8 ms·

Product of Negatives (2010)

94 points by polyphonicist 7y ago

saagarjha 7y ago
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 7y ago
  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 7y ago
    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.
yunruse 7y ago
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 7y ago
  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 7y ago
    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 7y ago
      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]
throwfermat 7y ago
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/
- lidHanteyk 7y ago
  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.)
- akalin 7y ago
  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.)
  - 7y ago
kidintech 7y ago
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 7y ago
  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.
  - deleted 7y ago
    [deleted]
  - kidintech 7y ago
    > 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.
- brainscdf 7y ago
  > 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."
  - kidintech 7y ago
edtechdev 7y ago
This site had more intuitive explanations for things like this, including imaginary numbers, calculus, etc https://betterexplained.com/articles/rethinking-arithmetic-a-visual-guide/ https://betterexplained.com/articles/rethinking-arithmetic-a...
tromp 7y ago
summary: -1*-1 = -1*-1 + -1*1 + 1 = -1*(-1 + 1) + 1 = -1*0 + 1 = 1
- deleted 7y ago
  [deleted]
- mathnmusic 7y ago
  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).
- mantap 7y ago
  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.
- nabla9 7y ago
  in complex plane you can go e^[i(2nπ+π)] × e^[i(2nπ+π)] = ... e^[2i(2πn+π)] = 1
soVeryTired 7y ago
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 7y ago
  What does this analogy have to do with multiplication?
  - soVeryTired 7y ago
    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 7y ago
      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 7y ago
        > 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.
b0rsuk 7y ago
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 7y ago
  He had a dispute with his original camera man that resulted in some of his earlier videos being taken down.
wsxcde 7y ago
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 7y ago
  > 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 7y ago
    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.
cousin_it 7y ago
If both 1 * -1 = -1 and -1 * -1 = -1, then -1 / -1 has two solutions.
- throwaway2245 7y ago
  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.
dwheeler 7y ago
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").
dang 7y ago
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 know at hn@ycombinator.com. We catch a lot of abuse between software and moderation, but unfortunately not all. Vigilant users make a huge difference, and protecting the integrity of HN is a community effort. (Please don't post insinuations about abuse in the threads, though, since most suspicions don't end up leading to real evidence. Send them to hn@ycombinator.com. This is in the site guidelines: https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html)
- thaumasiotes 7y ago
  > All: if you notice fishy things like this, please let us know at hn@ycombinator.com What would we have noticed, in this case?
  - dang 7y ago
    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&query=susam&sort=byDate&type=comment https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...
- trevyn 7y ago
  Manipulate HN into talking about math? I don’t have the full picture, but the discussion this generated is better than many posts.
  - susam 7y ago
    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.

94 points by polyphonicist 7y ago

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 7y ago
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 7y ago
  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.

yunruse 7y ago

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 7y ago
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 7y ago
  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 7y ago
    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]

throwfermat 7y ago

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/

lidHanteyk 7y ago
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.)
akalin 7y ago
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.)
- 7y ago

kidintech 7y ago

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 7y ago
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.
- deleted 7y ago
  [deleted]
- kidintech 7y ago
  > 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.
brainscdf 7y ago
> 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."
- kidintech 7y ago

edtechdev 7y ago

This site had more intuitive explanations for things like this, including imaginary numbers, calculus, etc https://betterexplained.com/articles/rethinking-arithmetic-a-visual-guide/ https://betterexplained.com/articles/rethinking-arithmetic-a...

tromp 7y ago

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

deleted 7y ago
[deleted]
mathnmusic 7y ago
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).
mantap 7y ago
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.
nabla9 7y ago
in complex plane you can go e^[i(2nπ+π)] × e^[i(2nπ+π)] = ... e^[2i(2πn+π)] = 1

soVeryTired 7y ago

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 7y ago
What does this analogy have to do with multiplication?
- soVeryTired 7y ago
  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 7y ago
    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 7y ago
      > 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.

b0rsuk 7y ago

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 7y ago
He had a dispute with his original camera man that resulted in some of his earlier videos being taken down.

wsxcde 7y ago

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 7y ago
> 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 7y ago
  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.

cousin_it 7y ago

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

throwaway2245 7y ago
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.

dwheeler 7y ago

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").

dang 7y ago

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 know at hn@ycombinator.com. We catch a lot of abuse between software and moderation, but unfortunately not all. Vigilant users make a huge difference, and protecting the integrity of HN is a community effort. (Please don't post insinuations about abuse in the threads, though, since most suspicions don't end up leading to real evidence. Send them to hn@ycombinator.com. This is in the site guidelines: https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html)

thaumasiotes 7y ago
> All: if you notice fishy things like this, please let us know at hn@ycombinator.com What would we have noticed, in this case?
- dang 7y ago
  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&query=susam&sort=byDate&type=comment https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...
trevyn 7y ago
Manipulate HN into talking about math? I don’t have the full picture, but the discussion this generated is better than many posts.
- susam 7y ago
  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.