# An old chessnut -a*-b =?

1. Mar 11, 2006

### toasty

Here below is an exchange between me, the perplexed, and Dr Math.
Read it, and please if you can show me why -a*-b MUST NOT equal ONLY either +ab or -ab. What else can -a*-b be equal to?

>[Question]
>Dear Dr Math I long ago wrote you asking how to prove the convention,
>- * - = + and you replied with the webpage which illustrates the point
>and ends with these remarks.
>
>" For example, if we adopted the convention that (-1)(-1) = -1, the
>distributive property of multiplication wouldn't work for negative
>numbers:
>
> (-1)(1 + -1) = (-1)(1) + (-1)(-1)
>
> (-1)(0) = -1 + -1
>
> 0 = -2
>
>As Sherlock Holmes observed, "When you have excluded the impossible,
>whatever remains, however improbable, must be the truth."
>
>Since everything except +1 can be excluded as impossible, it follows
>that, however improbable it seems, (-1)(-1) = +1.
>
>[Difficulty]
>Well that just confuses me more, and I cannot get it straight in my
>mind, besides there has to be another more cogent way which is not
>hard to understand.
>
>[Thoughts]
>I suppose that - a * - b = - ab.
>
>But - a * - b can only = either + ab or -ab.
>
>And self evidently + a * -b = -ab
>
>From the supposition above and the last proposition
>-a * -b = + a * -b
>
>dividing across by -b
>
>then + a = -a ; this however is false,
>
>Therefore -a * -b MUST BE = +ab.
>
>This proof works for me, I hope it works for other people too.
>
>Thank You

This is enough for your own purposes, which is to see why -a * -b is
NOT -ab. I wouldn't call it a proof, because it is not really clear
that -a * -b has to be only ab or -ab; also, the method of
is not to really prove it, but to convince yourself that it makes sense.

2. Mar 11, 2006

### topsquark

Basically your question revolves around the question of -1*-1. This argument is going to be rather Abstract Algebraish, so some of this you might have to look up or take my word for.

First of all, all we really need to refer to here is the set of integers, but you can work with the set of real numbers as well.

So. Integers. The set of integers is {...,-3, -2, -1, 0, 1, 2, 3,...} with two defined binary operations on it, addition and multiplication. Blah, blah, blah. The point is that the integers with these two operations form what is known as a "ring." (Actually, it's an "integral domain.") This means we have definitions that state things like multiplication distributes over addition and that the additive inverse of an integer is unique.

The additive inverse of an integer x is defined by the equation x + (-x) = 0. For a given x there is only one such integer -x.

All of this is prelude. I want to calculate -1 + -1*-1.
So:
-1 + -1*-1
= -1*1 + -1*-1 (Multiplicative identity property)
= -1*(1 + -1) (Distributive property of multiplication over addition)
= 0 (Multiplicative property of 0)
So -1 + -1*-1 = 0. But this means, by definition, that -1*-1 is the additive inverse of -1. But the additive inverse of -1 is 1 because the additive inverse of 1 is -1 (coming from the fact that addition is commutative, that is: a + b = b + a so -1 + 1 = 1 + -1 = 0.) Since the additive inverse of an integer is unique we conclude that -1*-1 = 1.

I have to say if you don't know basic Algebra (not the kind they teach in HS, the kind they teach in College) you might find this argument rather hard to swallow. All I can say is that everything in the above is taken directly from definiton or capable of being proven. This is the simplest line of argument I can think of that doesn't do any "hand waving."

-Dan

Last edited: Mar 11, 2006
3. Mar 11, 2006

### toasty

"This is enough for your own purposes, which is to see why -a * -b is
NOT -ab. I wouldn't call it a proof, because it is not really clear
that -a * -b has to be only ab or -ab; also, the method of
is not to really prove it, but to convince yourself that it makes sense"

Ok for somebody already familiar with it, your proof does work! but that is not what I want to discover.

In the quotation above Dr Math writes;

' it is not really clear
that -a * -b has to be only ab or -ab '

If so, what else could -a * -b BE equal to? in a static equation of the form 'y=z'; as opposed to the dynamic 'y + z'.

4. Mar 12, 2006

### VietDao29

It can be anything like:
(-a) (-b) = ab, or (-a) (-b) = ba.
There's no guarantee that (-a) (-b) must equal either -ab or ab, UNLESS you prove it. If you have not done anything to prove your claim, then it (i.e your claim) may just be plain wrong. In mathematics, everything must be proved in other to be true (unless it's an axiom). And once it's correctly proved, it remains true for eternity. This is a special aspect of mathematics that neither physics nor chemistry has.
(-a) (-b) + (-a) b = (-a) (-b + b) = 0, so (-a) (-b) is the additive inverse of (-a) b.
(-a) b + ab = (-a + a) b = 0, so (-a) b is the additive inverse of ab. And we have already proved that (-a) (-b) is the additive inverse of (-a) b.
So (-a) (-b) = -((-a) b) = ab (Q.E.D)

Last edited: Mar 12, 2006
5. Mar 12, 2006

### topsquark

I don't really see what are you looking for, then. As VietDao29 says, -a*-b could theoretically be anything. I'll add to that though, we could DEFINE -a*-b to be something else, but what we would get would NOT be the set of integers I was working with above. If we are working with the integers (or real numbers for that matter) -a*-b can ONLY be a*b because of the proof I showed above.

-Dan

6. Mar 12, 2006

### toasty

Hmmm I think you left something out of your expression. It is an unfinished static statement' -a*-b='. NOT '-a*-b', which is a complete dymnic expression.

It assumes the observer already knows what '=' means, it assumes that we are talking about numbers, i.e. that with which we count things. It assumes the observer knows what '+', and its opposite '-', means.

I also think you are implying that 'a' is not the same thing as '+a', which in some dymanic math may be the case; but here we are talking about an equivalence - AND it does not matter what you want to call the things which are expressed in that statement, i.e. numbers. In short the idea that there is something else that could be put into such a statement merely begs the question as to how it could count anything.

To me, then, equivalence is soley about quantites of things and thus there are only, or soley, MORE or LESS of them. If more, then the number has to be '+' either stated or implied; however, if less, then the number MUST always be '-'. I also define multiplication as repetitive addition and division the opposite. All higher operations share this same characteristic.

In short, it is implied but not shewn that there IS some other 'signing' of numbers in statements of equivalence BUT it appears to me as if people assume everybody else knows what these are. And its no use talking about 'absolute' values here since they too MUST sooner or later be reckoned as either '+' or '-', in order to discover the value of elements in such statements of equivalence.

A proof offered above states

"It can be anything like:
(-a) (-b) = ab, or (-a) (-b) = ba ; here the writer specifies a to the power of b and the reverse"

but does not go on to shew that such is possible under very specific conditions. Ie it doesn't work at all!

I think folks versed in commutative and distributative logic lose sight of the very complex nature of numbers per se, and sometimes draw in axioms and or logic by implication THAT IS NOT AT ALL CLEAR even to one versed in other forms of the study of numbers and their properties.

For example, today it is assumed students are already familiar with the number line analogy, and that they reckon with such a measure in their brain. Well, not everybody does, and often for the very sound reason that the number line itself is a paradox of such enormity that all kinds of bizarre conclusions may be extracted from it's logic. Thus between each point must lie a measured space, but the space itself is of points; thusly, either the line is or it isn't made of 'number points. But it is both and neither which is absurd.

An analogy yes, but a logical tool never! and I really do mean NEVER.

Second proof from Dan! Very nice if leaving out some steps so leaving me the reader more confused than ever... ouch, did it. But it doesn't. So
tell me how you can get from

-1 + -1*-1
to
= -1*1 + -1*-1

in a universal proof which could have used the number 2; and so would fail right there at that step?

And you state

' But the additive inverse of -1 is 1 because the additive inverse of 1 is -1 '

You probably wanted to state that the INVERSE of - 1 HAS to be +1.

If so then my assertion that the result of '? * ?= must always equal EITHER ' + ?' or ' - ?', is valid, and my proof universal.

I.e either ' - a * - b = + ab OR -a * -b = -ab and anything else MUST either be + or - even if its a power of either value and so on'

In fact now that I reflect on your proof Dan, I can see that my original school math version works far far better, and in so saying I now believe Dr Math simply did not understand what I submitted. OC I will gladly take it all back if you can create some new method of combining or dismantling numbers that is NOT ultimately either addition or subtraction.

Last edited: Mar 12, 2006
7. Mar 12, 2006

### matt grime

let's not. let's actually do maths instead with the integers Z, as we should. If we are doing things 'with which which we count things' then you are perfectly entitled as what on earth -1 counts, and let's not go there. let's do maths instead.

that is an unnecessary pussy footing around. lets use addition and subtraction, one of which is the inverse of the other.

no comment.

nonsense... florid nonsense, though.

.

no, the additive inverse of 1 is minus 1, that is its definition.

8. Mar 12, 2006

### Hurkyl

Staff Emeritus
That's not how proof works. If you wish to assert that the only possibilities are:

-a * -b = -(a * b)
-a * -b = a * b

then it's your job to prove it. It isn't our job to to come up with a counterexample.

The rest of us know that you're right, we know the proof that -a*-b = a*b, when a and b denote "numbers", and - and * denote the "usual" operations upon "numbers".

(So you are clearly correct, although superfluous, when you state that the only possibilities are those two formulas above)

I strongly suggest you try to get it straight in your mind. Not only is algebra very important to mathematics, it also tends to give the clearest proofs. Though, I'm not sure why Dr.Math doesn't simply prove (-1)*(-1) = 1, rather than showing why the alternatives are impossible. The following is taken from Jacobson's Basic Algebra I:

(He has already proven a0 = 0 = 0a in any ring)
--------------------------------------------
We have the equation

0 = 0b = (a + (-a)) b = ab + (-a)b

which shows that

(-a)b = -(ab)

Similarly, a(-b) = -(ab); consequently

(-a)(-b) = (-a)(-b) = -(-(ab)) = ab
--------------------------------------------
Where I've added some parentheses to make his notation more clear. Note that he has also used the fact that if a + b = 0, then a = -b, which is proven as follows:

a + b = 0
(a + b) + (-b) = 0 + -b
a + (b + (-b)) = -b
a + 0 = -b
a = -b

The point is that if we accept these five formula:
1: a*0 = 0
2: a + (-a) = 0
3: (a+b)c = ac + bc
4: a + 0 = a = 0 + a
5: a + (b + c) = (a + b) + c

then we must accept (-a)(-b) = ab, because we have a proof of it from these five formulae.

'-a*-b=' is not an "unfinished static statement" -- it is a collection of symbols that does not correspond to any grammatic element of the language of mathematics.

'-a*-b' is not a "complete dymnic expression" -- it is a term in the language of mathematics, consisting of two variable symbols, and three function symbols.

Mathematical language assumes nothing. It may have been your intent that all formulae are to be interpreted in, say, a model of the theory of integer arithmetic, but that intent is not expressed by using those symbols.

The best you can rely upon is the unspoken agreement that when otherwise unspecified, that's what we mean.

It is clearly not -- the first is a string consisting of one symbol, and the latter is a string consisting of two symbols. Of course, in the theory of arithmetic, we can prove that "a = +a", and in any particular, 'a' and '+a' denote the same element in any model of arithmetic.

I have no idea what you're trying to say here. It doesn't seem mathematical, though.

If you want to study numbers in some non-mathematical manner, then do so in a non-mathematical forum. :tongue:

Only by those who have no clue what they're doing. The clueless can extract paradoxes and bizarre conclusions from anything.

9. Apr 9, 2006

### toasty

'+' , '-' in your world are 'functions'. Hmmm seems to me you reason through a sieve! Primarily ' +' is the act of adding.

A function is a complex expression of several actions such as ^2XLog;base3 158743598734... and so on, when applied to values or implied values in equations.

That Algebraic logic does not ground Number math proofs can easily be demonstrated by supposing there is no such thing as Number or Value.

Yes classes of things can be roughly related by more or less - ness, but nothing is exactly more or less than anything else. Thus a + b cannot = anything at all.

Now thats a completely DYMANIC ism for you, which is a good way to begin the study of circular and linear comparrison.

:rofl:

Last edited: Apr 9, 2006
10. Apr 9, 2006

### arildno

Look up on the concept of a function before you make a statement of what they are.

11. Apr 9, 2006

### Hurkyl

Staff Emeritus
You're in a mathematics forum. So, you're going to get mathematical responses. Mathematically, '+' might be a binary function, a morphism, a function symbol, or even just a function symbol, it is certainly not "the act of adding".

If you want to talk about some sort of correspondence between mathematical formalism and the "real world", then go to a philosophy forum.

Anyways, since you seem to have absolutely no interest in doing anything mathematically, I see no point in leaving this thread open.