# How can you multiply two negative numbers?

Mentor

#### LukeD

Is this true? Do you have a source or is this an opinion?

So if we take the mass of an object, and say said object is accelerating at some rate and we multiply the mass and the acceleration, we would not have the force? Is this due to an error in formula (perhaps we should subtract?), or numerical inaccuracy in measuring mass and acceleration?
I may have been exaggerating when I said every piece of physics, but every physicist I've spoken to or heard speak has expressed that they don't believe that our best physics theories are exactly correct (as in down to every detail and accurate on all scales).

If some mathematical structure happens to describe what is going on with a physics theory, but the theory doesn't work everywhere, then the math only approximately describes the situation. Therefore, the math is an idealization, so it wouldn't be considered "real". That's why I'm not considering any physics theories that are only approximately correct. What I meant is that every physicist I've heard speak has expressed that every theory that we have is only approximately correct.

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By the way, what you said is just the definition of a force. There's no physics in a definition; it's just naming something. I take it that what you meant then is conservation of momentum. Maybe conservation of momentum is correct; it is one thing that few people who think could be violated. It would certainly surprise most people. So that might be one thing that most people do think is correct.

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#### Dark Fire

we can assume i^2 = -1
I'm not really into math atm, however, doesn't that also mean that I^3=1.
But cuberoot of 1 is only 1, not the sqrt of -1.

My take is that world have no choice, but to follow math. And while we can treat math as a "game of our own making" math in fact does exist on its own.
And either way is there no way to prove either, at least not closer than the distant future, so discussing either theory is, in my opinion, of the same level of discussing Christianity versa Buddhism; you're going nowhere.

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#### CRGreathouse

Homework Helper
I'm not really into math atm, however, doesn't that also mean that I^3=1.
But cuberoot of 1 is only 1, not the sqrt of -1.
No, that means i^3 = -i and i^4 = 1. The four fourth roots of 1 are 1, i, -1, and -i. (There are four roots thanks to the fundamental theorem of algebra.)

#### Dark Fire

No, that means i^3 = -i and i^4 = 1. The four fourth roots of 1 are 1, i, -1, and -i. (There are four roots thanks to the fundamental theorem of algebra.)
That told me nothing but to memorize that it's ^4 and not ^3: I did not become any wiser.

#### CRGreathouse

Homework Helper
I did not become any wiser.
Is that because you already knew the fundamental theorem of algebra, or because you declined to learn it? I think it's a radically important part of mathematics.

#### Dark Fire

Is that because you already knew the fundamental theorem of algebra, or because you declined to learn it? I think it's a radically important part of mathematics.
It's because I decided not to google it-- rather pray for you to explain.

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#### LukeD

Dark Fire: I see that you still haven't looked up how exponentiation works, but I'll be happy to refresh you.

If n > 0 is an integer, then x^n means x*x*x*...*x (with n terms).
Exponentiation is then extended continuously to most real numbers (except for 0^0 which is usually undefined and 0^(-a) where a is any positive real number) via the following properties which you can prove in the case where we had x^n and n > 0 was in integer

(x^a)*(x^b) = x^(a+b)
(x^y)^z = x^(y*z)
(x^z)*(y^z) = (x*y)^z
x^0 = 1 (as long as x is not 0)

We then just assume that as long as x is not 0, then we have at least one number defined by x^y

By the first property, if x is not 0, we have that (x^a)*(x^(-a)) = x^0 = 1, so this tells us that x^(-a) = 1/(x^a)

By the second property, we have that if x is not 0, then (x^(1/a))^a = x^(a/a) = x^1=x

In particular, this means that x^(1/2) is one of the square roots of x

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i^2 = -1 means that i*i=-1
Then multiplying by i again gives
i^3 = -1*i = -i
Then we can evaluate i^4 as
i^4 = i^(2+2)=(i^2)*(i^2) = (-1)*(-1)=1

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By the way, I said before that we just assumed that we had some number x^y that satisfied the required properties, but at least in the case where x and y are real numbers and x is non zero, we can prove not only that at least one such number exists, but we can prove that x^y is continuous in both x and y

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#### Dark Fire

i^2 = -1 means that i*i=-1
Then multiplying by i again gives
i^3 = -1*i = -i
I know the general of how exponents work, which is why I've never bothered to look it up, correct.
We never learned about exponents which is less than 1, including negative numbers, however I've seen exponents with negative numbers earlier (not @ school), but never {0, 1} or [0, 1] I forgot which one is which (never learned this @ school either, I haven't done high-school).
I just haven't done school/math in a couple of years, but yes, I realize that it was wrong twice, it should be ^4, actually without even reading the whole sentence, though my mistakes were irrelevant in both cases, so I honestly think you were going off-topic (you in plural).
i^4=1
but
1^0.25!=i
where ! means not equal to, and don't go off-topic again if 1^0.25 is wrong :P
If it's wrong, then please tell me if you want, but add a reply to what's on-topic:
I^4 is 1 but 1 4root isn't only i but 1, meaning i can also be 1 since 1*1*1*1=1 right?
2root = squareroot, 3root = cuberoot, 4root = unnamed.

#### DavidWhitbeck

Well.. let's say that two negatives multiplied gave another negative:

(-1)(-1) = -1 and at the same time 1*(-1) = -1
That would imply that (-1)(-1) = 1*(-1) <=> 1 = -1 which is obviously not true.
It probably is a convention, although a convention that makes sense as we've just seen!
You can prove that (-1)(-1) = 1 just by saying that R with the + and x operators form a field (as part of the definition of R).

(-1)(-1) - 1 = (-1)(-1) + (-1)(1) by the defn of 1
= (-1)(-1 + 1) by distributive law
= -1(0) since -1 is the additive inverse of 1
= 0 by the defn of 0
but
1 - 1 = 0 since -1 is the additive inverse of 1

but now we have shown that
(a) (-1)(-1) is the additive inverse of -1 and
(b) 1 is the additive inverse of -1.

But additive inverses are unique. Therefore (-1)(-1) = 1. Is there flaw in that argument? I'm weak with fields so if you poke holes in it I won't be sad.

#### CRGreathouse

Homework Helper
It's because I decided not to google it-- rather pray for you to explain.
A polynomial of degree n with real (or complex) coefficients has exactly n roots, counting multiplicity. So there are six roots for x^6, x^6 + 3x^4 - 2x, and pi * x^6 - e * x^5 + 3.132432567. The fact that there are n nth roots of unity is just a special case of this. In the case of roots of unity, they are always distinct (unlike (x - 2)(x - 2) which has two roots, both 2).

#### LukeD

I know the general of how exponents work, which is why I've never bothered to look it up, correct.
We never learned about exponents which is less than 1, including negative numbers, however I've seen exponents with negative numbers earlier (not @ school), but never {0, 1} or [0, 1] I forgot which one is which (never learned this @ school either, I haven't done high-school).
I just haven't done school/math in a couple of years, but yes, I realize that it was wrong twice, it should be ^4, actually without even reading the whole sentence, though my mistakes were irrelevant in both cases, so I honestly think you were going off-topic (you in plural).
i^4=1
but
1^0.25!=i
where ! means not equal to, and don't go off-topic again if 1^0.25 is wrong :P
If it's wrong, then please tell me if you want, but add a reply to what's on-topic:
I^4 is 1 but 1 4root isn't only i but 1, meaning i can also be 1 since 1*1*1*1=1 right?
2root = squareroot, 3root = cuberoot, 4root = unnamed.
Like I said before, x^y isn't exactly a function, it can have multiple, or even infinite values. It's just that a certain specific one called the principle value is taken as the value that people usually mean when they write x^y.

This means that x^y is not invertible. So we might have that a^2 = b^2 but not that a=b.
For instance, 1^2 = (-1)^2, but 1 is not -1.

Also, as you pointed out i^4 = 1, but 1^(1/4) = 1 (again, this is just a convention, it could just as well be -1, i, or -i)

#### elarson89

I haven't read all 5 pages but i understand the question. I wanna restate it because I am a bit confused on it too, but in more of a general sense.

It's more intuitive to use the natural numbers in math, but somewhere along the line multiplication was generalized to include decimal, irrational, imaginary, and negative numbers. How was this set generalized, specifically with negative numbers.

Another question I have, how was exponentials generalized to include the negative numbers. (its easy to see the imaginary generalization thanks to Euler) Why is it that a negative exponent is just the reciprocal with a positive exponent?

#### Egor50

To DarkFire Specifically:

We are of like mind on the conceptual swallowing of a negative multiplier. Like you I see how negative values can add, subtract, and divide. But in multiplying I am conceptually snagged to see past zero as a mutiplier no matter how I try to construct it. You just can't multiply negatively. All negative values really say to me is, "its always some positive amount of negative value! No matter how I think of it I can't see past zero as the multiplier dead end. And this follows all of the math examples that try to show me otherwise. Now that said...Can I ignore my own inner concept that you can't multiply anything in this whole universe less than 1 and follow the "rule as explained", yes. But I am not at ease with it. It seems to violate some basic thought rule in me. And its one of those conceptual things I coined the "perceptual snag". I think the world is full of them. And due to natures constraints in the way we think, we are blind to them. And these perceptual snags remain invisible corrupting lots of areas of recorded thought. I think the idea of a negative multiplier is strong enough to be a "LAW". "There can be no multiplier less than the smallest fraction toward infinity. anything beyond (less than) that is ZERO! And how can you go negative (if the concept of negative is - = < 1) times "something" counted? Furthermore if you look at whats there in all negative math references you will see that even though math calls it negative, the value can always be interpreted as some Positive amount to a negative value. this of course really only applies to the "real world" value of negatives. But I can argue this point in its simplest form against any negative value.

#### Egor50

To DarkFire Specifically:

We are of like mind on the conceptual swallowing of a negative multiplier. Like you I see how negative values can add, subtract, and divide. But in multiplying I am conceptually snagged to see past zero as a mutiplier no matter how I try to construct it. You just can't multiply negatively. All negative values really say to me is, "its always some positive amount of negative value! No matter how I think of it I can't see past zero as the multiplier dead end. And this follows all of the math examples that try to show me otherwise. Now that said...Can I ignore my own inner concept that you can't multiply anything in this whole universe less than 1 and follow the "rule as explained", yes. But I am not at ease with it. It seems to violate some basic thought rule in me. And its one of those conceptual things I coined the "perceptual snag". I think the world is full of them. And due to natures constraints in the way we think, we are blind to them. And these perceptual snags remain invisible corrupting lots of areas of recorded thought. I think the idea of a negative multiplier is strong enough to be a "LAW". "There can be no multiplier less than the smallest fraction toward infinity. anything beyond (less than) that is ZERO! And how can you go negative (if the concept of negative is - = < 1) times "something" counted? Furthermore if you look at whats there in all negative math references you will see that even though math calls it negative, the value can always be interpreted as some Positive amount to a negative value. this of course really only applies to the "real world" value of negatives. But I can argue this point in its simplest form against any negative value.

#### epkid08

To DarkFire Specifically:

We are of like mind on the conceptual swallowing of a negative multiplier. Like you I see how negative values can add, subtract, and divide. But in multiplying I am conceptually snagged to see past zero as a mutiplier no matter how I try to construct it. You just can't multiply negatively. All negative values really say to me is, "its always some positive amount of negative value! No matter how I think of it I can't see past zero as the multiplier dead end. And this follows all of the math examples that try to show me otherwise. Now that said...Can I ignore my own inner concept that you can't multiply anything in this whole universe less than 1 and follow the "rule as explained", yes. But I am not at ease with it. It seems to violate some basic thought rule in me. And its one of those conceptual things I coined the "perceptual snag". I think the world is full of them. And due to natures constraints in the way we think, we are blind to them. And these perceptual snags remain invisible corrupting lots of areas of recorded thought. I think the idea of a negative multiplier is strong enough to be a "LAW". "There can be no multiplier less than the smallest fraction toward infinity. anything beyond (less than) that is ZERO! And how can you go negative (if the concept of negative is - = < 1) times "something" counted? Furthermore if you look at whats there in all negative math references you will see that even though math calls it negative, the value can always be interpreted as some Positive amount to a negative value. this of course really only applies to the "real world" value of negatives. But I can argue this point in its simplest form against any negative value.
Who says you have to think of numbers like you do money? Why not think of it like vector space?

#### Diffy

If you can conceptually swallow how negative value's can divide, then think of multiplication as division.

Let a, and b, be negative numbers,

Then a x b = a / (1/b)

#### Egor50

No you are not feeling my snag. Its not that I can not conceptualize the math rules for negative numbers. Even in vector space, its only use is as a reference below some scalier reference. At best as a concept, I still see it as non multiplier in any sense unless there is a corresponding negative value. Do you get me? Its deeper to me. Just "CONCEPTUALLY" a stand alone concept. Negative values can not be used as the multiplier. Multiplicand? Yes!

#### BSMSMSTMSPHD

Proof that (-1)(-1) = 1

Let a be any real number. Then, a = (a + 1) - 1, and so, a2 = [(a + 1) - 1]2.

Now, just expand both sides and simplify:

a2 = (a + 1)2 - 2(a + 1) + (-1)(-1)

a2 = a2 + 2a + 1 - 2a - 2 + (-1)(-1)

0 = -1 + (-1)(-1)

1 = (-1)(-1).

#### arildno

Homework Helper
Gold Member
Dearly Missed
No you are not feeling my snag. Its not that I can not conceptualize the math rules for negative numbers. Even in vector space, its only use is as a reference below some scalier reference. At best as a concept, I still see it as non multiplier in any sense unless there is a corresponding negative value. Do you get me? Its deeper to me. Just "CONCEPTUALLY" a stand alone concept. Negative values can not be used as the multiplier. Multiplicand? Yes!
Your "concept" about distinctive roles of multipliers and multiplicands is totally WORTHLESS, since multiplication is commutative.

Your "concept" is just another misconception.

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