why're two negs mult pos?

[Gale]

Ok, so in discussion today we were talking about combinatonics and counting. We were going over binomial coeffients, and i didn't like the explanation, and i related it to "its like, when you're learning to multiply, and they tell you two negatives equal a positive because when you put one minus sign on top of the other it makes a plus sign." But moving the lines around doesn't actually explain why two negatives equal a positive, its just a conveinient way of learning it so you can do the problems right.

...but then someone asked me after class, "why DO two negatives equal a positive?" and i was like "err... cause multiplication is the addition of groups and when you have a negative and you put it into negative groups... Hmmph! i don't actually know dammit!"

So, i've been chewing it over, and i think i've made some sense thinking about it in terms of negative meaning direction, and multiplying meaning you increase the magnitude... but i dunno.... i want a better explanation. Please.

[whozum]

If a and b are two negative integers they can be written as two positive integers c and d such that

[itex] ab = (-1)c(-1)d = (-1)(-1)cd = 1cd = cd [/tex] where a and c are equal in magnitude, as are b and d.

edit: itex is prettier.

[TD]

Perhaps this (http://mathforum.org/dr.math/faq/faq.negxneg.html) may help :smile:

[Gale]

but why is neg one times neg one positive? just an identity? is there a better conceptual way to think of it?

how about some number theory or set theory or something that describes what happens to negative numbers?

[whozum]

read the page he sent you.

[TD]

It contains multiple examples to make it 'understandable' as well as some mathematical details I believe.

[Gale]

ya, ok, i guess that works... i have a better question to ask anyway...

[Werg22]

-1x=opposite of x. -1(-x)=opposite of a negative, a positive.

[Gokul43201]

The integers are designed so that they form what is known as a ring (http://mathworld.wolfram.com/Ring.html). A ring must have (among other things) an additive identity and an additive inverse. The additive identity is the number we call 0, which has the property that 0 + x = x + 0 = x. Next we define the additive inverse of 1 as that element which when added to 1, gives the sum 0, or if x + 1 = 0, then x is the additive inverse of 1. We use the symbol "-1" to represent this number. Next we make use of the distributive property and the definition 0 = 1 + (-1) to write
0 = 0*(-1) = {1 + (-1)}*(-1) = 1*(-1) + (-1)*(-1)
Since 1 is the multiplicative identity, we know that 1*x = x*1 = x, for all x, and hence, 1*(-1) = -1. So we have 0 = -1 + (-1)*(-1). But we know that 0 = -1 + 1, therefore, from the uniqueness of addition (-1)*(-1) = 1.

Note : The operation of subtraction is merely a shorthand for adding a negative number.

[hypermorphism]

ya, ok, i guess that works... i have a better question to ask anyway...
Or the other way around from the page is to accept the field axioms (http://mathworld.wolfram.com/FieldAxioms.html), then prove yourself that (-1)(-1)=1 using them. You just need to use more precise terms of what a negative refers to and what rules multiplication follows.

[eNathan]

We debated this a while back on this thread http://physicsforums.com/showthread.php?t=82997.

My approch was to use English grammer exmaples to explain why -- = +. Basicly, any negative statment such as "not" or "didnt" counts as a -1.

Well that kinda gets into the very logic of what a negative is. Think of it as logic "not" operations. Think about using these in sentences and you will get the idea.

-1*-1=1 ; No + No = Yes
1*1=1 ; Yes + Yes = Yes
-1*1=-1 ; No + Yes = No
1*-1=-1 ; Yes + No = No

For instance, if I said...
I did not not go to the store. That really means, you DID go to the store. -1 * -1 = 1 ; No + No = Yes ... and so forth.

You kinda see how there is a relationship between mathematics and grammer? lol Im not sure if this is an official way to present it, but I just though it up a while back.

[Hurkyl]

So, i've been chewing it over, and i think i've made some sense thinking about it in terms of negative meaning direction, and multiplying meaning you increase the magnitude... but i dunno.... i want a better explanation. Please.

You've almost got the geometric picture, methinks. It's a nice one, and is important to know when you deal with the complex numbers.

In the real case, multiplication by a positive means you leave the direction unchanged, and multiplication by a negative means that you flip the direction. So, if you have a negative, and multiply by another negative, the result is a positive.

In the complex case, numbers can be seen as having a magnitude and a direction in the complex plane. Multiplying by a complex number that lies at an angle θ means rotation by θ. (Angles are measured counterclockwise from the positive x axis)

[ktoz]

... multiplication is nothing more than a shorthand for repeated additions and in that light, there's no way to justify -1 * -1 = 1. Despite the convoluted logic the conventional explanation for multiplying two negative numbers amounts to this

for negatives
-1^2 = -1 + 2
-2^2 = -2 + 6
-3^2 = -3 + 12
-4^2 = -4 + 20

whereas for positives
1^2 = 1 + 0
2^2 = 2 + 2
3^2 = 3 + 6
4^2 = 4 + 12

So we're supposed to buy the explanation that, through some magical process, the interval between a negative and it's square is always larger then the interval between the positive of the same number and it's square?

Doesn't smell right.

[ktoz]

Well that kinda gets into the very logic of what a negative is. Think of it as logic "not" operations. Think about using these in sentences and you will get the idea.

-1*-1=1 ; No + No = Yes
1*1=1 ; Yes + Yes = Yes
-1*1=-1 ; No + Yes = No
1*-1=-1 ; Yes + No = No

For instance, if I said...
I did not not go to the store. That really means, you DID go to the store. -1 * -1 = 1 ; No + No = Yes ... and so forth.

In the sentence examples you stated, the logic follows, but your logic table could also equal this:

-1*-1=1 ; No + No = Emphatic No
1*1=1 ; Yes + Yes = Emphatic Yes
-1*1=-1 ; No + Yes = Terminating Yes
1*-1=-1 ; Yes + No = Terminating No

[HallsofIvy]

ya, ok, i guess that works... i have a better question to ask anyway...


Well, what is the better question?? We're waiting! :smile:

[hypermorphism]

... multiplication is nothing more than a shorthand for repeated additions and in that light, ...

That loses meaning when one looks beyond integers. What does it mean to repeat addition 1/2 times, or Pi times ? i times ?
If you just want to stay within the integers, you first have to define what a negative integer is. In that case, it is the number that when added to the corresponding positive integer, returns 0 (called the additive inverse). Multiplication defined your way makes sense for positive multiples. A negative multiple would have to be translated to a positive multiple multiplied by -1, in which you would apply your repeated addition definition, then have the negative sign applied. A negative multiplied by a negative gives us the following problem then: (-a)*(-b) = (-1)*(-1)*a*b, so we now have the problem of figuring out (-1)*(-1). Now, we know from our definition that 1 + (-1) = 0, so we have (-1) + (-1)*(-1) = 0. It is easy to prove that each integer has only one additive inverse (uniqueness). Thus, (-1)*(-1) must be the additive inverse of -1, which is 1.

[SteveRives]

...
for negatives
-1^2 = -1 + 2
-2^2 = -2 + 6
-3^2 = -3 + 12
-4^2 = -4 + 20

whereas for positives
1^2 = 1 + 0
2^2 = 2 + 2
3^2 = 3 + 6
4^2 = 4 + 12

Doesn't smell right.

That's because you are not following the rules!

It would be like me saying,

1 + 1 = 2 + 0

but

2 + 2 = 5 - 1

And then declaring that I have shown something smells fishy.

In reality, however, you have established a false opposition. It is not as if
-1^2 = -1 + 2
in contrast to
1^2 = 1 + 0

To prove it, I could just as easily say:
-1^2 = 1 + 0
and
1^2 = -1 + 2

The fact is, -1^2 = 1^2

Why? Because multiplication by -1 is like 180 degree rotation (flipping) -- note that 90 degree rotation would be multiplication by i. Someone has already pointed us in this direction.

[Gokul43201]

So we're supposed to buy the explanation that, through some magical process, the interval between a negative and it's square is always larger then the interval between the positive of the same number and it's square?

Doesn't smell right.
For k>0
neg interval = (-k)2 - (-k) = k2 + k
pos interval = k2 - k

So, neg interval = pos interval + 2k, and since k>0, neg interval > pos interval.

See ? No magic ! No bad smell !

[Hurkyl]

I feel the need to comment that -1^2 = -1. Exponents before multiplication.
:tongue2: (Of course, I know both of you meant (-1)^2)

[arildno]

a) multiplication is nothing more than a shorthand for repeated additions
b) Doesn't smell right.
a) False
b) Correct, your assertion in a) stinks.

[arildno]

Gale17:
Several of the posters have already provided you with sufficient information to get this, so my post is probably redundant:

We'll show this in 3 steps:
1. Proposition: a+(-1)*a=0 ("a" is an arbitrary number)
Proof:
a=a*1.
Hence, a+(-1)*a=a*1+(-1)*a=a*(1+(-1))=a*0=0 Q.E.D.
(The last step, a*0=0 requires some additional justification; I'll skip that for now)

2. Proposition: (-a)=(-1)*a
Proof:
a+(-a)=0 (by definition of the negative)
Hence, by adding (-1)*a to both sides, we get:
a+(-a)+(-1)*a=0+(-1)*a
We shuffle about the left hand side, add together a+(-1)*a (getting 0 from Prop. 1)), and remember that for any number "b", b+0=b.
Thus, we get the desired relation:
(-a)=(-1)*a Q.E.D

3. Proposition: (-1)*(-1)=1
Proof:
1+(-1)=0
Squaring both sides yields:
1+(-1)+(-1)+(-1)*(-1)=0
Identifying 1+(-1)=0 on the left-handside, we have:
(-1)+(-1)*(-1)=0
Adding 1 to both sides yields:
(-1)*(-1)=1 Q.E.D.

Thus, we have as a consequence:
(-a)*(-b)=((-1)*a)*((-1)*b)=((-1)*(-1))*a*b=1*a*b=a*b, where a,b are arbitrary numbers.

[Hurkyl]

Werg: that makes no sense at all.

[Werg22]

Here is a very simple proof:

Let:

n - a = b
n - b = a

Replace a in the first equation;

n - (n - b) = b
-(-b) = b
-1(-1)b = b
-1(-1) = 1

Any multiplication of two negative number can be written as:

(-a)(-b); where a and b are positive,

(-1)(-1)ab

Since we have prooven that (-1)(-1)=1, and ab is positive, the product of two negative number is positive.

Q.E.D., as they say.

[Hurkyl]

Ah, that makes sense. (I peeked at the source before you deleted it)

[eNathan]

I feel the need to comment that -1^2 = -1. Exponents before multiplication.
:tongue2: (Of course, I know both of you meant (-1)^2)

As far as I know, when you raise a number to e (or square it), you are mulitplying it itself e number of times. You dont mulitply the absolute value of it. So how is -1^2 = -1 OR -1 \cdot -1 = -1? Unless you are saying that -1^2 is actaully -1 * abs(-1) or something of that sort :uhh: which is the only way you could derive a product of -1.

Im sure you know what im saying, there's probally just some sort of misunderstanding as to what you meant to say.

Edit :: Did you mean - (1^2) = -1?

[Hurkyl]

Yes, -1^2 means -(1^2).

[Hurkyl]

Here is a very simple proof:

I will make one complaint: you've not shown that -b = (-1)b.

[Werg22]

This is by definition I beleive. But perhaps you can show it?

[motai]

One of my teachers put it this way:

The Friend of my Friend is my Friend | + * + = +
The Friend of my Enemy is my Enemy | + * - = -
The Enemy of my Friend is my Enemy | - * + = -
The Enemy of my Enemy is my Friend | - * - = +

While not exactly a mathematical proof, it makes sense.

[hypermorphism]

This is by definition I beleive. But perhaps you can show it?
-b represents the additive inverse of b, ie., the number "-b" such that b + (-b) = 0. -1 is simply the additive inverse of 1. It is not defined that -1*a is the additive inverse of a, but it is trivial to prove. :smile:

[Hurkyl]

It's good for a heuristic proof, but if you want a really formal, rigorous proof, these things can be tricky to get right, because we're so used to assuming all of the niceties of arithmetic!

(Of course, it also depends on from what axioms and definitions you start! You could start with the definition -b := (-1)b, but then you'd have to prove that b + (-b) = 0)

[Hurkyl]

I guess the real question is if Gale is happy with the responses!

[eNathan]

One of my teachers put it this way:

The Friend of my Friend is my Friend | + * + = +
The Friend of my Enemy is my Enemy | + * - = -
The Enemy of my Friend is my Enemy | - * + = -
The Enemy of my Enemy is my Friend | - * - = +

While not exactly a mathematical proof, it makes sense.

Well put, motai! This example is excellent for people to grasp the concept of a negative :approve:

[Werg22]

Hurkyl if we keep it that way we will eventually have to prove that we exist. :rofl:

[Gale]

I guess the real question is if Gale is happy with the responses!
am i ever really happy? i was basically satisfied after the first link... but you guys are really having fun with it... so lets prove it in at least... 3 more different ways.

[HallsofIvy]

Hurkyl if we keep it that way we will eventually have to prove that we exist. :rofl:

I can prove (to my satisfication) that I exist. I'm not so sure about the rest of you.

[roger]

I can prove (to my satisfication) that I exist. I'm not so sure about the rest of you.


Hallsoivy, please could you tell me how to prove the existence of the uniqueness of addition ?

In summary , what is the minimum number of definitions or axioms one needs to prove that -a*-b = a*b ?

[HallsofIvy]

First, I didn't say I could prove the existence of anything but myself!

Second, you don't mean "prove the existence of the uniqueness of addition", you mean simply "prove that the result of addition is unique".

Exactly now you do that depends on what number system you are using.

The simplest is the natural numbers where you take as axioms, the existence of a number "1" and the existence of the "successor" of any number as a one-to-one, onto function from the natural numbers to the natural number without "1" and take the "principle of induction" as an axiom (If a set X contains the number 1 and, whenever it contains the number k, it also contains the successor of k, then X is the set of all natural numbers"
We then define addition by "n+ 1 is the successor of n. If m is not 1, then it is the successor of some number, say p. We define n+ m as "the successor of n+ p".
Now we can prove that m+ n is unique:
Let n be a counting number and let X be the set of all m such that n+ m is unique. n+ 1 is, by definition, the successor of n. That is unique since "successor" is a function. Suppose k is in X. Then n+ successor of k is, by definition,the successor of n+ k. n+ k is unique since k is in X. The successor of n+ k is then unique because "successor" is a function. Therefore the succesor of k is in X. Therefore X is all natural numbers: For every natural number, n+ m is unique for all natural numbers m.

Once we have that we can define the integers by: Let NxN be the set of all pairs of natural numbers. We say that two such pairs, (a,b) and (c,d) are equivalent if and only if a+d= b+ c. It's easy to show that that is an equivalence relation and so partitions the set of pairs into equivalence classes. An "integer" is such an equivalence relation.
We define addition of integers by: If x and y are two integers, that is, two such equivalence classes, choose two "representative" pairs, (a,b) from x, and (c,d) from y. "x+ y" is the equivalence class containing the pair (a+c, b+d). In order to show uniqueness here, we must show that choosing different pairs from x and y would give the same equivalence class for x+ y.
That is, suppose (a,b) and (a',b') are both in equivalence class x and (c,d) and (c',d') are both in equivalence class y. That is, a+b'= b+a' and c+d'= d+c'. Adding those two equations, a+b'+c+d'= b+a'+d+c' so (a+c)+(b'+d')= (b+d)+ (a'+c') showing that (a+c, b+d) is equivalent to (a'+c', b'+ d').

[Robokapp]

I got the easiest answer. My math teacher shoed this.

(-1)*(-1)=1 Given
1(-1)*1(-1)=1 1 multiplied to any factor equls that factor.
(1*1)*[(-1)*(-1)]=1 Association property

1*1=1 so substituting that above...

1*[(-1)*(-1)]=1

as stated in second line, 1 times any number = that number, so we can write

(-1)*(-1)=1 and that has to be true due to last form of the expression. Imagine the (-1)*(-1)=x

solve 1*x=1

there you go!

[jcsd]

-a*-b = a*b folows on form the follwing properties:

additve identity, additive inverse, additive associativty and distributivity

so it is true in any ring.

[Hurkyl]

In summary , what is the minimum number of definitions or axioms one needs to prove that -a*-b = a*b ?
Just 1: take as an axiom that -a*-b=a*b. :smile:

It may seem like a cheap trick, but these sorts of silly things are very useful.

In any case, I think you meant to ask about trying to weaken the axioms of arithmetic as much as possible -- I just wanted to make this explicit.