Product of Negatives (2010)(susam.in)
susam.in
Product of Negatives (2010)
https://susam.in/blog/product-of-negatives/
78 comments
Manipulate HN into talking about math? I don’t have the full picture, but the discussion this generated is better than many posts.
[deleted]
> All: if you notice fishy things like this, please let us know at [email protected]
What would we have noticed, in this case?
What would we have noticed, in this case?
I wish I could spell it all out, but unfortunately that would help spammers.
Here's one thing though: multiple accounts submitting, commenting, and promoting the same person's sites, articles, and (importantly) repos.
https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...
Here's one thing though: multiple accounts submitting, commenting, and promoting the same person's sites, articles, and (importantly) repos.
https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...
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.
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.
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.
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.
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.
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]
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]
You're describing some sort of type mismatch between two concepts I think, and I really don't understand it: and I feel like as someone who occasionally teaches these things I really would like to. Could you elaborate?
For my part, I do like to think of adjoining numbers onto an existing system, but that immediately becomes matrices.
So you decide to adjoin an ε such that ε² = 0. Your numbers are now vectors (a, b) and the action of ε is to map this to (b, 0) so that it is represented by the matrix
When you adjoin to the reals an i such that i² = -1 these matrices have the shape of the 2D scaled rotations, so that is what complex numbers just “are” to me. The representation as a tuple is to me the same as representing the above matrix as (a, b) or (a, b, a+b) to save time or space... The whole thing is a matrix but the entries are indeed redundant and so you don't need to store all of them at once.
So that's why I do not understand what you mean by these being separate concepts. Can you elaborate?
For my part, I do like to think of adjoining numbers onto an existing system, but that immediately becomes matrices.
So you decide to adjoin an ε such that ε² = 0. Your numbers are now vectors (a, b) and the action of ε is to map this to (b, 0) so that it is represented by the matrix
[ 0, 0 ]
[ 1, 0 ]
[ a, 0 ]
[ b, a ]
[ a, b ]
[ b, a + b ]
When you adjoin to the reals an i such that i² = -1 these matrices have the shape of the 2D scaled rotations, so that is what complex numbers just “are” to me. The representation as a tuple is to me the same as representing the above matrix as (a, b) or (a, b, a+b) to save time or space... The whole thing is a matrix but the entries are indeed redundant and so you don't need to store all of them at once.
So that's why I do not understand what you mean by these being separate concepts. Can you elaborate?
I'd like to understand what you're doing, but I'm missing something.
1. Adjoining ε such that ε² = 0.
If the numbers have the form a + bε, isn't the action of ε to map (a, b) to (0, a)? Did you mean to say that ε = [0,1; 0,0]?
2. Adjoining φ = (1 + √5)/2 = [0,1; 1,1].
I'm fascinated by the idea of determining that a 2x2 matrix is equal to a real number.
I see that if you take successive powers of the real number φ, and express them in the form aφ + b, the coefficients a and b will take on values from the Fibonacci sequence. So far so good.
I don't follow the claim that the matrix [0,1; 1,1] actually represents φ. This would imply that the formula F_n = [1, 0] · φ^(n+1) · [1, 0] means that F_N equals the real number 1 (= 1 + 0φ), times φ^{n+1}, times 1 again. But this isn't true. It certainly is true that [1,0][0,1; 1,1]^{n+1}[1,0] is equal to F_n, but I don't follow the interpretation as adjoined numeric values (as opposed to as coefficients of φ).
3. Equivalence of representations [a,b] and aI + bM, where M is a 2x2 matrix representing any adjoined number.
This looks to me like the claim that if [a,b]M = [c,d], then (aI + bM)M = cI + dM. I tried to work this out algebraically and I'm pretty sure it isn't true in general. I'm open to being told that I'm wrong about this. Have I interpreted the idea correctly? What are the conditions under which the equivalence holds?
1. Adjoining ε such that ε² = 0.
If the numbers have the form a + bε, isn't the action of ε to map (a, b) to (0, a)? Did you mean to say that ε = [0,1; 0,0]?
2. Adjoining φ = (1 + √5)/2 = [0,1; 1,1].
I'm fascinated by the idea of determining that a 2x2 matrix is equal to a real number.
I see that if you take successive powers of the real number φ, and express them in the form aφ + b, the coefficients a and b will take on values from the Fibonacci sequence. So far so good.
I don't follow the claim that the matrix [0,1; 1,1] actually represents φ. This would imply that the formula F_n = [1, 0] · φ^(n+1) · [1, 0] means that F_N equals the real number 1 (= 1 + 0φ), times φ^{n+1}, times 1 again. But this isn't true. It certainly is true that [1,0][0,1; 1,1]^{n+1}[1,0] is equal to F_n, but I don't follow the interpretation as adjoined numeric values (as opposed to as coefficients of φ).
3. Equivalence of representations [a,b] and aI + bM, where M is a 2x2 matrix representing any adjoined number.
This looks to me like the claim that if [a,b]M = [c,d], then (aI + bM)M = cI + dM. I tried to work this out algebraically and I'm pretty sure it isn't true in general. I'm open to being told that I'm wrong about this. Have I interpreted the idea correctly? What are the conditions under which the equivalence holds?
(1) This gets a little into difficult notation but e.g. in Mathematica-style notation,
(2) So the case for φ is very similar to the case for ε: you want to start with the constitutive relation, in this case φ² = φ + 1, to build the matrix with first column {{0},{1}} (multiplying 1 by φ gives φ) and the second column {{1},{1}} (the above constitutive relation).
The isomorphism is then that if X represents this adjoined unit then the point {{a}, {b}} becomes a I + b X one way [or say if you have a cubic constitutive relation then {{a},{b},{c}} becomes a I + b X + c X² similarly] and M becomes M {{1},{0}} [or say M {{1},{0},{0}} etc].
I would definitely agree that this is probably a much narrower statement than your “if [a,b]M = [c,d], then (aI + bM)M = cI + dM” as the structure of X is very tight. It is always [e_2 e_3 ... e_n c] where e_i is a unit column vector with a 1 in the i’th place and 0s everywhere else, and c is the vector embodying the constitutive relation.
But if you want a condition, the isomorphism condition is probably the best place to go.
So the claim is that φ^n = F_{n-1} + F_n φ.
{{0,0},{1,0}} {{a},{b}} = {{0}, {a}}.
(2) So the case for φ is very similar to the case for ε: you want to start with the constitutive relation, in this case φ² = φ + 1, to build the matrix with first column {{0},{1}} (multiplying 1 by φ gives φ) and the second column {{1},{1}} (the above constitutive relation).
The isomorphism is then that if X represents this adjoined unit then the point {{a}, {b}} becomes a I + b X one way [or say if you have a cubic constitutive relation then {{a},{b},{c}} becomes a I + b X + c X² similarly] and M becomes M {{1},{0}} [or say M {{1},{0},{0}} etc].
I would definitely agree that this is probably a much narrower statement than your “if [a,b]M = [c,d], then (aI + bM)M = cI + dM” as the structure of X is very tight. It is always [e_2 e_3 ... e_n c] where e_i is a unit column vector with a 1 in the i’th place and 0s everywhere else, and c is the vector embodying the constitutive relation.
But if you want a condition, the isomorphism condition is probably the best place to go.
So the claim is that φ^n = F_{n-1} + F_n φ.
I've heard the objection about complex number not being real many times. I think the sensible answer is to argue that the natural numbers don't "actually exist" either. They're an abstraction just like the complex numbers.
Arguably we might one day find out that the universe is discrete at which point we could begin to try to define the naturals as something that "exists", at least up to some maximum large number. But even then the numbers are probably still best thought of as just a helpful abstraction.
Arguably we might one day find out that the universe is discrete at which point we could begin to try to define the naturals as something that "exists", at least up to some maximum large number. But even then the numbers are probably still best thought of as just a helpful abstraction.
I have a much harder time beliving in the full set of real numbers than I do believing in the basic construction of complex numbers. The full set of real numbers requires me to accept things like the axiom of choice, and to believe that non-computable numbers 'exist' on the same level as computable ones. That doesn't sit right with everyone.
Basic complex numbers, on the other hand, just require me to expand what I accept as the solution of an equation. Note that I've already done this with fractions.
Fractions: Given integers a and b, ax + b = 0 has a meaningful solution
Complex numbers: The equation x^2 + 1 = 0 has a meaningful solution
Basic complex numbers, on the other hand, just require me to expand what I accept as the solution of an equation. Note that I've already done this with fractions.
Fractions: Given integers a and b, ax + b = 0 has a meaningful solution
Complex numbers: The equation x^2 + 1 = 0 has a meaningful solution
The construction of real numbers does not need the axiom of choice in the sense that there are constructions of models of real numbers that do not need it. One example of such construction is described on the Wikipedia page [1], look for "Edudoxus reals" there.
[1] https://en.wikipedia.org/wiki/Construction_of_the_real_numbe...
[1] https://en.wikipedia.org/wiki/Construction_of_the_real_numbe...
At which point does the construction of the reals require the axiom of choice (AC)? I'm not familiar what exactly happens without AC, but defining R as the set of rational Cauchy-sequences modulo zero-sequences does not seem to use it?
The moment where you want a Cauchy sequence of real numbers to converge.
You have a Cauchy sequence of real numbers (a_i)_i. You pick a representative (b_ij)_j (a Cauchy sequence of rational numbers) for each sequence member a_i (axiom of choice!) and then produce the diagonal sequence (b_ii)_i which is the constructed limit of the Cauchy sequence (a_i)_i.
You have a Cauchy sequence of real numbers (a_i)_i. You pick a representative (b_ij)_j (a Cauchy sequence of rational numbers) for each sequence member a_i (axiom of choice!) and then produce the diagonal sequence (b_ii)_i which is the constructed limit of the Cauchy sequence (a_i)_i.
One could argue, easily, that the real numbers are not, in fact, "real," either, that they are rather an abstraction needed to make any measures "commensurable." As far as the magnitude goes, for example, there may be physical limits on how small things can be. (By the same token, even large positive integers can easily be seen as unreal, if you consider a number that is greater than the total count of things in the universe.)
I’m not sure that I would rely on a physical manifestation to show that a number ‘exists’. They exist because we’ve defined them as part of the rule set we use for the game we call math. Regardless of their applicability to nature, they still ‘exist’ in that sense.
Unfortunately (or perhaps fortunately?) we depend upon the a+bi notation to do the right thing algebraically with complex numbers. I do agree, though, that we should always teach that ALL numbers are really complex numbers, although the phase is usually 0 or pi (and that a negative real part is just a shorthand to save us having pi all over the place in most simple math.)
what does “exists” mean? We made the whole thing up.
In a way, sure. In math, existence means consistency, i.e. being free of logical contradiction. Often, "a proof of existence" is done by construction based on a system whose "existence" is assumed. In particular, the complex numbers are usually constructed from the reals in the form (a, b).
A number is just something that acts like a number. Polynomials are numbers. True and False are numbers.
complex/negative numbers don't actually exist in contrast to natural numbers that don't actually exist in a different way
In fact, it's an interesting (and non-trivial) philosophical topic, to consider all different ways in which things do not exist.
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).
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.
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.
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.
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.
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/
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/
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.)
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.)
> 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.
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.
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.
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.
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.
> 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.
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'.
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'.
In a finite numeric field, then, negatives are the same as positives. (For example, in Z mod 5, you get 2 + 3 = 0 and, sure, 2 + (-2) = 0, too.)
Yes?
In Z mod 5 using your notation, you have that 3 = -2. It doesn't make sense to distinguish two classes of "negative" and "positive" numbers in that case, but it still makes sense to talk about -2.
In Z mod 5 using your notation, you have that 3 = -2. It doesn't make sense to distinguish two classes of "negative" and "positive" numbers in that case, but it still makes sense to talk about -2.
That was precisely the original point - the distinction only makes sense in an ordered set; otherwise, calling -2 "negative" may be devoid of any meaning (e.g. when -2 = 3).
The typical example of a field is the collection of rational numbers. These are still numeric, so they might not seem too exotic. Similarly, the typical example of a ring is the collection of single-variable polynomials with ringed (integer) coefficients. In both of these examples, the product of negatives is positive.
A more interesting example: If R is a ring, then R-valued square matrices of fixed size also give a ring, using addition and multiplication of matrices. Matrices aren't just positive, negative, or zero; they can have a mix of positive and negative entries. In these "matrix rings", the product of negatives isn't exactly positive, although I bet that somebody can make this more rigorous. (Come to think of it, this applies to the rings of polynomials, too.)
A more interesting example: If R is a ring, then R-valued square matrices of fixed size also give a ring, using addition and multiplication of matrices. Matrices aren't just positive, negative, or zero; they can have a mix of positive and negative entries. In these "matrix rings", the product of negatives isn't exactly positive, although I bet that somebody can make this more rigorous. (Come to think of it, this applies to the rings of polynomials, too.)
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.
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.
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.
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.
[deleted]
> 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.
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.
> or tell the reader your assumptions
It is right there in the first section of the article.
"In this discussion, we assume that we already know some basic properties of arithmetic operations such as the distributive property of multiplication over subtraction, existence of the additive inverse of real numbers, etc."
It is right there in the first section of the article.
"In this discussion, we assume that we already know some basic properties of arithmetic operations such as the distributive property of multiplication over subtraction, existence of the additive inverse of real numbers, etc."
Okay, I either didn't see it or that was added after my comment. Either way, that is exactly what I wanted.
> I highly doubt they were there when I wrote the comment.
Hi! I am the author of this blog post. The assumptions in the third paragraph were there at least since 16 Feb 2019. See https://github.com/susam/susam.in/commits/master/content/blo... for the change history of this post.
Hi! I am the author of this blog post. The assumptions in the third paragraph were there at least since 16 Feb 2019. See https://github.com/susam/susam.in/commits/master/content/blo... for the change history of this post.
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...
summary:
-1*-1 =
-1*-1 + -1*1 + 1 =
-1*(-1 + 1) + 1 =
-1*0 + 1 =
1The 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).
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).
This would be clearer if the -1 was in parentheses (I can't do it because I'm on mobile). I think you are missing a plus sign on line two.
[deleted]
in complex plane you can go
e^[i(2nπ+π)] × e^[i(2nπ+π)] =
...
e^[2i(2πn+π)] = 1
e^[i(2nπ+π)] × e^[i(2nπ+π)] =
...
e^[2i(2πn+π)] = 1
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
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
What does this analogy have to do with multiplication?
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.
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.
> 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.
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.
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
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.
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.
He had a dispute with his original camera man that resulted in some of his earlier videos being taken down.
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.
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.
> 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?
Well, sure, if you change the definition of something, then it may end up having different properties. What's your point?
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.
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.
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.
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.
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.
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.
> 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 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.
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 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.
> 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.
Of course, when we say they are field axioms we mean those properties hold true for the elements of the field. If you see those properties, they talk about distributivity over the elements of the field and additive inverses of the elements also belong to the field, so the distributivity automatically applies to additive inverses too.
After that with a little algebra, "product of additive inverses of two elements is equal to the product of the two elements" comes out as a result (not a definition).
Of course, by "product" we mean whatever * represents. It is not necessarily the multiplication operator we see in numbers.
Of course, when we say they are field axioms we mean those properties hold true for the elements of the field. If you see those properties, they talk about distributivity over the elements of the field and additive inverses of the elements also belong to the field, so the distributivity automatically applies to additive inverses too.
After that with a little algebra, "product of additive inverses of two elements is equal to the product of the two elements" comes out as a result (not a definition).
Of course, by "product" we mean whatever * represents. It is not necessarily the multiplication operator we see in numbers.
While this product of additive inverses property does follow from the field axioms, there is a short discussion in the blog comments at https://susam.in/blog/product-of-negatives/comments/ which shows that this property holds true for all rings too.
So it is not just numbers for which this property holds true but for all elements of fields and rings too. Quite simply, (-a)(-b) = (a)(b) in all rings where (-a) and (-b) are the additive inverses of a and b respectively.
So it is not just numbers for which this property holds true but for all elements of fields and rings too. Quite simply, (-a)(-b) = (a)(b) in all rings where (-a) and (-b) are the additive inverses of a and b respectively.
> 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.
This is backward reasoning. The chosen definition of multiplication is not to keep things "intuitive". If you start with the field axioms, the chosen definition of multiplication is pretty much dictated by the axioms. If you choose another definition of multiplication, you would end with contradictions like 1 = 0 and such nonsense! And mathematicians abhor contradictions!
"Product of additive inverses of two elements is equal to the product of the two elements" is dictated by the field axioms in all fields.
This is backward reasoning. The chosen definition of multiplication is not to keep things "intuitive". If you start with the field axioms, the chosen definition of multiplication is pretty much dictated by the axioms. If you choose another definition of multiplication, you would end with contradictions like 1 = 0 and such nonsense! And mathematicians abhor contradictions!
"Product of additive inverses of two elements is equal to the product of the two elements" is dictated by the field axioms in all fields.
> Also, this post conflates the unary negation operator with negative numbers.
-a is a standard way to represent additive inverse of an element in field.
The point about "unary negation operator" seems irrelevant.
In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number.
The additive inverse of 3.14 is -3.14. Unary negation operator applied to 3.14 gives us -3.14. I don't see how conflating unary negation operator with negative numbers here is any issue here.
-a is a standard way to represent additive inverse of an element in field.
The point about "unary negation operator" seems irrelevant.
In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number.
The additive inverse of 3.14 is -3.14. Unary negation operator applied to 3.14 gives us -3.14. I don't see how conflating unary negation operator with negative numbers here is any issue here.
> In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number.
This is a fact that follows from the definition of +. But + needs to be defined before you can start making assumptions about what the additive inverse is. The set over which the field is defined (Z or R) already contains -3, -2 etc. and -3 * -2 or -3 + -2 needs to be defined when you're constructing the field. It then turns out that -3 is the additive inverse of 3. You can't use this when arguing about the definition of why applying * on negative 2 and negative 3 gives you the result positive 6. Because you need to define * over all members of the field before you construct a field in the first place.
This is a fact that follows from the definition of +. But + needs to be defined before you can start making assumptions about what the additive inverse is. The set over which the field is defined (Z or R) already contains -3, -2 etc. and -3 * -2 or -3 + -2 needs to be defined when you're constructing the field. It then turns out that -3 is the additive inverse of 3. You can't use this when arguing about the definition of why applying * on negative 2 and negative 3 gives you the result positive 6. Because you need to define * over all members of the field before you construct a field in the first place.
> The set over which the field is defined (Z or R)
What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is!
> already contains -3, -2 etc.
Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning.
> It then turns out that -3 is the additive inverse of 3.
Are you making this all up with your original research or do you have any proper literature written by a professional mathematician to back it up?
What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is!
> already contains -3, -2 etc.
Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning.
> It then turns out that -3 is the additive inverse of 3.
Are you making this all up with your original research or do you have any proper literature written by a professional mathematician to back it up?
To be fair, you can have a field that only has integers. For example, Z mod 5 is a field.
I am aware. It is typically represented as Z_5. They are called prime fields.
I highly doubt wsxcde meant prime fields in their comment though. wsxcde seemed to be talking about the set of all integers and the set of all real numbers in their comment. Only the latter is a field (and a ring) whereas the former is only a ring.
And (-a)(-b) = ab holds in rings (and thus fields).
I highly doubt wsxcde meant prime fields in their comment though. wsxcde seemed to be talking about the set of all integers and the set of all real numbers in their comment. Only the latter is a field (and a ring) whereas the former is only a ring.
And (-a)(-b) = ab holds in rings (and thus fields).
If both 1 * -1 = -1 and -1 * -1 = -1, then -1 / -1 has two solutions.
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.
Parallel to your statement: If both 1 * 0 = 0 and 2 * 0 = 0, then 0 / 0 has two solutions.
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 ; 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").
If you want an absolutely rigorous proof, you can view this Metamath proof: 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").
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