# Dividing by 0

1. Nov 28, 2007

### l46kok

We are told over and over of the fact that you just cannot divide by 0. Why? Yes, I know it doesn't make sense to divide by 0, but really, there has to be a mathematical reasoning of why this cannot be accomplished.

2. Nov 28, 2007

### shuh

Because this is what may happen in the real world

Last edited by a moderator: May 3, 2017
3. Nov 28, 2007

### dodo

Funny : )

l46kok, think on how do you define division, in the first place. When you divide two numbers, say a / b, you are looking for another number such that, multiplied by b, gives you a.

Now imagine b is zero. What number, multiplied by zero, gives you some particular number a?

If a is not zero too, then there is not such a number. No number multiplied by zero gives non-zero, that's why a / 0 has no result if a is non-zero.

And if a is also zero, then any number would do. 0 / 0 is considered undefined, since it has not an unique solution.

Hope this helps.

4. Nov 28, 2007

### arildno

Sure it is:

1. Having no reference to division at all, we may prove by standard properties of real numbers that for any real "a", a*0=0.

2. If therefore, for real numbers b and for real numbers c, the division b:c shall equal a number k so that c*k=b, it follows that for any non-zero b, we cannot in general let c=0,
since we already have proven that..c*k=0.

Therefore, in order to maintain consistency of our mathematical operations we see that division by 0 is prohibited

5. Nov 28, 2007

### Hurkyl

Staff Emeritus
Because division is defined to operate on nonzero denominators -- your question is like asking why we use the word "blue" to decribe light with a wavelength of 465 nm.

There are lots of reasons why this definition is a useful one -- the other posts demonstrate some bad things that we would face if we attempted to define a new division operator that allowed division by zero.

6. Nov 28, 2007

### Zurtex

To be fair that's not entirely true. While you can look it that way, just dryly, saying that "because that's the way it's defined".

A more interesting way to look at it is to consider the Field of real numbers, which have a certain set of properties that must obey. Then it's non-trivial to show that 0 has no multiplicative inverse, you tackle it much the way arildno has.

7. Nov 28, 2007

### arildno

I think Hurkyl was very fair, Zurtex; because he is right.

Although we shouldn't throw out the idea of mathematical consistency, I'm sure we could throw out the distributive law of multiplication, say, and preserve a "division" concept applicable for zero.

(But then, multiplication is no longer what it once was..)

Essentially, therefore, the ultimate justification for why division by 0 is prohibited is DEFINING the operation in such a way that it remains consistent/meaningful with whatever other axioms are held to be true.

If we choose to change those axioms themselves, we might not need to exclude 0 from being a divisor in our definition of division.

Last edited: Nov 28, 2007
8. Nov 28, 2007

### Zurtex

I know it's true just to say "division isn't defined for 0", I'm not arguing otherwise.

My quible, if any, was to do the the analogy:

'your question is like asking why we use the word "blue" to decribe light with a wavelength of 465 nm.'

The reason it's called blue is arbitrary, we could call it green, we could it elephant, we could call it awoejhropwiheroiuwheifuwheoifugwoeifugilqhwb4ihfbghfwiherbfwihebfiwhegbf. The reason our definition of division doesn't allow for for divide by 0 isn't arbitrary, 0 has special properties in a field and it wouldn't make sense for us to choose some other qualitatively different number.

9. Nov 29, 2007

### JonF

Because that’s how fields are defined.

10. Nov 29, 2007

### Zurtex

I know you can define a field like that, but I was always taught to define a field by these:

http://mathworld.wolfram.com/FieldAxioms.html

Where devision is defined through the multiplicative inverse, the Field Axioms then imply division by 0 isn't possible because 0 has no multiplicative inverse. Sorry I may seem a little over the top about this, I just don't feel in this case it's a very sufficient answer for someone who is trying to learn something new.

It can also nicely be shown that a field becomes inconsistent if you allow 0 to have a multiplicative inverse.

11. Nov 29, 2007

### JonF

I think my answer is the best one to this question because it lets him go as deep as he wants it has the: “Because that’s how we define it” element but if he wants to know WHY we define it that way he need to study fields.

12. Nov 29, 2007

### CRGreathouse

Let's try to define a field-like object with a multiplicative inverse. To start with, let's give it two distinct elements, the 0 and the 1.

0 + 1 = 1 + 0 = 1 (additive identity)
Thus 1 + 1 = 0 because otherwise there is no additive inverse.
Distributing, (1 + 1) * 0 = (1 * 0) + (1 * 0).
1 * 0 = 0 (multiplicative identity)
So (1 + 1) * 0 = 0 + 0 = 0 (additive identity for the last step)
Of course we also know that 1 + 1 = 0, so (1 + 1) * 0 = 0 * 0.
If 0 has a multiplicative inverse, it must be 0 (1 doesn't work by multiplicative identity). Thus 0 * 0 = 1.

So 0 = (1 + 1) * 0 = 1. But we know that 1 and 0 are distinct.

13. Dec 3, 2007

### rbj

i dunno what the big deal is. when we say:

$$Q = \frac{N}{D}$$

we are saying that quotient Q is such a number that when multiplied by denominator D, you get numerator N. there is a clear problem with this, if N is not zero, yet D is zero, no?

do you accept first that zero times anything is zero? it has to be if the distributive property is true. so if you accept that

$$0 \times Q = 0$$

then there is no way for

$$D \times Q = N$$

if N is not 0 and D is 0. you get to a self-contradictory state, given such a premise (that we can divide by zero). hence the premise must be an unsupportable premise.

14. Dec 4, 2007

### CRGreathouse

Let's try another way. Consider working with subsets of the real numbers instead of real numbers directly. Let $A+B$ be $\{a+b:a\in A,b\in B\}$, $A-B$ be $\{c:c+b=a,a\in A,b\in B\}$, $A\cdot B$ be $\{ab:a\in A,b\in B\}$, and $A/B$ be $\{c:cb=a,a\in A,b\in B\}$. Then $\{1\}+\{1\}=\{2\}$, and in general
$$\{a\}\oplus\{b\}=a\oplus b$$ where $\oplus$ is +, -, or *.

Then $\{a\}/\{0\}=\mathbb{R}$ if $a=0$ and $\{a\}/\{0\}=\emptyset$ otherwise.

15. Dec 4, 2007

### rbj

this is potentially true (but i always worry about what it is in the limit, but there doesn't appear to be a limit here), but there is no way we can tell what particular real number $\{a\}/\{0\}$ is. it could be anything, making $\{a\}/\{0\}$ pretty meaningless whether a is zero or not.

16. Dec 4, 2007

### morphism

?

What CRGreathouse did was define operations on sets. What he wrote down isn't "potentially true", it's absolutely true. Under his definitions, {a}/{0} is not a real number, it's a set.

17. Dec 4, 2007

### rs1n

I prefer to use layman's terms in this case. Dividing is essentially a partition into equal parts. So, if you have A/B, you are essentially dividing A into equal parts of size B, and A/B represents the number of these equal parts. For a concrete example, take 16/2:

If you partition 16 into equal parts of size 2, then you would need 8 such equal partitions. However, what happens when you try to form partitions of size 0. How many would you need? An infinite number of them (and intuitively, even an infinite number of them may not be enough, which leads to indeterminate forms -- but that's for another rainy day). Unfortunately, infinity isn't a number -- it's a concept (at least as far as I am concerned; though it's considered a number in the extended reals).

18. Dec 4, 2007

### CRGreathouse

Yeah, my operations are just direct and indirect images in disguise. I hoped the form of it would show that even in that extension there can be no single element that will work.

You could try adding in ur-elements, but I think you'd find the same problems -- it you have division by zero your other properties fall apart. You could have x / 0 = 3 or x / 0 = $\mathcal{U}\!\mathcal{R}_1$ (some arbitrary ur-element), but then you have to abandon the idea that 0x = 0 for all x, which means you can't have additive inverses. By fixing up a small hole in multiplicative inverses you blow a huge one in the additive inverses -- is it worth it?

19. Dec 4, 2007

### Diffy

This is not a mathematical argument, and it assumes too much. It may be easier to think about it in terms of metaphors, but that is not mathematics, it is philosophy!

Why use "laymen's terms" when we have an exact construction and contradiction in proof by CRGreathouse?

20. Dec 4, 2007

### rs1n

I do not claim that this is a mathematical argument; hence "in layman's terms." Metaphors? Partitions are well known notions in mathematics (e.g. combinatorics). Given a/b, where a and b are positive real numbers, division is equivalent to partitioning a into equal parts of size b, and then counting the number of parts. There is nothing metaphorical about this approach. Philosophy? To someone lacking the mathematical background (esp. someone without the mathematical background needed to understand that the $$\mathbb{R}$$ is a field), the other approaches naturally raise the question: why are fields important?

My point is that one can deduce inconsistencies with lesser tools if division by zero is allowed. I have nothing against rigorous proofs, and as a mathematician, encourage rigorous arguments. However, proofs are generally the end result of mathematical understanding. (How often is it that you solve a problem by jumping straight to the proof?) I prefer to use "layman's terms" because the OP's question can be addressed without resorting to more abstract objects such as fields; basic intuition about partitions and positive real numbers is sufficient for showing that inconsistencies occur if we extend the notion of division to include 0. After that comes a proof.