# Logic/Math philosohy of zero

1. May 12, 2007

### crxyem

Well here's the discussion that always arises with my father and I.

It has to do with the multiplication of 1 and 0 and the Commutative Law

Given: In ordinary (real number) arithmetic

$$x * y = y*x$$

The argument is that
$$0*1 = 0$$ and $$1*0=0$$ should not be Commutative.
in a purely mathematical sense this seems very logical. But, in a philosophical sense of tangible objects it does not.

For example.
I have nothing or zero apples and I multiply this by 1 apple, the result I still have nothing as there was nothing to multiply. This logic is acceptable.

I have one apple and I multiply it by zero apples, which seems to possess some logic. Although this is supposed to lead to a result of having zero apples. But this is not the case the result is I still have my original apple. No one took it away synonymous with subtraction.

So what are your thoughts ???

Last edited: May 13, 2007
2. May 13, 2007

### honestrosewater

The two operations that you describe, as mathematical operations, are equally mathematical. And the two sets of philosophical ideas that you have about the operations are equally philosophical.

You just have two different operations.

3. May 13, 2007

### -Job-

When i think of A * B, i think of it as producing a B number of As.
From this perspective 1 * 0 should yield 0. The result of 1 * 0 is zero ones, which is 0.

4. May 13, 2007

### honestrosewater

Oh, also, did you want to set 1*0 != 0*1, or would you be happy setting 0 = 1? You could leave * commutative that way and have 1*0 = 0 = 0*1 = 1.

Or perhaps I missed the point and you are seeking "the right way", or the way that makes the most sense? I don't see why there must be one best way.

Last edited: May 13, 2007
5. May 13, 2007

### out of whack

First, if you multiply apples then the result is not apples but apples squared, whatever this might be. In both cases you will end up without any apple squared at all. Well, this makes sense since I have never seen an apple squared. To correct this, you need to get rid of one of these "apple" units on one side of the multiplication. You have four possibilities:

0 * 1 apple = 0 apple
0 apple * 1 = 0 apple
1 * 0 apple = 0 apple
1 apple * 0 = 0 apple

Now your problem with the last case is solved: you have one apple and you multiply that fact by zero, which essentially nullifies the fact (as if someone ate your apple) and you end up with no apple.

But that's not exactly the nature of your problem. The problem is that you disagree with the interpretation that multiplying something by zero is equivalent to subtracting the same value so that you end up with nothing. But that's just a matter of accepting the definition or not. Multiplying by 0 indeed, by common understanding of this operation, means that you end up with nothing, in the same right as multiplying by 1 means that nothing happened. If you disagree with this then you should not use multiplication but use something else that you will define for yourself. It's pretty straightforward in the end, either you accept the common understanding of the operation or you create one of your own. But then nobody else will understand what you are doing until you publish your new rules.

I can't believe I wrote so much on this subject...

6. May 13, 2007

### crxyem

I myself accept the mathematical proofs (Riemann sphere, l'Hôpital's rule) that exist to prove such cases that you can divide by zero or at least rationalize using zero in the denominator.

I like the argument of "out of whack" has stated that when you multiply say $$x * x$$ the result is $$x^2$$ and I myself have never seen a square apple either.

So trying to be less structured and mathematical and more philosophical during a conversation was a more interesting question, what does it mean to multiple something by zero

7. May 13, 2007

### honestrosewater

I don't know what you are referring to here, but whether or not some operation is defined for some input is just a matter of definition. There is nothing to prove in the usual sense.

You wouldn't see a square apple because squares are 2-dimensional. You might see a cube-shaped apple someday, though. I would think they would be more efficiently transported and stored.

But anyway, I don't think that was even the intended word. "Squaring" there was meant as an operation, not a geometric figure (a quadrilateral with sides of equal length or whatever).

Perhaps this kind of activity is super and useful in its own right, but it is not philosophical thinking. Philosophical thinking requires, at least, some rigor and precision. What you are doing seems more like some kind of free-flow superficial word- or concept-association.

8. May 14, 2007

### JonF

We know 0 + 0 = 0 by the definition of zero

We then know 0*1 + 0*1 = 0 by the definition of one

We know that (0+0)1 = 0 by the law of distribution

But since 0+0 = 0 by the definition of zero we know then know (0)*1 = 0.

The 1*0=0 case is similar symmetric

The properties that 0*1=0 follows by logical necessity from distribution. And if you throw distribution out the window, you are talking about a different type of operation than “+” in the sense it is used combine apples.

Last edited: May 15, 2007
9. May 15, 2007

### honestrosewater

You have it already right there. I assume that, by the definitions of 0 and 1, you mean as the additive and multiplicative identities, respectively. You can have right-identities (x1 = x) and left-identities (1x = x). For the most familiar number systems, it is both, or two-sided (1x = x1 = x).

10. May 15, 2007

### JonF

Yup normally, but he was essentially asking for a reason why identities are commutative. Not all algebras have THE identity, some have a left-sided and a right-sided which are not equal. My proof showed that as long as we have the distributive law 1*0=0*1 by necessity.

11. May 15, 2007

### honestrosewater

Haha, my bad. I totally missed the point there.

(By the bye, I think you have a typo, if you want to fix it: "(0)*1 = 1")

12. May 15, 2007

### Staff: Mentor

Yes, but you saved me the time of writing out, which has value (at least to me right now). I owe you one....

apples squared :rofl:

13. May 15, 2007

### honestrosewater

Hm, something is still not right here. You want to prove that a left-identity for multiplication combined with some other assumptions makes your left-identity two-sided, right? And then do the same for a right-identity? Did I miss something? You don't appear to have done that.

You didn't specify your premises, so I'll just list a bunch a see what gets used. I might try later. I looked quickly and the way I saw didn't work.

Assume:
There exists some 0 and 1 such that, for all x, y, and z,
a) (x + y) + z = x + (y + z)
b) (x * y) * z = x * (y * z)
c) 0 + x = x
d) x + 0 = x
e) 1 * x = x
f) x * (y + z) = (x * y) + (x * z)
g) (y + z) * x = (y * x) + (z * x)
h) x * 0 = 0
i) 0 * x = 0
j) For all x, there exists some -x such that x + -x = 0
k) For all x != 0, there exists some x-1 such that x * x-1 = 1.

Prove:
For all x, x * 1 = x

14. May 15, 2007

### JonF

nope, that wasn't what i intended. All I proved was the special case of 1*0 = 0*1 as long as we have distribution. I didn't prove in general 1*x=x*1 because that wasn't the question.

your assumptions kinda undermine his question because he is essentially asking: do the left and right identity need always be the same? If there is a left identity must there be a right identity? Are 1*0 and 0*1' necessarily equal (where 1 is the left identity and 1' is the right identity)? I only addressed the last question.

Also note: a few of your assumptions can be derived from other assumptions you make and others aren't necessary for the conclusion you seek....

Last edited: May 15, 2007
15. May 15, 2007

### honestrosewater

Oh, I guess I misunderstood your explanation.
Heh, that's why I said:
(I was actually trying to find out what precisely you meant by "the definition of one", etc.)

You're right, though, that I didn't see that interpretation of the OP's question. I thought they wanted * to not be commutative generally. Thanks.

16. May 15, 2007

### honestrosewater

Sorry, this is still eating at my brain. Something seems superfluous to me again.
If 1*x = x and x*1' = x for all x, then 1*x = x*1' as they're both equal to x, without the need for distribution or restriction to x = 0. Or where am I confused?

17. May 15, 2007

### Hurkyl

Staff Emeritus
I disagree.

1left = 1left * 1right = 1right

18. May 15, 2007

### crxyem

Thanks for the discussion on the topic, I generally can handle a good philosophical debate until it comes to debating math, I have a terrible memory for rules and theory of mathematics. Yet I can solve a Laplace Transform without trouble.

I was just stating that the equation 0*1 = 1*0 , shouldn't be commutative, because the left doesn't necessarily equal the right, which is my fathers argument and I've always pretty much accepted the fact. But if somehow I could prove his case it could be quite enlightening.

Wow, I never looked at the equation that way before but it makes sense to. After the discussion/arguments I've seen. I can now bring this up with my father again and really prove that it is in fact the same solution.

Also when I was talking about l'Hôpital's rule I was referring to it's use with limits and indeterminate form which allows a solution to be found when dividing by zero

Last edited: May 15, 2007
19. May 16, 2007

### honestrosewater

Heh, you're not the only one. I find myself sometimes lost as to where to start a mathematical argument or even make sense of a definition. The problem is usually just in getting started, and I think it might be needing to assign meaning to enough of the parts so that you can actually work in a model. (That might be why I sometimes try to resort to mechanical computation.) Some researchers have proposed two different styles of thinking, one that works more step-by-step and one that works more like building a web. It seems sometimes that I prefer web-style and mathematics is often presented step-style. The research was actually on differences between males and females. I could try to find it again if anyone is interested. Anywho...

Well, by the usual definition, that specific equation is true if * is commutative (which I guess is why I jumped to that), so you would have to make * non-commutative or at least change the definition to include this one exception.

I imagine that if it makes sense to you, you can find the definitions that work. You only have to worry about your theory being inconsistent, and if you aren't mistaken in your mental model of the system, then inconsistency isn't a problem. You can't imagine an inconsistent system any more than you can imagine a square circle. Well, that's maybe saying it too loosely, but you get my drift, right?

I liked Hurkyl's even better.

20. May 16, 2007

### JonF

eh, i musta be remembering something wrong from my group theory course. I was almost certain that the professor said there were algebras with left sided and right sided that were distinct but your proof is valid so i must be confused or forgetful.