Hey! Sign in to get help with your study questionsNew here? Join for free to post
 You are Here: Home >< Maths

# why does negative*negative=positive?

Announcements Posted on
Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter 24-10-2016

1. weeping
2. Let's stop with the rest of this nonsense. No complex numbers, no vectors, no calls to analogous intuitions in other subjects or explanations using debts.

This is a simple and necessary consequence of the rules (axioms) we take for granted in arithmetic:

Put these two together:

Expand the brackets:

Add 1 to both sides (Written as 1 times 1, for reasons you'll soon see):

Factorise the first two terms

A.K.A:

If I hear one more person suggest that this is an axiom or a rule that mathematicians decided then there will be a rampage.

And yes, this is the same thing as @notnek was suggesting, just written out in painstaking detail. Yes you can replace some terms with and if you want to make it more general but there's no point doing that if you're trying to explain this in the first place and it follows incredibly easily from this anyway.
3. (Original post by ValerieKR)
Any complex number can be written in the form re^i(theta), for negative numbers theta = (2n+1)pi (r is the (positive) magnitude of the negative number)
When you multiply two numbers:
re^i(2n+1)pi*se^i(2m+1)pi = rse^i(2n+2m +2)pi = rse^i(2k)pi
e^ik(2pi) = 1
therefore = rs*1 = rs
no negative sign any more
Too complex.
(Original post by p29)
yh
OK. Let's word this:

Sanjay is going to roll a special dice.
The dice has six sides, numbered: 1, 2, 3, -1, -2, -3.

If the number 1 is rolled, then that is good, and will have happened once. Giving Sanjay a good/positive score. 1*1=1

If the number -1 is rolled, then that is bad, and will have happened once. Giving Sanjay a bad/negative score. -1*1=-1

If the number 1, 2, or 3 is rolled, then the number -1 has not been rolled, which is good. If 1*x happens when x is rolled. Then -1*y happens when y is not rolled. So if -1 is not rolled, -1*-1=1 because not rolling -1 is a good thing.

Make sense?
Spoiler:
Show
EDIT: Before you all criticise me for not using binary numbers representing events occurring correctly, I know. I was trying to explain it with a simple problem. I don't think OP will eventually get to the stage of binary numbers representing events so relax.
4. (Original post by 04MR17)
Too complex.OK. Let's word this:

Sanjay is going to roll a special dice.
The dice has six sides, numbered: 1, 2, 3, -1, -2, -3.

If the number 1 is rolled, then that is good, and will have happened once. Giving Sanjay a good/positive score. 1*1=1

If the number -1 is rolled, then that is bad, and will have happened once. Giving Sanjay a bad/negative score. -1*1=-1

If the number 1, 2, or 3 is rolled, then the number -1 has not been rolled, which is good.

Make sense?
Spoiler:
Show
EDIT: Before you all criticise me for not using binary numbers representing events occurring correctly, I know. I was trying to explain it with a simple problem. I don't think OP will eventually get to the stage of binary numbers representing events so relax.
"
If 1*x happens when x is rolled. Then -1*y happens when y is not rolled. So if -1 is not rolled, -1*-1=1 because not rolling -1 is a good thing. "
this part confused me
5. 2 negatives make a positive yet 2 wrongs don't make a right...
6. (Original post by Lissy14)
2 negatives make a positive yet 2 wrongs don't make a right...
mind boggling....isn't it?
7. (Original post by ValerieKR)
Translating A and B one unit up changes the 'area' the way you've calculated it,
You beat me to it. I was coming back to make that very comment myself.

[edit: in fact, on second thoughts, translating up does *not* change the area since I've defined the sides in terms of signed displacements. The -2 vector still points -ve however we translate it - but the argument needs more work for sure, to make sense]

so you can't claim their congruency means they have the same 'area' without already understanding that neg*neg=pos so that horizontal shifts won't change it the way you've calculated it.
I'm not quite sure what you're getting at here, but yes, the "(signed) area invariant under translation" justification breaks too easily. Needs some more thought.
8. (Original post by RichE)
Well amongst other things you selectively use congruence in your argument to make a case for C and D having the same overall area as A and B. But D is congruent to A and yet has different "area" in your argument.
That's true - the congruency argument breaks quite easily (though I think that I can handwave that specific problem away with a little effort.)

In fact, the OP's problem is one that I mull over occasionally (i.e. how can you give a nice intuitive argument that -1 x -1 = 1 that satisfies even a raw beginner, and holds up well mathematically) and I've being trying to come up with one based on area - I thought this looked pretty good this morning but it's back to the drawing board for me.

BTW, another justification that I've seen is a "continue the pattern" approach:

3 x -2 = -6
2 x -2 = -4
1 x -2 = -2
0 x -2 = 0
-1 x -2 = ?

noting that as we decrease the 1st column by 1 each time, we increase the last column by 2
9. (Original post by p29)
lel.
It is a rule in maths.
If you want the story behind it, you have to travel through time.
10. (Original post by atsruser)
BTW, another justification that I've seen is a "continue the pattern" approach:

3 x -2 = -6
2 x -2 = -4
1 x -2 = -2
0 x -2 = 0
-1 x -2 = ?

noting that as we decrease the 1st column by 1 each time, we increase the last column by 2
This is an approach I've used in the past when introducing this topic.

Interestingly, the only child who ever asked, "why should the pattern continue" turned out to be one of my best ever students
11. (Original post by DJMayes)
If I hear one more person suggest that this is an axiom or a rule that mathematicians decided then there will be a rampage.
That's a bit of an overreaction, since it is not at all obvious at first that this can be deduced from the other axioms (most notably the axiom of distributivity applying to negative numbers) and isn't itself an axiom.
12. Here's what I remember from Primary:

friend= + enemy= -

A friend of a friend is a friend- + * + = + (positive x positive = positive)
A friend of an enemy is an enemy- + * - = - (positive x negative = negative)
An enemy of an enemy is a friend- - * - = + (negative x negative = positive)
13. (Original post by LDS16)
Here's what I remember from Primary:

friend= + enemy= -

A friend of a friend is a friend- + * + = + (positive x positive = positive)
A friend of an enemy is an enemy- + * - = - (positive x negative = negative)
An enemy of an enemy is a friend- - * - = + (negative x negative = positive)
Ha, that's kinda cool actually.
14. (Original post by IrrationalRoot)
Most of the posts here provide intuition as a reason, which isn't very precise.
The intuition is what made mathematicians decide that . They define it like that; it's axiomatic.
Do you not believe that an intuitive understanding is desirable?
15. (Original post by AlexLawrence1453)
Do you not believe that an intuitive understanding is desirable?
Never said that; I said people are giving intuition as the explanation why (-1) x (-1) = 0 as if it's some part of nature. I thought it was an axiom but apparently it can be deduced from other axioms (including the distributivity of negative numbers, which is just as hard, if not more, to explain...), so intuition is definitely not a precise reason why (-1) x (-1) = 1.
16. (Original post by p29)
im learning so much
ha ha ha ha ha ha !!!!
17. (Original post by ValerieKR)
Any complex number can be written in the form re^i(theta), for negative numbers theta = (2n+1)pi (r is the (positive) magnitude of the negative number)
When you multiply two numbers:
re^i(2n+1)pi*se^i(2m+1)pi = rse^i(2n+2m +2)pi = rse^i(2k)pi
e^ik(2pi) = 1
therefore = rs*1 = rs
no negative sign any more
use latex
18. (Original post by IrrationalRoot)
That's a bit of an overreaction, since it is not at all obvious at first that this can be deduced from the other axioms (most notably the axiom of distributivity applying to negative numbers) and isn't itself an axiom.
Either an over-reaction or a joke, but every time this question comes up (which is not uncommon on this forum alone let alone elsewhere) somebody gives a response to the effect of:

" and not because mathematicians chose it to be"

Your earlier response of:

(Original post by IrrationalRoot)
Most of the posts here provide intuition as a reason, which isn't very precise.
The intuition is what made mathematicians decide that . They define it like that; it's axiomatic.
Is indeed what prompted me to say that. It's a common response that is incredibly frustrating, because:

1) It is wrong
2) It is completely unhelpful to whomever asks the question

The correct response to this entire question, that's unfortunately probably a bit deeper than the average person asking it wants, is that when you extend the naturals to the integers, this is a necessary condition to retain the same "nice" structure that you have for addition and multiplication of positive numbers, which are the real axioms.

This deduction isn't really that difficult in my opinion and there are substantially better mathematicians than myself out there, which is why I get annoyed whenever I see it suggested that mathematicians just decided this would be so. I will give it to you that it's not obvious enough at the level this question is asked though.
19. (Original post by Copperknickers)
Maths is a description of a real phenomenon, or rather a range of phenomena which are defined by the laws of physics.
Noooooo! Please do not demote mathematics to the level of (lower in this case) physics. Maths has nothing to do with the laws of physics; the laws of physics use maths, but not a single part of maths depends on the laws of physics. Not a single law of physics is even proven to be true.
20. (Original post by DJMayes)
...
Yeah I get that it's frustrating but it was a misunderstanding on my part (I did look it up before posting it just to make sure; apparently I found the wrong sources/misinterpreted them).

On a separate note, I guess the follow-up question OP might ask is why distributivity (of both positive and negative numbers) is true, in which case my answer would apply. So I think they should understand that ok, this specific fact can be deduced from a couple of axioms but those axioms are taken to be true without proof.

Write a reply…

Submit reply

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
that username has been taken, please choose another Forgotten your password?
2. this can't be left blank
this email is already registered. Forgotten your password?
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
your full birthday is required
1. Oops, you need to agree to our Ts&Cs to register
2. Slide to join now Processing…

Updated: August 30, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.