0
$\begingroup$

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: $$ab = ba$$

$\blacksquare$

$\endgroup$
4
  • 1
    $\begingroup$ At your first step, I got $a(b+b)=(a+a)b$, not $b(a+a)$. $\endgroup$ Feb 19 '19 at 17:06
  • 1
    $\begingroup$ "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 '19 at 17:06
  • 1
    $\begingroup$ 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 '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 '19 at 17:22
3
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ You mean I performed 2 steps? One swapping $2$ and $b$, and then associating them together into $(22^{-1})$? $\endgroup$
    – caveman
    Feb 19 '19 at 17:02
  • 2
    $\begingroup$ @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 '19 at 17:04
1
$\begingroup$

Two things. If you don't have commutativity we can't 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 commutivity.

....

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.)

$\endgroup$
6
  • $\begingroup$ 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 '19 at 17:42
  • 1
    $\begingroup$ 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 '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 '19 at 17:56
  • $\begingroup$ In your first line, I think you mean "we can't claim". $\endgroup$
    – Théophile
    Nov 4 '20 at 17:35
  • $\begingroup$ "In your first line, I think you mean "we can't claim". It's been so long I can't tell what I meant but... yeah. Leaving ought the "not" is a common for me typo (an infuriating one as it fumdamentally changes the mean, but irritatingly hard to proof as I pychologically just don't see it because it is so fundamental). And this does make utterly no sense otherwise. $\endgroup$
    – fleablood
    Nov 4 '20 at 17:53

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.