# Can we logically derive a value for 0÷0? [closed]

I have a "proof" that 0÷0 = 2: 0÷0 = (100 - 100) ÷ (100 - 100)
= (10⋅10 - 10⋅10) ÷ (10⋅10 - 10⋅10)
= (10² - 10²) ÷ 10(10 - 10)
= (10 + 10)(10 - 10) ÷ 10(10 - 10)
= (10 + 10) ÷ 10
= 20 ÷ 10
= 2

It's obviously wrong mathematically (0÷0 is not 2), but this seems like a logically valid proof. Every rewriting step is valid:

• 0 becomes 100 - 100
• 100 becomes 10⋅10
• a⋅a becomes a²
• ab - ac becomes a(b - c)
• a² - b² becomes (a + b)(a - b)
• ac ÷ bc becomes a÷b
• 10 + 10 becomes 20
• 20÷10 becomes 2

Is this answer wrong both logically and mathematically, or only wrong mathematically and possible logically? If it's possible logically, does it have any mathematical meaning?

• @wizzwizz4: Gah, you're right, don't know what I was thinking! Deleted my comment. Jul 24, 2022 at 12:48
• The derivation fails on the fifth line where you try to cancel "(10-10)" in both numerator and denominator. That would be canceling out an undefined 0/0 term which is not allowed. Jul 24, 2022 at 13:11
• @WenuraRavindu In line 5, cancelling out 10 - 10 in the numerator and denominator implies you assume that 0/0 = 1. First, you cannot assume any value since you are trying to prove it has one. Second, your assumption that 0/0 = 1 implies that 0/0 = 2 is false. Thus, your proof is inconsistent. Jul 24, 2022 at 16:07
• @MauroALLEGRANZA Why did you replace the tags with reference-request? It is a philosophy-of-mathematics question, despite most mathematicians learning a particular philosophical stance in school. Jul 26, 2022 at 14:07
• @wizzwizz4 - do you really think that the question about 0/0=2, which is a big mathematical mistake at intermediate level school is a question worth of "philosophical" discussion? Why the OP speak of "logical possibility"? What does it mean? As per comments above he made a mistake in applying the arithmetical "rules of the game": that's all. The purported proof above shows simply why the "official" answer is UNDEFINED: the OP probably didn't notice that the same computation can produce EVERY result. Jul 26, 2022 at 14:13

There is also some Circular Reasoning here, which is straightforwardly a logical mistake, in addition to any mathematical errors.

On line 5, you cancel out the equal divisor and numerator values (10 - 10). What is your justification for this move, given that the purpose of the exercise is determining the value of 0/0?

• If you assume that cancelling out (10 - 10) / (10 - 10) gives you 0/0, you're instead left with 0/0 = 2×0/0, which suggests 0/0 = 0. Jul 26, 2022 at 14:05
• If you remove the circular reasoning step like that, you still get a value for 0/0. Why is that not a valid thing to do? Aug 13, 2022 at 12:16
• @wizzwizz4, when taking algebraic steps like this, the "cancelling out" effect appeals to an understanding that x/x=1. But if you're trying to make an argument about the value of 0/0, you can't assume that 0/0=1 in the course of your proof. Aug 13, 2022 at 18:16
• Or, to put it another way, in your edited argument, you have asserted that the step "ac ÷ bc becomes a÷b" is valid for all c, but can you produce a proof for that which includes 0 in the domain of possible values for c? Aug 13, 2022 at 18:20
• You don't need to turn "ac ÷ bc" into "a÷b"; you can turn it into "a÷b × c÷c" which is fairly uncontroversial. You can then treat 0/0 as opaque (instead of applying the hidden 0/0=1 assumption to cancel it out), and solve for 0/0. However, that still assigns a value to 0/0; either 0/0=0 is valid, or there's another problem with the argument. Aug 13, 2022 at 20:03

This proof is disguising a trick that seems like it would work mathematically but actually doesn't. The factoring of the 100 - 100 into different polynomials is just to make the problem seem more valid.

A simpler equivalent would be this:

``````0 = 1 * 0 = 2 * 0

0   2 * 0
- = -----
0   1 * 0
``````

Cancel the zeroes to get 0/0 = 2/1

The only issue is that you cannot cancel zeroes. This problem just hides that by cancelling (10 - 10) instead. As Speakpigeon said in a comment, cancelling the (10 - 10) would imply that 0/0 = 1, which contradicts the statement that 0/0 = 2.

• Is very first line of the "proof" not also incorrect, since it states an equality between two expressions that are undefined? (0/0 and (10-10)/(10-10) ) Jul 13 at 7:37

As you've noticed, compared to +, - and ×, division is special. There are three schools of thought around questions like this:

## 0÷0 is undefined

Whenever you try to define it, you end up with a contradiction. Therefore, the division operation is undefined for 0 and 0.

## x÷0 is a multi-valued function with all possible values

For every value, you can prove that value is equal to 0÷0. Therefore, it is equal to 0÷0.

What is division, anyway? a÷b means to separate a into b equally-sized groups; when the denominator is 0, there are no groups, so every statement about the group size is vacuously true.

Since it has all values, we don't have a contradiction when we say it has a particular value. Narrowing a multifunction down to a single value is not always a valid thing to do; if you've got a contradiction because you discarded solutions, all that means is you made an algebra mistake.

Here's a simpler example of such an algebra mistake: I'll "prove" that -6 = 0.

• -3 × -3 = 9
• Square-rooting both sides, we get √(-3 × -3) = √9
• √(a × a) = a, so √(-3 × -3) = -3
• √9 = 3
• Therefore, -3 = 3
• Subtract 3 from both sides: -6 = 0

To resolve this issue, we have to either treat define the square root as the positive square root (so √(a × a) = |a|), or treat the square root consistently like a multifunction (√(a × a) = a or -a):

• -3 × -3 = 9
• √(-3 × -3) = {-3, 3}
• √9 = {3, -3}
• Therefore, {3, -3} ⊥̷ {-3, 3}

## 0÷0 is defined as having a specific value

Sometimes, it's inconvenient to make division anything other than a total function, so we just define 0÷0 as something, usually 0 or 1. This does mean there are a few algebraic tricks we can't do any more, since they'd lead to a contradiction if we tried to do them on 0÷0.

These different definitions of 0÷0 affect the validity of cancelling out the (10 - 10). Instead of viewing that as a single operation, it might be clearer to think of it as three operations:

• ac ÷ bc = (a÷b) × (c÷c)
• c÷c = 1
• (a÷b) × 1 = a÷b

If these three claims are true, we can "cancel top and bottom". We can prove the third equality quite easily, but the second and (to a lesser extent) first equalities depend on the definition of 0÷0!

• If 0÷0 is undefined, c÷c = 1 iff c≠0. But, c = 10 - 10, so it's not valid to cancel.
• If 0÷0 is a multi-valued function, then 1 is always a valid solution for c÷c. However, if c=0, there are other solutions too. Discarding some of those solutions is an extra premise in your argument – or, if you'd prefer, a sloppy argument-by-cases –, and you need to keep that in mind if you end up with a contradiction, because that doesn't necessarily tell you that your original premise was faulty.
• If 0÷0 is defined as having a specific value:
• If c÷c = 0, c÷c = 1 if c≠0, otherwise 0. Following your proof with this rule, we conclude that 0÷0 = 2×0, which isn't a contradiction.
• If c÷c = 1, then the second rule is always valid, but ac ÷ bc = (a÷b) × (c÷c) is only a valid operation when c≠0 or a=b. (This is why I don't like defining 0÷0 = 1; I like the first rule!)