
Product of Negatives (2010) - polyphonicist
https://susam.in/blog/product-of-negatives/
======
dang
We've banned the submitter, the site, and dozens of other accounts, including
susam, for using a ring of accounts to manipulate HN. Such abuse is not
tolerated.

All: if you notice fishy things (as a user did in this case), please let us
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(Please don't post insinuations about abuse in the threads, though, since most
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hn@ycombinator.com. This is in the site guidelines:
[https://news.ycombinator.com/newsguidelines.html](https://news.ycombinator.com/newsguidelines.html))

~~~
trevyn
Manipulate HN into talking about math? I don’t have the full picture, but the
discussion this generated is better than many posts.

~~~
susam
I did use a voting ring to manipulate HN. I am sorry for doing so. Such
behaviour is harmful to the forum and disrespectful to other users who are
participating in good faith as well as to the moderators who work very hard to
keep this forum clean and wonderful.

I have sent an email to dang with an apology and a promise to not repeat this
again. If he can forgive this offence and give me a second chance, I would
like to contribute further to the discussions in the community the right way,
i.e., as myself only, for all future discussions.

------
yunruse
An important thing about numbers in general is that whenever somebody says
“complex/negative numbers don’t actually exist”, they are somewhat right, in a
sense. What exists is magnitude and phase

Does that mean we should abandon them? Absolutely not. Encoding phase (or in a
much more common subset, parity) is so absolutely useful it’s no wonder we
bake 90° intervals (-, i) into our notations: they can be intuitively dealt
with. It’s still somewhat easy to skip over the property, however; as a
student at least I seem to need to backtrack over signs at least once an hour
when working with anything rigorous enough. I wonder if 2-tuple notation, eg
(+, 23) or (-i, x²), would be more intuitive by making parity/phase explicit
rather than implicit.

Complex numbers are a little more nuanced, but no less useful. I imagine you
could develop an alternative notation to make things more intuitive, but
thankfully it’s generally taken for given nowadays that they’re intrinsic to
how we’ve explored nature.

~~~
stan_rogers
Complex numbers, the way they're used in most cases, _is_ a tuple notation.
They're a handy way of keeping your chocolate separate from your peanut
butter, so to speak, as that little "times _i_ " makes it difficult to
accidentally get things mixed up. And that's the way I always explained it to
my students: there are imaginary numbers in the original sense of fake roots
that will go away if you ignore them long enough, and there are imaginary
numbers in the sense that it makes some kinds of calculations easier to keep
straight. I've never been convinced that they are the same thing. One is an
annoying but temporary consequence of arithmetic, while the other is just a
convention, when all's said and done.

~~~
TheOtherHobbes
They're the same thing in the sense they have the same roots.

The most confusing thing about complex numbers is the language. First you're
told negative numbers can't have roots, then you're told they so can too, but
you have to call the roots "complex" or "imaginary."

This sets up cognitive dissonance which can be harder to deal with than the
math. (What even is an "imaginary number"? What are those words supposed to
mean?)

In reality complex numbers are a way of moving from the number line to a
number circle. (Which eventually generalises to a 3-sphere when you get to
quaternions.)

That's all they are. Instead of linear arithmetic - which is about combining
magnitudes in one dimension - you can now do arithmetic that combines
magnitudes with rotations.

The extra dimension makes it possible to solve equations with solutions that
don't exist on the basic number line. It also makes it easier to do
calculations that combine magnitude with phase - which includes pretty much
anything that rotates or processes linear combinations of sine waves, and
which a straight vector tuple can't handle.

If someone had told me this when I was learning complex numbers the cognitive
dissonance wouldn't have hurt quite as much.

~~~
mokus
I found so much of math I had learned previously in the “it’s weird but this
works if you take it on faith” sense was suddenly blatantly obvious after
learning some abstract algebra. I wish I had learned that stuff way earlier.

In the case of complex numbers, I find the “paradox” disappears when you think
of it in terms of fields abstractly. To put it maybe a bit overly simply -
instead of focusing on the idea of “square roots of negative numbers”, instead
step back and consider that number-like operations make sense for things that
aren’t numbers at all in the traditional sense. One particularly useful
example is 2d vectors, which you can add in the usual sense and “multiply” in
polar form by multiplying “r” and adding “theta”. It turns out that these
vectors with these operations act a LOT like numbers, and it also turns out
that that weird multiply operation is actually super useful. One easy
interpretation is combined scale and rotation transforms, with
“multiplication” implementing composition.

Once you do that, it also turns out that solving equations like “what
transformation composed with itself equals a 2x scaling with 180° rotation?”
also make sense (i.e. “solve x^2 = -2”), and when you solve polynomials in
this new system you get more solutions than you did for regular numbers. And
that the thing you just invented IS the field of “complex numbers”.

[Sorry for the verbosity and probably poor organization, I’m in a bit of a
hurry IRL and didn’t have time to edit it down. I did edit a bit for clarity
and to fix typos, etc., though]

------
soVeryTired
I've always thought the best way to explain this was by analogy with the '90s
TV show "The Crystal Maze" [0].

Contestents are put in a dome filled with gold and silver tickets being blown
around by fans. For every gold ticket they collect, they get a point. For
every silver ticket, they lose a point. If they collect enough points, they
win a prize.

Sorting through the team's collection of tickets and throwing away a silver
ticket (minus a -1) is just as good as adding another gold ticket (+1).

Not sure the kids these days are down with the crystal maze though. More loss
to them - Richard O'Brien was a national treasure.

[0]
[https://en.wikipedia.org/wiki/The_Crystal_Maze](https://en.wikipedia.org/wiki/The_Crystal_Maze)

~~~
creddit
What does this analogy have to do with multiplication?

~~~
soVeryTired
It shows that subtracting a minus one is equivalent to adding a plus one. The
one logical leap that isn't explicitly spelled out is that subtracting X is
the same as adding (-1)X. But I'm pretty sure that's the definition of integer
multiplication.

~~~
creddit
I see how it’s intuition for addition/subtraction but that doesn’t tell us
much about multiplication. You’re asserting that negative one times X is
itself negative which is in fact what the article is attempting to prove in
the first place so by explicitly supposing that, your analogy isn’t useful.

~~~
thaumasiotes
> You’re asserting that negative one times X is itself negative which is in
> fact what the article is attempting to prove in the first place

Absolutely not. The article explicitly postulates this:

> We also take for granted the fact that the product of a positive real number
> and a negative real number is a negative real number

You're right that that's the interesting part of the question, but as far as
the article is concerned, it's just an uninteresting assumption.

------
saagarjha
An alternative, "common sense proof" would be that you're undoing the taking
away of things, meaning you have more than you started (i.e. a positive
result).

~~~
throwfermat
Here is a nice one I read sometime back.

Create a video of your friend walking 3 metres. Now play the video 4 times.
Your friend walks 12 metres in the video. Play the video in reverse 4 times.
Your friend walks 12 metres backwards in the video.

Create another video of your friend walking backwards 3 metres. Now play the
video 4 times. Your friend walks 12 metres backwards in the video. Play the
video in reverse 4 times. Your friend walks 12 metres forwards in the video.

~~~
dorchadas
As a math teacher, I've found the best luck with teaching negatives by putting
them in terms of direction. When kids start seeing "right" as "positive" and
"left" as "negative", it makes a bit more sense to them. Then couple in
positive/negative with forward/backward and it just generally clicks.

It also even helps explain division when you talk about it in terms of
direction and how many times you have to move to get to 0. It also helps them
understand why you can't divide by 0. e.g. 3/0 would be explained like
"Starting at 3, how many times can you move 0 to get to 0," They can clearly
see it's impossible, and it helps give them at least some basic intuition into
it.

I've also found this extends to at least some properties of complex numbers as
well, as you can easily extend from the number line to a coordinate plane.

------
b0rsuk
I've seen a better explaination in this Mathologer video. In a bizarre twist
it is now private (?!). Maybe it will work for you. But I suspect it was a
takedown notice because he used a short clip from a movie famous among
teachers. [https://www.youtube.com/watch?v=ij-EK-
MZv2Q](https://www.youtube.com/watch?v=ij-EK-MZv2Q)

The first number represents the amount of something. If it's negative, you
have a debt. The second number represents either a gain (if it's positive) or
a loss (if it's negative).

From that point you can explain it to yourself using plain english. So, -4 *
(-3) can be understood as "Lose a debt of 4, three times". If you have -4 * 3,
you could be said to "gain a debt of 4 three times". 4 * -3 means (Lose 4
three times).

In the video Mathologer criticized exactly the kind of proofs like in this
video. Just saying it's intuitive doesn't make it so. Fundamental things
shouldn't be proven using a number of laws. They should be understood on the
intuitive level and a proof is just to double check.

~~~
empath75
He had a dispute with his original camera man that resulted in some of his
earlier videos being taken down.

------
kidintech
I strongly dislike these kinds of articles/posts due to one reason:

if you're going to prove such a fundamental thing, can you please provide the
axioms that we start from? I.e. "we know" that a - a = 0, multiplication is
distributive, and a x - b = - a x b. These seem arbitrary properties and
"equally" fundamental to -a x -b = ab. Either start from peano and prove
everything along the way, or tell the reader your assumptions. Don't just
divine things along the way.

EDIT: Assumptions are in the third paragraph of the post. I highly doubt they
were there when I wrote the comment. Either way, my concern has been resolved.

~~~
yori
Like mentioned in another comment on this thread, the assumptions are well
known field axioms. They form a good starting point.

And why start with Peano axioms? They seem like a bad starting point because
it would take pages upon pages of proof and it won't easily extend to other
algebraic structures like rings and fields.

~~~
kidintech
> the assumptions are well known field axioms. They form a good starting
> point.

I gave Peano as an example. I don't mind the assumptions, as long as they're
reasonable and presented before the proof. Another comment pointed me to the
fact that they were mentioned in an earlier paragraph, so my issue is
resolved.

------
dwheeler
This is a decent intuitive explanation.

If you want an absolutely rigorous proof, you can view this Metamath proof:
[http://us.metamath.org/mpeuni/mulge0.html](http://us.metamath.org/mpeuni/mulge0.html)
; this has more far more steps, but is totally rigorous. It particular, its
only axioms are those of classical logic and ZFC set theory (not even numbers
are presumed, the system first proves "numbers exist and have these
properties").

------
tromp
summary:

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

~~~
mathnmusic
The explanation that appeals to me:

If number N is an arrow on the number line from 0 to N, then multiplying N by
-1 flips the arrow with the result -N. Multiplying by -1 again would be
another flip, taking you back to N. So a flip followed by a flip is same as no
change (i.e. multiplicative identity 1).

------
edtechdev
This site had more intuitive explanations for things like this, including
imaginary numbers, calculus, etc

[https://betterexplained.com/articles/rethinking-
arithmetic-a...](https://betterexplained.com/articles/rethinking-arithmetic-a-
visual-guide/)

------
cousin_it
If both 1 * -1 = -1 and -1 * -1 = -1, then -1 / -1 has two solutions.

~~~
throwaway2245
Sure, but having a unique solution is not a required property of integer
division.

Parallel to your statement: If both 1 * 0 = 0 and 2 * 0 = 0, then 0 / 0 has
two solutions.

------
throwfermat
There is a discussion in this post's comments section⁽¹⁾ that this works for
fields and rings too.

I know there are precise definitions for fields and rings but can someone here
give me some good examples of fields and rings? Being a non-mathematician, I
find it easy to manipulate examples than manipulate definitions.

Are the set of integers a field? I guess not because the multiplicative
inverse of 2 is not present in this set.

Is the set of integers a ring? I think, yes.

For prime p, is Z_p = {0, 1, ..., p - 1} a field? I think, yes.

Are there any non-numeric rings where product of negatives is positive?

⁽¹⁾ [https://susam.in/blog/product-of-
negatives/comments/](https://susam.in/blog/product-of-negatives/comments/)

~~~
akalin
There's a subtle point to keep in mind when generalizing to rings/fields. The
concept of 'positive' and 'negative' are defined in terms of an order
relation, e.g., 'positive' means >0 and 'negative' means <0\. The integers /
real numbers have the usual order relation such that the additive inverse of a
positive number is negative and vice versa, but an arbitrary ring or field
might not even have an order relation.

For example, the integers mod n is a ring, so (-a) * (-b) = a * b holds, but
it doesn't make sense to call a number mod n positive or negative, since -a
mod n effectively means n - a mod n.

(posted an earlier version of this comment on susam.in.)

~~~
thaumasiotes
> The concept of 'positive' and 'negative' are defined in terms of an order
> relation, e.g., 'positive' means >0 and 'negative' means <0.

I thought the concept of "negative" was defined by reference to an operation.
"Negative 5" is whatever value Q satisfies the equation 5 + Q = 0.

That definition immediately tells you that the negative of a negative is a
positive. Once we know 5 + Q = 0, we ask what the negative of Q is. It's the
value V such that Q + V = 0. But by the definition of Q (and the commutativity
of addition), we already know V = 5.

Once you define negatives this way, it's trivial to show that negatives obey
the standard ordering. But that ordering wasn't necessary in order to define
them.

Summing up, the product of negatives is positive because negation is a kind of
inversion (additive inversion), and two successive inversions always cancel in
any context.

~~~
akalin
That's precisely the subtlety I'm talking about, in which additive inverses
and real numbers less than 0 can both be referred to as 'negatives', and that
the operation of taking additive inverses and real numbers less than 0 both
use the symbol '-'.

It's pretty standard, though, that a 'negative number' is one that is less
than 0, and a 'positive number' is one that is greater than 0, where a
'number' is an element of some subring of the reals.

~~~
thaumasiotes
I don't understand the point you're trying to make. You claimed that negative
and positive are defined by absolute reference to zero. I claimed that they
aren't. But you're presenting an example that assumes I'm right and you're
wrong!

> For example, the integers mod n is a ring, so (-a) * (-b) = a * b holds, but
> it doesn't make sense to call a number mod n positive or negative, since -a
> mod n effectively means n - a mod n.

If negative numbers were defined by reference to a comparison to zero, then
the expression (-a) * (-b) would be meaningless nonsense in Z mod 5 -- as you
point out yourself, Z mod 5 is not ordered in that way. But it isn't nonsense,
and you're not saying it is -- instead, you assume it's obviously valid when
you observe that the equality (-a)(-b) = ab holds.

~~~
akalin
I guess I'm not being too clear, so I'll try again. There are two concepts:

1) Positive and negative numbers (defined in terms of comparison to 0)

2) The negation of a number (i.e., the additive inverse)

They're related in that when both concepts are defined, a negative number is
the negation of a positive number. However, the two concepts don't coincide.
I'm sure you know this, but even over the reals '-x' is the negation of a
number, but not necessarily a negative number.

(-a) * (-b) = a * b is an equation about #2, and it holds in any ring/field,
even ones where #1 doesn't make sense, e.g. Z mod 5. If #1 makes sense, then
this immediately implies that the product of two negative numbers is positive.

My original point was that the blog post is talking about real numbers, for
which #1 and #2 are both defined. However, if it's generalized to arbitrary
rings/fields, where only #2 is defined, then you can't really refer to the
equation '(-a) * (-b) = a * b' as 'the product of two negative numbers is
positive'.

------
wsxcde
This is not a proof of why the product of negative numbers is positive. The
reason why the product of negative numbers is positive is that we define
multiplication to be that way.

Also, this post conflates the unary negation operator with negative numbers.
The two are not the same. In so far as this post constitutes a proof (which
IMO it does not), it is a proof about the behavior of the negation operator.

A good question to ask is why we made this specific choice of definition. Why
should multiplication be defined such that -2*-3 = 6? This is a question that
the post does shed some light on. If we'd chosen some other definition of
multiplication, a lot of the "intuitive" properties of multiplication that
hold over the natural numbers (such as the distributivity of multiplication
over addition and subtraction) would no longer be true over the integers.

~~~
thaumasiotes
> If we'd chosen some other definition of multiplication, a lot of the
> "intuitive" properties of multiplication... would no longer be true

Well, sure, if you change the definition of something, then it may end up
having different properties. What's your point?

~~~
wsxcde
My point is that you cannot prove something that is true by definition. The OP
trying to prove that the product of two negative numbers is positive is like
asking to prove that 0 + 1 = 1 in Peano arithmetic.

The OP thinks that his "proof" is showing why multiplying negative values
yields a positive result. But the proof is a load of nonsense because it
assumes facts like distributivity of multiplication over addition and
subtraction. It is literally impossible to prove that $\forall a, b, c \in Z.
(a - b) * c = (a * c - b * c)$ -- distributivity of multiplication over
subtraction -- without having already defined the meaning of a * b for all
integers! This leads to a circular reasoning loop that the OP's "proof" can't
get out of.

The thing to realize is that multiplication is not some magic operation handed
down to us by god. It is just a binary total function defined over the
integers. What the OP is trying to confusedly get at is the following:

1\. There is an intuitive definition of multiplication as repeated addition
over natural numbers.

2\. It is not clear what the corresponding definition of multiplication over
negative numbers is.

3\. If we want to define multiplication as a total function over the integers,
we need to define what the result should be when multiplying negative
integers.

4\. Specifically, with (3), we are taught in school that the result of
multiplying two negative numbers should be positive, but it is not clear why
this seemingly arbitrary choice was made.

Unfortunately, the OP is going about this all backwards. One cannot prove what
the OP wants to prove. What one can instead do is argue that the specific (but
seemingly arbitrary) definition that one has chosen for multiplication is a
"good" choice because it has the same properties (distributivity etc.) as
multiplication over natural numbers. At its core, this is a stylistic appeal
about the "naturalness" of the definition.

~~~
yori
Not every arithmetic property needs to be proved from Peano axioms. One can
but it is tedious and unnecessary. A much better starting point is the set of
field axioms where the distributivity property is already available as an
axiom.

The assumptions made in the article are perfectly fine as per field axioms.
Granted it would have been nicer if distributivity over addition was used
instead of distributivity over subtraction. But it is not a big leap to derive
distributivity over substraction from field axioms by distributing
multiplication over a positive number and the additive inverse of another
positive number.

Wherever you see an assumption made about negative number, just mentally
replace it with additive inverse of a positive number and you would be fine.

~~~
wsxcde
You are wrong in several ways.

1\. You literally cannot prove this fact from the Peano axioms because Peano
arithmetic operates on natural numbers, not integers.

2\. As I said in my original post at the top, negative numbers are different
from the unary subtraction operator (the additive inverse in the field). The
number -2 is an entity that exists by itself regardless of whether you've
defined an additive inverse. It turns out that the additive inverse of every
positive integer is the corresponding negative integer, but this follows from
the definition of +, not the other way around.

3\. Even if you give OP the benefit of the doubt regarding his dodgy proof, it
is saying something about the additive inverse and its relation to
multiplication. It is not saying why the result of multiplying two negative
values must be positive.

4\. The multiplicative operator over the field must already be defined for you
to be able to prove distributivity over addition. You can't assume
distributivity over an operator that is only partially defined.

\--

Think about how you'd define a field. First, you need a set (let's call it Z),
then you need two total operators over the set (+ and _), and two elements of
the set (0 and 1) and each of these must satisfy specific properties (aka the
field axioms). In particular, + and_ must be defined for all members of Z, not
just Z+ and further + and * must be distributive. These are all facts you need
to prove about Z, *, +, and 1 and only then do you have a field. You cannot
work backward by assuming the field axioms (which are unfortunately named
because they are not axioms at all but properties) to derive the definition o
the field operators.

~~~
carloswilson
> You literally cannot prove this fact from the Peano axioms because Peano
> arithmetic operates on natural numbers, not integers.

Sure you can. With definitions! Define integers from natural numbers. Define
rationals from integers. And so on. And so on.

> The number -2 is an entity that exists by itself regardless of whether
> you've defined an additive inverse.

I see a serious misunderstanding of this topic. Please read upon the field
axioms and ring axioms if you haven't so already. Then please check
[https://math.stackexchange.com/a/878844](https://math.stackexchange.com/a/878844)
which is arguably more rigorous than this post. But the essence is the same.
This is more rigorous because the subtraction operator is not used anywhere.
Only addition, multiplication and additive inverses have been used. Like
another commenter said, if you just replace subtraction with addition with an
additive inverse in the OP's post, things fall in place.

