Why does 0x0 not equal 0/0?

Suppose that we have a bunch of boxes, and a bunch of apples. I put three apples in each box, and give you two boxes. How many apples do you have? You have six: 2 x 3 = 6.

Now I put five apples in each box, and give you three boxes. How many apples do you have now? Fifteen: 3 x 5 = 15.

Now I empty out all of the boxes so they all have zero apples, and give you no boxes. How many apples do you have now? What is 0 x 0?

Here is a multiplication table:



—————————————————————————
| 1 | 2 | 3 | 4 | 5 | 6 |
—————————————————————————
| 2 | 4 | 6 | 8 | 10| 12|
—————————————————————————
| 3 | 6 | 9 | 12| 15| 18|
—————————————————————————
| 4 | 8 | 12| 16| 20| 24|
—————————————————————————
| 5 | 10| 15| 20| 25| 30|
—————————————————————————
| 6 | 12| 18| 24| 30| 36|
—————————————————————————


I’ve drawn this out for multiplying any two values between 1 and 6.

Note that, within any row or column, moving in any particular direction, each step changes the value by the same amount. For example, in the column that starts with a 4, each step down adds 4 and each step up subtracts 4. Or in the top row, each step right adds 1 and each step left subtracts 1.

This is a very nice pattern, and mathematics is in general the study of patterns. Many phenomena in the world adhere to patterns, and so by understanding such patterns better abstractly, we can also understand the phenomena that model them.

Suppose we wished to extend this table to the left in such a way as that we maintained all these patterns. What should go to the left of the 1 in the corner there? Well, each step left in that row subtracts 1, and so going left we should end up with 1 - 1 = 0.

Indeed, each step left in the 2 4 6 8 10 row subtracts 2, and so going to the left of the 2 we should end up with 2 - 2 = 0. Each step left in the 3 6 9 12 row subtracts 3, and so going to the left of the 3, we should end up with 3 - 3 = 0. Our extension becomes like so:



—————————————————————————————
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
—————————————————————————————
| 0 | 2 | 4 | 6 | 8 | 10| 12|
—————————————————————————————
| 0 | 3 | 6 | 9 | 12| 15| 18|
—————————————————————————————
| 0 | 4 | 8 | 12| 16| 20| 24|
—————————————————————————————
| 0 | 5 | 10| 15| 20| 25| 30|
—————————————————————————————
| 0 | 6 | 12| 18| 24| 30| 36|
—————————————————————————————


In the same way, we can extend up. Each step up in the 1 2 3 column subtracts 1, so above that 1, we should get a 1 - 1 = 0. Each step up in the 2 4 6 column subtracts 2, so above that 2, we should get a 2 - 2 = 0. And each step up in the column of 0s? Well, it’s a column of 0s; each step up leaves it unchanged and makes another 0. So preserving these natural patterns again, when we extend our table up, it looks like this:



—————————————————————————————
| 0 | 0 | 0 | 0 | 0 | 0 | 0 
—————————————————————————————
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
—————————————————————————————
| 0 | 2 | 4 | 6 | 8 | 10| 12|
—————————————————————————————
| 0 | 3 | 6 | 9 | 12| 15| 18|
—————————————————————————————
| 0 | 4 | 8 | 12| 16| 20| 24|
—————————————————————————————
| 0 | 5 | 10| 15| 20| 25| 30|
—————————————————————————————
| 0 | 6 | 12| 18| 24| 30| 36|
—————————————————————————————


So now we have a multiplication table for values between 0 and 6. And, lo and behold, what does it tell us 0 times 0 is? It says it’s 0! Indeed, it tells us 0 times any of these values is 0. It’s just what we get by following the natural patterns. We define multiplication so as to make these patterns hold.

(And if we kept going, extending these patterns once more left and up, we’d how -1 times -1 = 1. This is often a source of confusion for students, but is just what falls out of these patterns. Again, as mathematicians, we’re interested in clean patterns; we define multiplication so as to make these patterns hold.)

For all the OP’s self-admitted ineptness with math, the above snippet DOES make some sense.

x/0 (for any non-zero x, for example 6/0) and 0/0 are both undefined, but for distinctly different reasons.

Division is defined as the inverse of multiplication: The statement
a / b = c
means exactly the same thing as:
a = b × c

But try constructing some more specific examples of this, where you have some zeros in there:
a / 0 = c
means exactly the same thing as:
a = 0 × c
What if a is any non-zero number, for example:
7 / 0 = c
This must mean:
7 = 0 × c
What value of c could you possibly put there that would work?
Answer: There is no possible value of c that works here.
Thus,
a / 0 = c
for any non-zero numerator a, must be undefined because there is no possible value of c that can be the solution to this.

When you divide 0 by 0,
0 / 0 = c
you get a very different situation. This means exactly the same as:
0 = 0 × c
What possible values could c be in this case? This time, any number you put in for c will work! There are infinitely many values (namely, all possible values) for c that will work here.
Thus, we must conclude that 0/0 does not define any particular number, and must therefore be considered undefined. Note that this is very different from the non-zero numerator case.

The OP’s first few sentences suggest that OP is really totally lost on this subject. BUT, the OP’s last few sentences suggest that OP may actually, in fact, have a clue.

As they say: If you’re confused, it probably means you’ve been paying attention.

That’s not true at all. As you’re well aware, zero factorial is equal to 1, which is definitely not the same as 0 times 0. :slight_smile:

(runs away)

It isn’t declared unsolvable. Indeed there are plenty of problems that have in a sense an infinite number of answers. x[sup]2[/sup] + y[sup]2[/sup] - 1 = 0 has an infinite number of solutions. (They are the points on a circle of radius one.)

But if your answer is: all possible numbers, it is pretty much useless. (It isn’t almost infinite, it is infinite. If there were a hole in the space of answers, it could quite possibly be useful.) Your solution is anything you want it to be. Which whist being most simple answer you can imagine, is also the least useful.

Allowing this answer to remain in a problem contaminates the problem from then on in, as you always have all possible numbers as your answer. So the entire process becomes futile. Indeed if you do let the contamination persist, you can prove anything you want. There are a number of amusing mathematical tricks that seek to show that some obviously wrong proposition is true that rely upon sneaking a divide by zero into the manipulations where no-one notices. Asking what 0/0 - 0/0 is is part of the problem. You have “all possible numbers” - “all possible numbers”. The answer is still “all possible numbers” - even though you might hope it is 0.

Thanks, Dr. Strangelove, I was resisting the urge to make that joke myself.

Senegoid, I’ve actually seen different words used for the situation with 1/0 and 0/0. 1/0 (or x/0 where x is any nonzero number) is undefined. 0/0, however, is undetermined.

I believe that’s improper as regards the concept of division; the operation is simply undefined for a/b when b=0. But when considering limits, a limit that converges to 0/0 is considered indeterminate, for precisely the reason that it could be any answer, depending upon how the function being limited is constructed.

I don’t understand some things in this thread, but I think I learned some things. 0 x 0 = 0 because multiplying by zero is the same as not multiplying by anything; the zero remains unchanged. Any number divided by zero can’t equal zero because zero is a number and not nothing. Any number divided by zero = no number.

This.

Or, your car is moving at 0 miles per hour, and you have spent 0 hours on the road so far. How far have you gone? 0 miles/hr x 0 hr = 0 miles.

p.s. If your car moved 2 miles in 0 hours, then its speed is 2/0 = undefined. You can interpret that as “impossible” or “infinity.”

If your car moved 0 miles in 0 hours, then your speed is 0/0 = undefined, i.e. you don’t know.

No, multiplying by 1 is the same as not multiplying by anything. Multiplying by 0 always gives you 0, no matter what you were multiplying it by (even by another 0).

Sure you can. There are an infinite number of integers, but a finite number with certain properties (if you define the properties to have an end point). Let’s say there’s an infinite number of worlds, but gravity or some other physical law decays from the “center of the universe” so that after you go a finite distance out, life is impossible.

If I give you five bags, and each bag contains no apples, how many apples have I given you? No apples. 5 x 0 = 0.

If I have bags that each contain five apples, but I give you no bags, how many apples have I given you? No apples. 0 x 5 = 0.

If I have bags that contain no apples, and I don’t give you any bags, how many apples have I given you? No apples. 0 x 0 = 0.

Now let’s try the inverse. You have no apples. I gave you some bags and each bag contained no apples. How many bags did I give you?

You have no apples. I have some bags that contain the same number of apples, but I didn’t give you any of those bags. How many apples are in each bag?

You have no apples. I didn’t give you any bags, and there were no apples in the bags anyway. What are we even talking about now?

When I read this earlier today, I thought of the unit cancellation method to illustrate the ‘mechanics’ of solving the problem. Not that you need to do that in this case, but since you’re dealing with miles and hours, I thought it might be useful to show how you cancel Hours, leaving Miles. Then I thought, maybe not.

But since it’s in my head, I’ll post of unit cancellation instructional video that may be helpful to people having trouble with unit conversions, and just to illustrate the logic of mathematics.

FWIW: While I got 'A’s in my math classes up through Calculus 2 (never did take Calc 3), math is not my best subject. (Better than Chemistry, though!) It took a lot of work, and I can sympathise with people who struggle with it.

I feared this would happen. You better run! And I don’t want to see any comments about “run factorial” either!!

You are postulating special circumstances. If we can go there then we can fabricate just about any rationale we want.

In this universe we hold to the Cosmological Principle which states that the universe is homogeneous and isotropic. Till someone proves that wrong we should probably stick with that when discussing this.

Can somebody explain why it makes sense to say that 0! = 1?

Also, if you could put your explanation in terms of bags and apples that would be great.

Because, when it comes to multiplication, 1 is the “starting point,” and you go on from there. If you started with 0, you’d never escape 0.

Or because there are several useful formulas involving factorials that work if 0! is defined to be 1 but not if it’s defined to be 0 or anything else. For example, n! = n x (n-1)! for any whole number n. (For example, 5! = 5 x 4!. So we want 1! = 1 x 0!.)

Or, if you really want bags and apples: n! represents the number of different ways you could put n apples into n different bags (one to a bag). If you have 0 apples to put into 0 bags, there’s 1 way to do this: don’t put any apples into bags.

You could start by reading Empty product on Wikipedia.

Suppose you have three apples. How many ways are there to arrange them in a line? You can pick any of the 3 for the first position, then any of the remaining 2 for the second position, which leaves only one choice for the last. There are total of 321 = 3! possibilities. With two apples, there are 2! possible arrangements. With one apple, there is only 1! = 1 possible arrangement. Now suppose you have zero apples. How many ways are there to arrange “them”? The most useful answer is there is one way, the arrangement which consists of no apples, so 0! = 1.

This is how it was explained to me but not with apples or bags.

5! = 54321 = 120
4! = 5!/5 = 120/5 = 24
3! = 4!/4 = 24/4 = 6
2! = 3!/3 = 6/3 = 2
1! = 2!/2 = 2/2 = 1
0! = 1!/1 = 1/1 = 1