b$ becomes only $(1+1)b$ in the next step? $\endgroup$ – caveman Feb 19 at 17:42
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    $\begingroup$ No, because $b = 1*b = b*1$. That's an axiom of the indentity. Identities and inverses commute. Now you can have rings that have left and right identities and left and right inverses that don't commute. But if we assume all the field axioms except multiplicative commutivity, We have $1$ commutes by axiom. $\endgroup$ – fleablood Feb 19 at 17:53
  • $\begingroup$ And if $1$ commutes, then $1+1$ and $1+1+1$ and $(1+2)^{-1}$ and $(1+1+1+1+1)*(1+1)^{-1}$ also commute. But a "non-commutative field" without a general commutivity axiom, may have elements that can't be expressed in that way that don't commute. I think. ... So far as I know, there isn't much use or purpose in studying "non-commutative fields". $\endgroup$ – fleablood Feb 19 at 17:56
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The step you justified by associativity cannot be justified that way; it is actually a case of commutativity (you swapped the order of $2$ and $b$ on the left, and the order of $2$ and $a$ on the right).

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  • $\begingroup$ You mean I performed 2 steps? One swapping $2$ and $b$, and then associating them together into $(22^{-1})$? $\endgroup$ – caveman Feb 19 at 17:02
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    $\begingroup$ @caveman You could think of it that way; associativity is so thoroughly baked into our notation that I tend not to think of invoking it as a "step". $\endgroup$ – Ian Feb 19 at 17:04

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