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It is an accepted knowledge in mathematics that division by zero is undefined. But this is not intuitive to me. I'm not a mathematician, but how about if there is an axiomatic definition that any division by zero is zero?

Going back to the most accepted answer that $\frac{x}{0}$ is undefined:

"the equation that defines it has no solutions: for something to be equal to $\frac{x}{0}$, with $x ≠ 0$, we would need $0z = x$. But $0z = 0$ for any z, so there are no solutions to the equation. Since there are no solutions to the equation, there is no such thing as "$\frac{x}{0}$". So $\frac{x}{0}$ does not represent any number."

I get that $ z = \frac{x}{y}$ is the same as $ x = yz $:

Step 1: $ z = \frac{x}{y}$

Step 2: $ y ⋅ z = y ⋅ \frac{x}{y}$

Step 3: $ yz = x $

Step 4: $ x = yz $

If $ y = 0 $ and $ x ≠ 0 $, then there are no solution to the equation $ x = 0z $ because:

Step 5: $ x = 0z $

Step 6: $ x = 0 $

Therefore $ x = 0 $ for for any value of $z$ and conflicts with the requirement that $ x ≠ 0 $.

But if we define $ \frac{x}{0} = 0 $ as an axiom, then we do not even arrive at the equation $x = 0z$.

Step 1: $ 0 = \frac{x}{0}$ ____________ as an axiom

This already reduces to $ 0 = 0 $, but if we go further:

Step 2: $ 0 ⋅ 0 = 0 ⋅ \frac{x}{0}$ ________multiply both sides by zero

Step 3: $ 0 = 0 $ ____________any number multiplied by zero is zero

Therefore the equation $ x = 0z $ (and consequently $ x = 0 $ since $ z = 0$) simply cannot be derived.

But if we continue with the re-arrangement and have the equation $ x = 0z $, then because it passed a step where $ y $ is a divisor, therefore, it implies that $ y ≠ 0 $ under this axiom, so we can't make further assumption that $ y = 0 $ after step 4, so the equation $ x = yz $ is only valid where $ y ≠ 0 $.

Also, there would be no problem also for $\frac{0}{0}$ as described here, for now, the result of $\frac{0}{0}$ could not be anything but zero.

What would be the problems if division by zero is defined as an axiom as defined above?

Noble_Bright_Life
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    It's difficult to address your full query here, but division isn't given by any kind of axiom; what mathematicians call division is just a certain function $\mathbb R\times \mathbb R\setminus{0}\to\mathbb R$. You are free to define other functions, but (generally speaking) it wouldn't be sensible to call those functions "division", since that word already has a reserved meaning. – Joe Jan 27 '24 at 09:41
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    What exactly do you want to do? Do we extend the definition division of rational (real) numbers to include $x/0$? Is the "usual" division left intact? – AlvinL Jan 27 '24 at 09:53
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    Note that your axiom implies that $\frac{1}{0} = \frac{2}{0}$ (since both sides are equal to zero), and hence by multiplying both sides by $0$ we get $0 \cdot \frac{1}{0} = 0 \cdot \frac{2}{0}.$ Since the cancellability property $y \cdot \frac{x}{y} = x$ now gives us $1 = 2,$ we lose the cancellability property. There are probably other arithmetical/algebraic manipulations we would also lose. I suppose one could try to keep track of all "problematic manipulations" so that when they're used, we have to check that division by zero isn't involved, (continued) – Dave L. Renfro Jan 27 '24 at 09:54
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    but it seems to me that's pretty much what we already do. That said, the algebraic structure wheels is one way of allowing division by zero to be defined. – Dave L. Renfro Jan 27 '24 at 09:54
  • @AlvinL: Yes, kept the usual division unchanged and only put a special definition for divisions by zero. – Noble_Bright_Life Jan 27 '24 at 10:00
  • After saying "..I get that $z=\frac{x}{y}$ is the same as $x=yz$.." you proceed with some sort of 'proof' containing 4 steps. However it does not concern a theorem here but a definition. We define $\frac{x}{y}$ as the real number that - multiplied by $y$ results in $x$. Definitions cannot be proved. – drhab Jan 27 '24 at 10:04
  • @Dave L. Renfro; If we multiply both sides by $0$ we get $ 0⋅\frac{1}{0}=0⋅ \frac{2}{0}$. But before we could apply cancellation, there is already a rule to reduce divisions by zero to zero, so it will be reduced to $ 0⋅0=0⋅0$. – Noble_Bright_Life Jan 27 '24 at 10:05
  • You could define $x/0 =:0$ , but then none of the usual manipulation tricks with fractions apply for this quantity. – AlvinL Jan 27 '24 at 10:07
  • @AlvinL: But wouldn't that apply as well if division by zero is undefined? – Noble_Bright_Life Jan 27 '24 at 10:08
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    Yes, because usually $x/0$ is undefined. But you insist on it being well defined, which is fair enough, but then you can't treat it as any other fraction or you will run into problems, some of which have been pointed out as well. .. but if you can't treat $0$ as a fraction, why write it as a fraction in the first place? Just a waste of chalk/ink. – AlvinL Jan 27 '24 at 10:10
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    But before we could apply cancellation --- The issue is that cancellation cannot always be applied, not whether in the specific example I gave we could avoid cancellation. You asked What would be the problems if division by zero is defined, and I gave one such problem, namely whenever applying cancellation in an algebraic manipulation we would need to check that division by zero is not involved. – Dave L. Renfro Jan 27 '24 at 10:22
  • Speaking of wheels, if we do the "wheelification" of $\mathbb R$, then $1/0$ would be $\infty$ (in this instance $\infty = - \infty$). – AlvinL Jan 27 '24 at 10:27
  • I do not think that exotics like the extended real line or rings with zero divisors are helpful for a non-mathematician. It has been demonstrated that within the real numbers division by $0$ cannot be defined consistent with the usual arithmetic properties. Considering the given context of this question , I think , nothing else should be mentioned. – Peter Jan 27 '24 at 12:43
  • About intuition... If we cut a cake in two parts (division by 2) we get two halfs; if we cut it into one part (division by 1) we get the cake untouched. What about dividing the cake "by zero"? Why your intuition says that the cake must disappear? – Mauro ALLEGRANZA Jan 30 '24 at 09:40
  • @Mauro ALLEGRANZA: Well, I don't think of it that way. I think of zero as representing a null on the number line, so addition and subtraction keeps the other number, while multiplication and division results in a zero or null. If you take your example (one cake), can you explain by intuition why multiplying 1 with a zero is zero and not 1? Multiplication also starts with a number. – Noble_Bright_Life Jan 31 '24 at 01:05

2 Answers2

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If $\frac x 0 = 0$, then it has to be the neutral element of the summation. But $$ 0 + 1 = \frac x 0 + \frac 1 1 = \frac {x \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac x 0 = 0 $$ which never holds, unless in the zero ring.

Federico T.
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This results in a problem. Lets start with your axiom.

$\forall{x}.0 = \frac{x}{0}$

Substitute 1.

$0 = \frac{1}{0}$

Rearrange according to the rule $z = \frac{x}{y}$ is the same as $x = yz$

$1 = 0 \times 0 $

This is a contradiction so the axiom doesn't work.

Q the Platypus
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  • I mentioned in the original question that re-arrangement would not apply because 1/0 at the very start is already 0, therefore, it could not reach that stage where the equation is re-arranged. – Noble_Bright_Life Jan 27 '24 at 09:48
  • However $x= yz$ is the definition of division. So it stops being division. – Q the Platypus Jan 28 '24 at 02:33