Someone recently asked me why a negative $\times$ a negative is positive, and why a negative $\times$ a positive is negative, etc.

I went ahead and gave them a proof by contradiction like so:

Assume $(-x) \cdot (-y) = -xy$

Then divide both sides by $(-x)$ and you get $(-y) = y$

Since we have a contradiction, then our first assumption must be incorrect.

I'm guessing I did something wrong here. Since the conclusion of $(-x) \cdot (-y) = (xy)$ is hard to derive from what I wrote.

Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?

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    $\begingroup$ You haven't proven that -xy = (-x)y. $\endgroup$ – Qiaochu Yuan Nov 12 '10 at 2:24
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    $\begingroup$ It is not an uncommon question, and it's never not easy to show. $\endgroup$ – J. M. is a poor mathematician Nov 12 '10 at 4:06
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    $\begingroup$ J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive. $\endgroup$ – Jonas Kibelbek Nov 12 '10 at 4:48
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    $\begingroup$ Does he understand why a negative times a positive is negative? And if so, how? $\endgroup$ – Paul Nov 8 '15 at 21:00
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    $\begingroup$ Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...? $\endgroup$ – Paul Nov 8 '15 at 21:03

40 Answers 40


Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.

Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.

Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.

With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.

Finally you'll have to try to picture what could $(-2)*(-3)$ mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.

I don't know if this will help but it's the only way I can think of this in some intuitive sense.

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    $\begingroup$ I can walk backwards. $\endgroup$ – user66081 Aug 7 '19 at 15:57

This is pretty soft, but I saw an analogy online to explain this once.

If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true. $$\text{backward} \times \text{backward} = \text{forward}$$ $$ \text{negative} \times \text{negative} = \text{positive}$$ It's not perfect, but it introduces the notion of the number line having directions at least.

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    $\begingroup$ This is a good answer, because the rate at which the individual is running in the video multiplied by the rate of the playback equals the apparent rate at which they run in the playback. That little piece of nuance is probably a bit tricky to explain to an 8-year-old though. $\endgroup$ – Dustan Levenstein Nov 9 '15 at 3:43
  • $\begingroup$ @MatthewGraves The video is gone - do you have another link? $\endgroup$ – max_zorn Feb 4 '18 at 1:21
  • $\begingroup$ @max_zorn youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse" $\endgroup$ – Matthew Graves Feb 12 '18 at 22:19
  • $\begingroup$ youtube.com/watch?v=N0EN6URYc0w and youtube.com/watch?v=8xTTDlVAv4c currently work; a thing to search for if those links fail is "people running backwards in reverse" $\endgroup$ – Matthew Graves Feb 12 '18 at 22:19

Informal justification of $\text{positive} \times \text{negative} = \text{negative}$

Continue the pattern:

$$ \begin{array}{r} 2 & \times & 3 & = & 6\\ 2 & \times & 2 & = & 4\\ 2 & \times & 1 & = & 2\\ 2 & \times & 0 & = & 0\\ 2 & \times & -1 & = & ? & (\text{Answer} = -2 )\\ 2 & \times & -2 & = & ? & (\text{Answer} = -4 )\\ 2 & \times & -3 & = & ? & (\text{Answer} = -6 )\\ \end{array} $$

The number on the right-hand side keeps decreasing by 2.

Informal justification of $\text{negative} \times \text{negative} = \text{positive}$

Continue the pattern:

$$ \begin{array}{r} 2 & \times & -3 & = & -6\\ 1 & \times & -3 & = & -3\\ 0 & \times & -3 & = & 0\\ -1 & \times & -3 & = & ? & (\text{Answer} = 3 )\\ -2 & \times & -3 & = & ? & (\text{Answer} = 6 )\\ -3 & \times & -3 & = & ? & (\text{Answer} = 9 )\\ \end{array} $$

The number on the right-hand side keeps increasing by 3.

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    $\begingroup$ Simple and just what I needed. Thank you. $\endgroup$ – John H Feb 18 '12 at 21:07
  • $\begingroup$ And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition. $\endgroup$ – Alfe Nov 12 '13 at 11:28
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    $\begingroup$ @Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-) $\endgroup$ – quapka Nov 20 '14 at 0:25
  • $\begingroup$ I find it nice to extend this to all the reals by drawing a graph. $\endgroup$ – Michael Anderson Jun 12 '15 at 7:41
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    $\begingroup$ Just to play devil's advocate, there is no reason why one should expect the "pattern" to always increase or decrease - unless one uses assumptions on the distributive properties of multiplication and similar. That in turn would be equivalent to proving the statement in the first place (therefore I find this explanation actually harder to understand). $\endgroup$ – gented Feb 27 '19 at 22:44

Someone sent this to me recently:

I give you three \$20 notes: +3 * +20 = +\$60 for you
I give you three \$20 debts: +3 * -20 = -\$60 for you
I take three \$20 notes from you: -3 * +20 = -\$60 for you
I take three \$20 debts from you: -3 * -20 = +\$60 for you


I think a lot of answers are either too simple or stray away from mathematics too much. Just remember that multiplication is repeated addition. When dealing with negative numbers, it becomes repeated subtraction.

I'd simply put it in this context:

  1. The equation: $$\begin{equation*}\begin{array}{c} \phantom{\times9}2\\ \underline{\times\phantom{9}2}\\ \phantom{\times9}4\\ \end{array}\end{equation*}$$ is just adding postive $2$, two times.

  2. The equation: $$\begin{equation*}\begin{array}{c} \phantom{\times999}2\\ \underline{\times\phantom{1}-2}\\ \phantom{\times9}-4\\ \end{array}\end{equation*}$$ is just adding positive $2$, negative two times, which means instead of adding in the positive direction, you add in the negative direction (subtraction).

    You could also just say that you're adding $-2$ (or subtracting $2$), two times.

  3. The equation: $$\begin{equation*}\begin{array}{c} \phantom{\times9}-2\\ \underline{\times\phantom{1}-2}\\ \phantom{\times999}4\\ \end{array}\end{equation*}$$ is adding $-2$, negative two times. Adding $-2$, two times, yields the diagram in (2). Since you have to add negative two times, you reverse the direction you are adding in.

    You could also say that you are subtracting $-2$, two times.

  • $\begingroup$ My memory of how I learned this is somewhat uncertain, because I was only 8 or 9 years old at the time, but I think this is very similar to how Isaac Asimov explained multiplication of negative numbers in either Realm of Numbers or Realm of Algebra. His explanation made sense to me at the time and I don't recall ever being confused as to why the product of negatives is positive. But I think it was a few years before this fact came up in the school curriculum. $\endgroup$ – David K Nov 9 '15 at 23:30
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    $\begingroup$ @homersimpson - I was thinking of this kind of logic but what knocks me out is how can we assume minus minus 2 is positive 2 - aren't we supposed to prove this? $\endgroup$ – KGhatak Jan 11 '16 at 8:00

$\overbrace{\bf\ Law\ of\ Signs}^{\rm\Large {(-x)(-y)}\ =\ xy} $ proof: $\rm\,\ (-x)(-y) = (-x)(-y) + \color{#c00}x(\overbrace{\color{#c00}{-y} + y}^{\Large =\,0}) = (\overbrace{-x+\color{#c00}x}^{\Large =\,0})(\color{#c00}{-y}) + xy = xy$

Equivalently, $ $ evaluate $\rm\,\ \overline{(-x)(-y)\ +\ } \overline{ \underline {\color{#c00}{x(-y)}}}\underline{\phantom{(}\! +\,xy}\, $ in $\,2\,$ ways, noting over/underlined terms $ = 0$

Said more conceptually $\rm (-x)(-y)\ $ and $\rm\:xy\:$ are both inverses of $\rm\ \color{#c00}{x(-y)}\ $ so they are equal by uniqueness of inverses: $ $ if $\,\color{#c00}a\,$ has two additive inverses $\,{-a}\,$ and $\,\color{#0a0}{-a},\,$ then

$$\rm {-a}\, =\, {-a}+\overbrace{(\color{#c00}a+\color{#0a0}{-a})}^{\large =\,0}\, =\, \overbrace{({-a}+\color{#c00}a)}^{\large =\,0}+\color{#0a0}{-a}\, =\, \color{#0a0}{-a} $$

Equivalently, $ $ evaluate $\rm\,\ \overline{-a\ \ +\!} \overline{\phantom{+} \underline {\color{#c00}{a}}}\underline{\ + \color{#0a0}{-a}}\, $ in $\,2\,$ ways, noting over/underlined terms $ = 0$

This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.

These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.

Remark $\ $ More generally the Law of Signs holds for any odd functions under composition, e.g. polynomials with all terms having odd power. Indeed we have

$\qquad\rm\ \ f(g)\ =\ (-f)\ (-g)\ = -(f(-g))\ \iff\ f(-g)\ = -(f(g)) $

$\qquad\rm \phantom{\ \ (-f)\ (-g)\ = -(f(-g))\ =\ f(g)\ }\!\!\overset{\ \large g(x)=x}\iff \ f(-x)\ = -f(x),\ $ ie. $\rm\:f\:$ is odd

Generally such functions enjoy only a weaker near-ring structure. In the above case of rings, distributivity implies that multiplication is linear hence odd (viewing the ring in Cayley-style as the ring of endormorphisms of its abelian additive group, i.e. representing each ring element $\rm\ r\ $ by the linear map $\rm\ x \to r\ x,\ $ i.e. as a $1$-dim matrix).

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    $\begingroup$ +1. The simplest way to prove why is by using the group axioms and thier consequences. $\endgroup$ – Mathemagician1234 Jun 13 '15 at 3:18
  • $\begingroup$ What's the logical thinking skills or tactic used to jump in one step from xy or (−x)(−y) to xy+x (−y)+(−x)(−y)? $\endgroup$ – Pacerier Feb 2 '18 at 7:15
  • $\begingroup$ @Pacerier It's a common way to prove uniqueness of inverses. I edited the answer to elaborate on that. $\endgroup$ – Bill Dubuque Oct 16 '18 at 17:48

Simple Answer: $$ (-a)b + ab = (-a)b + ab $$ $$(-a)b + ab = b(a-a) $$ $$(-a)b + ab = b(0) $$ $$(-a)b + ab = 0 $$ $$(-a)b = -ab $$ $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$ $$(-a)(-b) + (-ab) = (-a)(b-b) $$ $$(-a)(-b) + (-ab) = (-a)(0) $$ $$(-a)(-b) + (-ab) = 0 $$ $$*(-a)(-b) = ab $$ Hope this helps (Credit to Michael Spivak's Calculus)

~ Alan

  • $\begingroup$ Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple. $\endgroup$ – Mathemagician1234 Nov 20 '14 at 0:30
  • $\begingroup$ Easily the best and most satisfying answer here. $\endgroup$ – Derek 朕會功夫 Aug 17 '15 at 20:42

This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me” or “I was never good at it.” Usually the process is gradual, but for Ruth McNeill, the turning point was clearly defined. In an article in the Journal of Mathematical Behavior, she described how it happened:

What did me in was the idea that a negative number times a negative number comes out to a positive number. This seemed (and still seems) inherently unlikely — counter intuitive, as mathematicians say. I wrestled with the idea for what I imagine to be several weeks, trying to get a sensible explanation from my teacher, my classmates, my parents, any- body. Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something, and thus two negative numbers multiplied together should properly produce a very negative result. I have since been offered a moderately convincing explanation that features a film of a swimming pool being drained that gets run back- wards through the projector. At the time, however, nothing convinced me. The most commonsense of all school subjects had abandoned common sense; I was indignant and baffled.

Meanwhile, the curriculum kept rolling on, and I could see that I couldn’t stay behind, stuck on nega- tive times negative. I would have to pay attention to the next topic, and the only practical course open to me was to pretend to agree that negative times nega- tive equals positive. The book and the teacher and the general consensus of the algebra survivors of so- ciety were clearly more powerful than I was. I capitu- lated. I did the rest of algebra, and geometry, and trigonometry; I did them in the advanced sections, and I often had that nice sense of “aha!” when I could suddenly see how a proof was going to come out. Underneath, however, a kind of resentment and betrayal lurked, and I was not surprised or dismayed by any further foolishness my math teachers had up their sleeves.... Intellectually, I was disengaged, and when math was no longer required, I took German instead.

I will show in this answer that: negative $\times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.

Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.

Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.

enter image description here

Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$\rm Velocity=\dfrac{\rm Distance}{\rm Time}.$$ Solving for distance we get: $$\rm Velocity\times{\rm Time}={\rm Distance}.$$ Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$\rm positive\times positive=positive.$$

If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$\rm negative\times positive=negative.$$

If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$\rm positive\times negative=negative.$$

If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\rm\, negative\times negative=positive.\,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.

I hope this helps.
Best wishes, $\mathcal H$akim.

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    $\begingroup$ The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"? $\endgroup$ – Andreas Blass Jun 29 '14 at 21:43
  • $\begingroup$ @AndreasBlass Or in another form: "negative multiplication is reverse intensification". $\endgroup$ – Hakim Feb 22 '15 at 19:56
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    $\begingroup$ @Hakim, Doesn't Velocity×Time equal displacement instead of distance? $\endgroup$ – Pacerier Feb 2 '18 at 7:39

The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 \cdot (-40) = -120$ is added to your account.

The opposite of being billed would be billing someone else.

So, if you bill 3 people for 40 dollars each, $(-3) \cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.


Here's a proof. First, for all $x$, $x\cdot 0=x\cdot(0+0)=x\cdot 0 +x\cdot 0$. Subtracting $x\cdot0$ from each side, $x\cdot0=0$. Now, for all $x$ and $y$, $0=x\cdot0=x\cdot(-y+y)=x\cdot(-y)+x\cdot y$. Subtracting $x\cdot y$ from both sides, $x\cdot(-y)=-(x\cdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)\cdot(-y)=-(-x)\cdot y=-(-(x\cdot y))=x\cdot y$.

Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(x\cdot y)$ rather than simply $-xy$ to not be confused with $(-x)\cdot y$).

I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.


I have always viewed negative numbers as a "flip" on the number line.

For example, -2 is the same as 2, but flip-mirrored to the other side of the zero.

Multiplication then works as follows:

2 x 3 has no flips, so it's just 2x3 = 6.

-2 x 3 has one flip, so you start with 2x3 = 6 but with one flip so it becomes -6 instead.

2 x -3 is the same, only one flip, so it's 2x3 = 6 but flipped to -6.

-2 x -3 has two flips, so you start with 2x3 = 6. When you apply the two flips it gets you back to where you started because you flip to negative and then flip back. So -3 x -2 = 6.


I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!

Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.

Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.

So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.

One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!


Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones: If $x>a\ge0$ and $y>b\ge0$ then $$ (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b) $$ For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.


One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.

For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.

Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.

Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:

It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...

With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.

The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:

Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?

Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.

One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac\ ,\ a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.

But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.

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    $\begingroup$ This $(-a)(-b)=ab$ stuff started when we gave up on distinguishing multiplier and multiplied in a multiplication. I can have three unities of apples, but I can't have apple unities of threes. In this sense, it's possible to have 3 unities of an 1D vector backwards $3 \times -1$ but not -1 unities of a vector forwards $-1 \times 3$, unless we define that a negative multiplier change the sense of the vector. I think a negative-free algebra can be created by distinguish three kind of numbers: scalar (no sign), up and down. $\endgroup$ – Leonardo Castro Sep 9 '16 at 12:22
  • $\begingroup$ If you add an up and a down, you cancel out the amount they have in common keep the remaining up or down part. This would the subtraction of this algebra. Multiplying by $-1$ should be replaced by a simple inversion operator. Actually, $+1$ and $-1$ as multipliers are usually taken as the identity and inversion operators. In that way, they form a simple identity-inversion group and so $-1 \times -1 = 1$ since two inversions are equivalent to identity. I don't think, however, that algebra should necessarily be that way. $\endgroup$ – Leonardo Castro Sep 9 '16 at 12:33
  • $\begingroup$ I used up and down just to avoid all the assumptions about positive and negative, but they could be right and left, yin and yang, whatever... I wonder if yin times yin is yang or the other way around. $\endgroup$ – Leonardo Castro Sep 9 '16 at 12:35

Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $\pi$ radians. When you multiply by two negatives, you rotate by $2\pi$. :-)


Why don't we tell a story! The dastardly Dalton gang is on the loose but Al Catchem is hot on their trail and nearly catches them at their latest heist. When pulling out of the parking lot, there are a few possibilities:

  • Al is quick, up to the task and catches the Daltons,
  • Al's car stalls and the Daltons get away,
  • The Dalton's are tipped off early and escape, or
  • The Dalton's car stalls, they are captured and the city is saved.

In summary:

  • positive $\times$ positive: If a good thing happens to a good person, that's good! :)
  • negative $\times$ positive: If a bad thing happends to a good person, that's bad. :(
  • positive $\times$ negative: If a good thing happens to a bad person, that's also bad. :(
  • negative $\times$ negative: But, if a bad thing happens to a bad person, that's good!! :)
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    $\begingroup$ While I can understand the sentiment, I'm not sure this is the kind of morality we should be teaching 8-year-old kids. $\endgroup$ – MichaelS Nov 9 '15 at 3:32
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    $\begingroup$ Downvoted because this isn't really a mathematical explanation. $\endgroup$ – Daniel R. Collins Nov 9 '15 at 3:39
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    $\begingroup$ @DanielR.Collins That is funny - the question is how to explain the idea to an 8 year old. :) $\endgroup$ – Mark McClure Nov 9 '15 at 3:40
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    $\begingroup$ @MichaelS You are, of course, correct. When I've actually done this, which I have many times, I've tried to tell a story with it that (hopefully) makes the morality reasonably clear. Turns out, you can have great fun with it! $\endgroup$ – Mark McClure Nov 9 '15 at 4:10
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    $\begingroup$ Why is this immoral? From a naturalistic philosophical viewpoint, this is arguably the most moral, since babies have been shown to prefer "good" characters over "evil" ones. $\endgroup$ – March Ho Nov 9 '15 at 13:48

I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.

This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.

from the short story The Arnold Family

I faced a real difficulty with school mathematics several years after the multiplication table: it was necessary to leam that “minus multiplied by minus is plus” I wanted to know the proof of this rule; I have never been able to leam by heart what is not properly understood. I asked my father to explain the reason why (—1) • (—2) = (+2). He, being a student of great algebraists, S. O. Shatunovsky and E. Noether, gave the following “scientific explanation”: “The point is,” he said: “that numbers form a field such that the distributive law (x+y)z=xz+yz holds. And if the product of minus by minus had not been plus, this law would be broken”.

However, for me this “deductive” (actually juridical) explanation did not prove anything - what of it! One can study any axioms! Since that day I have preserved the healthy aversion of a naturalist to the axiomatic method with its non-motivated axioms.

The axiomophile Rene Descartes stated that “neither experimental tests that axioms reflect a reality, nor comparison of theoretical results with reality should be a part of science” (why should results correspond to reality if the initial principles do not correspond to it?).

Another thesis of Descartes’ theory and methods of education is even more peculiar and contemporary: “It is necessary to forbid all other methods of teaching except mine because only this method is politically correct: with my purely deductive method any dull student can be taught as successfully as the most gifted one, while with other methods imagination and even drawings are used unavoidably, and for this reason geniuses advance faster than dunces”.

Contrary to the deductive theories of my father and Descartes, as a ten year old, I started thinking about a naturally-scientific sense of the rule of signs, and I have come to the following conclusion. A real (positive or negative) number is a vector on the axis of coordinates (if a number is positive the corresponding vector is positively directed along this axis).

A product of two numbers is an area of a rectangle whose sides correspond to these numbers (one vector is along one axis and the other is along a perpendicular axis in the plane). A rectangle, given by an ordered pair of vectors, possesses, as a part of the plane, a definite orientation (rotation from one vector to another can be clockwise or anti-clockwise). The anti-clockwise rotation is customarily considered positive and the clockwise rotation is then negative. And lastly, the area of a parallelogram (for example, a rectangle) generated by the two vectors x and у (taken in a definite order) emanating from one point in the plane is considered to be positive if the pair of vectors (taken in this order) defines positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is negative).

Thus, the rule of signs is not an axiom taken out of the blue, but becomes a natural property of orientation which is easily verified experimentally.

from the short story Axiomatic Method

My first trouble in school was caused by the rule for multiplication of negative numbers, and I asked my father to explain this peculiar rule.

He, as a faithful student of Emmy Noether (and consequently of Hilbert and Dedekind) started explaining to his eleven-year-old son the principles of axiomatic science: a definition is chosen such that the distributive identity a(b+c)=ab+ac holds.

The axiomatic method requires that one should accept any axiom with a hope that its corollaries would be fruitful (probably this could be understood by the age of thirty when it would be possible to read and appreciate “Anna Karenina”). My father did not say a word either about the oriented area of a rectangular or about any non-mathematicai interpretation of signs and products.

This “algebraic” explanation was not able to shake either my hearty love for my father or a deep respect of his science. But since that time I have disliked the axiomatic method with its non-motivated definitions. Probably it was for this reason that by this time I got used to talking with non-algebraists (like L. I. Mandel’shtam, I. E. Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G. Gurvich) who treated an ignorant interlocutor with full respect and tried to explain non-trivial ideas and facts of various sciences such as physics and biology, astronomy and radiolocation.

Negative numbers I came to understand a year later while deriving an “equation of time”, which takes into account a correction for the length of a day corresponding to the time of year. It is not possible to explain to algebraists that their axiomatic method is mostly useless for students.

One should ask children: at what time will high tide be tomorrow if today it is at 3 pm? This is a feasible problem, and it helps children to understand negative numbers better than algebraic rules do. Once I read from an ancient author (probably from Herodotus) that the tides "always occur three and nine o'clock". To understand that the monthly rotation of the Moon about the Earth affects the tide timetable, there is no need to live near an ocean. Here, not in axioms, is laid true mathematics.

  • $\begingroup$ 2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs? $\endgroup$ – Mathemagician1234 Jun 13 '15 at 3:21
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    $\begingroup$ @Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject. $\endgroup$ – level1807 Jun 13 '15 at 7:36
  • $\begingroup$ With the interpretation in terms of oriented areas we would have $xy=-yx$ which is kind of problematic. $\endgroup$ – Sergio Feb 2 '18 at 22:55
  • $\begingroup$ @Sergio it wouldn't. In $ab$, $a$ is always measured along the x-axis and $b$ along the y-axis. When you then write $ba$, $b$ is measured along the x-axis. So the vectors themselves change when you commute the numbers. For instance, the orientation of $1*(-1)$ is the same as $(-1)*1$ because both require a clockwise rotation from the first vector to the second: $(0,1)\mapsto(0,-1)$ for the former and $(-1,0)\mapsto(0,1)$ for the latter. $\endgroup$ – level1807 Mar 12 at 18:56

Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.


As for the product of two negatives being a positive, simply consider the multiplicative inverse:

$$-a\cdot -b$$ $$(-1)a\cdot (-1)b$$ $$(-1)(-1)a\cdot b$$

Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.

$$(-1)(-1)^{-1}a\cdot b$$ $$(1)a\cdot b$$ $$=a\cdot b$$

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    $\begingroup$ Great answer! Can you explain how you came up with this? I mean it's pretty self explanatory that a number multiplied by it's multiplicative inverse is 1 but using it here is pretty clever. How did you think to use it here? $\endgroup$ – user500668 Jul 28 '18 at 13:27
  • $\begingroup$ @user500668 Good question. I used to be clever once upon a time. I believe at the time I was just making use of multiplicative inverses while solving "contest style" math problems. $\endgroup$ – daOnlyBG Oct 22 '18 at 14:53
  • $\begingroup$ For what it's worth, I wish whoever gave me a "-1" score would have the decency to justify their action. $\endgroup$ – daOnlyBG Oct 22 '18 at 14:54

A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.

See Lemma 7 starting at line 224 in http://www.dcproof.com/FieldLemmas.htm


I would explain it by number patterns.

First, to establish that a positive times a negative is negative: $3 \times 2 = 6, 3 \times 1 = 3, 3 \times 0 = 0$. Notice in each case, as we reduce the second factor by 1, the product is being reduced by 3. So for consistency the next product in the pattern must be $0 - 3 = -3$. Therefore we have $3 \times (-1) = -3, 3 \times (-2) = -6$, and likewise a negative for any other other positive times a negative.

Second, to establish that a negative times a negative is positive: we now know that $3 \times (-2) = -6, 2 \times (-2) = -4, 1 \times (-2) = -2, 0 \times (-2) = 0$. Notice in each case, as we reduce the first multiplier by 1, the product is being increased by 2. So for consistency the next product in the pattern must be $0 + 2 = 2$. Therefore we have $(-1) \times (-2) = 2, (-2) \times (-2) = 4$, and likewise a positive for any other other negative times a negative.

  • $\begingroup$ I tried explaining this to him and he looked even more confused than when I narrated the other examples, I just think this is a bit too complicated for someone like him to understand yet. $\endgroup$ – yuritsuki Nov 9 '15 at 7:20
  • $\begingroup$ Funny - this is the second argument that I typically present to elementary teachers and their students after my cops and robbers story! Note that it has been presented on this site before as well. $\endgroup$ – Mark McClure Nov 9 '15 at 15:16

I would go for the flipping explanation of the negative numbers: multiplying with a negative number flips from positive to negative and from negative to positive.
Imagine he understands that multiplying with 1 makes no difference, then it's very simple:
-1 * 1 = -1 can mean two things (for children, the fact that multiplication is commutative is obvious):
- either it keeps -1 as -1
- either it flips +1 (positive number) to the negative side

-1 * (-1) then simply flips it back from the negative to the positive side.

Good luck


It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.

When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.


If your son is clear on the concept of money and knows what a credit card is, this might be a good explanation:

Imagine you tell your son that you will buy him $\color{green}{\mathrm{seven}}$ gift vouchers worth £$\color{green}{5}$ each and pay for them using your credit card. Explain to your son that you now owe money, and say that it is $7 \times -5=-\mathrm{£}35$. Not part of your question; but this covers the $ \text{positive} \times \text{negative} = \text{negative}$ case.

You are having dinner with your best friend when the bill arrives from the credit card company, when your friend sees the bill he generously insists on paying it for you. You now have £$\color{blue}{35}$ worth of gift vouchers without having paid anything.

So you tell your son that your best friend $\color{red}{\fbox{took-away}}$ seven $\color{red}{\fbox{debts}}$ of £$5$ ($\color{red}{-7}\times\color{red}{-5}$) and this equals a gain of £$\color{blue}{35}$.

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    $\begingroup$ (+1) I've seen this taught particularly effectively by drawing the money in a "bag" (I imagine physical cards with + or - numbers on work well for the same reason). The basic premise is that if a bag contains 7, -3, and 5 (total = 9) we see that "taking away" or "crossing out" the -3 makes the new total 12, so "taking away -3 is the same as adding 3". If the bag contains two 6s, among other things, we can easily taking that "minus two lots of six from the bag is the same as minus 12 from the total". Or if the bag contains two -5s, among others, we see "minus two lots of -5 is the same as +10" $\endgroup$ – Silverfish Nov 9 '15 at 21:17

Your child may have been introduced to the minus sign by means of the word opposite. This is a great term to use in your conversations. Indeed, $a$ and $-a$ are opposites in that they are additive inverses. On the number line, opposite numbers are mirrored in their distances from zero, which provides a nice visual aid as well. We can use the term to describe arithmetic operations:

The opposite of three times five is the opposite of 15. $$-3 \times 5 = -15$$

The opposite of three times the opposite of five is the opposite of the opposite of fifteen... which is just fifteen. $$-3 \times -5 = 15$$

Okay, so we can use language to better cultivate our understanding of negative numbers. But what about a physical example? Well, here's a cute one that a friend once told me. It's a bit contrived, but I think it gets the point across. You can turn this into a demonstration too.

Suppose an ice cube lowers the temperature of a drink by $1$ degree.

Placing three ice cubes in a glass will lower the temperature by $3$ degrees, or $$ 3 \times -1 = -3$$

Removing two of the ice cubes will raise the temperature by $2$ degrees, or $$ -2 \times -1 = 2$$

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    $\begingroup$ Can you really warm up a drink by removing ice cubes from it? In reality, an ice cube in a drink absorbs some heat from the drink, and some of the ice may melt and become part of the liquid in the drink, then mix with the other liquid, lowering its temperature. In either case, when you remove the ice, you do not undo either of those effects; the drink remains cooler than before you put the ice in it. So you either have to make some complicated "averaging" argument or you have to rely on a false physical intuition to be "unlearned" later. $\endgroup$ – David K Nov 9 '15 at 13:49
  • $\begingroup$ @DavidK Just use the word "pretend." I doubt the child is ready for a talk about specific heat. $\endgroup$ – zahbaz Nov 9 '15 at 15:55
  • $\begingroup$ What should we pretend? Should we pretend that when you remove some of the ice, the liquid in the glass warms up? That's a "false intuition" of the sort I meant. There are so many analogies that actually work in real life, why not just use one of them instead of making up plausible (to a naive person) but incorrect physics? $\endgroup$ – David K Nov 9 '15 at 19:23
  • $\begingroup$ @DavidK No, we should suppose that the scenario is grossly simplified. Children can understand that sort of disclaimer. The point isn't to teach physics, but to paint a mental image of what opposites mean in terms of multiplication. We can use a metaphorical example as a heuristic, no? I get your point, but I find this example works well enough for pedagogical purposes.. If you want a physical example, then take the ratio of the mass of your ice cube and water to be approx 1:100. Then, yes, one ice cube will (eventually) lower the temperature of the drink by about 1 degree C. $\endgroup$ – zahbaz Nov 9 '15 at 20:42
  • $\begingroup$ Or if you will, replace the word ice cube by weight, glass by buoy, and temperature by height above sea level. It's just an architecture to express opposites. $\endgroup$ – zahbaz Nov 9 '15 at 21:05

One way to picture this is to imagine a number line. Then rotate it $180^{\circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{\circ}$ is the equivalent of multiplying by $-1$.

Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.

This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{\circ}$ counter clockwise to multiply by $i$.

But that's another story.

  1. Explain the definition of negative numbers.
  2. Point out that the definition of $-x$ implies that $-(-x) = x$.
  3. Explain that $-x = (-1)\times x$.
  4. Point out that (2) and (3) imply that $(-1)\times(-1) = 1$.
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    $\begingroup$ "Huh??? x is a letter! you can't multiply numbers and letters!" $\endgroup$ – Paul Sinclair Nov 9 '15 at 4:28
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    $\begingroup$ @PaulSinclair The children of today are used to using computers, they can easily be taught to think of $x$ as storing some arbitrary number. $\endgroup$ – Count Iblis Nov 9 '15 at 4:45
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    $\begingroup$ @CountIbis I wouldnt take that bet. I teach engineers and for many years this has been the assumption. Net result is that i get Bachelors level students that can not do anything with a computer, not even frigging graph a function. Some dont even know what a file is. $\endgroup$ – joojaa Nov 9 '15 at 5:08
  • $\begingroup$ @PaulSinclair Indeed! With kids of that age, a common response is that $x$ means 24. Because $a$ is one, and $b$ is two, and so on, and in that way the more inventive kids discover they can add numbers and letters. It's just that they've made the "wrong" extension into symbols. And explaining that $x$, or indeed $a$ might represent an arbitrary number is surprisingly hard if the kid hasn't developed that kind of abstraction yet. You probably would have more luck with "an empty box that a number can go in", rather than using a letter. $\endgroup$ – Silverfish Nov 9 '15 at 21:21

Symbology: $$\begin{align*} -a \times -b &= (-1 \times a)\times (-1 \times b) \\ &= -1 \times a \times -1 \times b \\ &= -1 \times -1 \times a \times b \\ &= (-1 \times -1) \times (a \times b) \\ &= a \times b. \end{align*}$$

What's going on?:
Consider the number line. When I multiply something by $2$, I double its distance from $0$. This happens whether the "something" is positive or negative. When I multiply something by $-2$, I double its distance from $0$ and flip to the other side of the number line. For instance $-2 \times 3 : 3 \rightarrow 6 \rightarrow -6$. If I start with a negative number, it's already on the negative half of the number line and will be flipped to the positive half: $-2 \times -3 : -3 \rightarrow -6 \rightarrow 6$.

That is, multiplication by a negative is the same as two steps: multiplying by the thing as if it had no negative, then applying the negative sign. That's what the symbology above says: multiplying by $-a$ is the same as multiplying by $a$ then by $-1$ and similarly for $-b$. But then the two "flip to the other half of the number line"s, the two "$-1$"s, cause two flips. But two flips takes anything away and then right back to itself, so two flips really does nothing. In this way $-1 \times -1$ is the same as doing nothing, or is the same as multiplying by $1$. That's why in the symbols above, we can drop the "$-1 \times -1$" -- they're the same as multiplying by $1$.

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    $\begingroup$ "When I multiply something by −2, I double its distance from 0 and flip to the other side of the number line": I like this answer a lot, but I think to serve as a good explanation for a kid at this level it would help if you could explain why multiplying works on the number line like this (it's safe to assume you can't use the symbology directly with the kid, although you're clearly trying to get across that train of logic). Until you can explain why negative multiplication reflects the number line, we have replaced one "why's does it work like that?" question by another. $\endgroup$ – Silverfish Nov 9 '15 at 21:14

Well.... this always made sense to me (but I've found it doesn't for others)

A) Multiplication is adding a number a bunch of times.

1) Negative numbers are numbers that are less than zero. The cancel out positive numbers. They are anti-numbers

So)i) positive x positive: add a bunch of positive numbers a positive number of times. Result: A big positive gain.

ii) positive x negative: add a bunch of negative anti-numbers. The result is a big amount of potential cancelling. Result: negative.

iii) negative x positive: take a bunch of positive numbers and take them away. Result: a loss; negative.

iv) negative x negative: take a bunch of anti-numbers and take them away. By taking away the taking away, what's left is putting things back. If you annihilite the annihitation the result is a net gain: Result: positive.


Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.).

We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency.

We wish to keep the property that 0*anything = 0, negative or positive.

We also want to keep the distributive property.

In order to keep the above two properties, we're forced to define the product of two negatives as a positive.

0*(-3) = 0

(5 + (-5))*(-3) = 0 (I'm plugging in 5+ (-5) for zero)

5*(-3) + (-5)*(-3) = 0 (distributive property)

add 5*3 to both sides. 5*3 cancels with 5*(-3)

(-5)*(-3) = 5*3


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