# Why is negative times negative = positive?

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?

• You haven't proven that -xy = (-x)y. – Qiaochu Yuan Nov 12 '10 at 2:24
• It is not an uncommon question, and it's never not easy to show. – J. M. is a poor mathematician Nov 12 '10 at 4:06
• 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. – Jonas Kibelbek Nov 12 '10 at 4:48
• Does he understand why a negative times a positive is negative? And if so, how? – Paul Nov 8 '15 at 21:00
• Have you shown him a sequence, to see a pattern? So for example $2×-3=-6$, $1×-3=-3$, $0×-3=0$, $-1×-3=3$...? – Paul Nov 8 '15 at 21:03

Why not look at a multiplication table? Let's make a little one, including some negative numbers. You could of course make it bigger to make the patterns clearer. Let's start with what we already know: $$\begin{array}{|c|c|c|c|c|c|} \hline &\textbf{-2}& \textbf{-1} & \textbf{0} & \textbf{1} & \textbf{2} \\ \hline \textbf{-2} & & & & &\\ \hline \textbf{-1} & & & & &\\ \hline \textbf{0} & & & 0 & 0 & 0\\ \hline \textbf{1} & & &0 &1 &2 \\ \hline \textbf{2} & & & 0& 2&4\\ \hline \end{array}$$ Now, let's just notice that the third row (i.e. the first filled in one) is constant - it's just a bunch of zeros, so we should extend it likewise. The fourth row, when read right to left is counting down $2$ then $1$ then $0$ - so we should keep counting down to fill in $-1$ and $-2$. The final row counts down by twos, so it should continue doing so to $-2$ then $-4$. Let's fill this in: $$\begin{array}{|c|c|c|c|c|c|} \hline &\textbf{-2}& \textbf{-1} & \textbf{0} & \textbf{1} & \textbf{2} \\ \hline \textbf{-2} & & & & &\\ \hline \textbf{-1} & & & & &\\ \hline \textbf{0} & 0 & 0 & 0 & 0 & 0\\ \hline \textbf{1} & -2&-1 &0 &1 &2 \\ \hline \textbf{2} &-4 & -2& 0& 2&4\\ \hline \end{array}$$ If we, in each column, do a similar thing, we can complete the table. Like, the first column is counting upwards by $2$ when we move up - it goes $-4$ then $-2$ then $0$ so we should continue counting this way to $2$ then $4$. If we apply the same reasoning to each column, we can fill in the whole table $$\begin{array}{|c|c|c|c|c|c|} \hline &\textbf{-2}& \textbf{-1} & \textbf{0} & \textbf{1} & \textbf{2} \\ \hline \textbf{-2} &4& 2& 0& -2& -4\\ \hline \textbf{-1} &2& 1& 0& -1& -2\\ \hline \textbf{0} & 0 & 0 & 0 & 0 & 0\\ \hline \textbf{1} & -2&-1 &0 &1 &2 \\ \hline \textbf{2} &-4 & -2& 0& 2&4\\ \hline \end{array}$$ And, if we trace back the steps that we used to generate this correct table, we can recover $(-1)\times (-1)=1$ as follows:

• Firstly, we note that one times something leaves that thing unchanged. So $1\times(-1)=-1$.

• Secondly, looking at the table again, we see that multiplying by $(-1)$ "reverses" the order of our usual counting - that is $(-1)\times 2$ is $-2$ then $(-1)\times 1$ is one more at $-1$ and $(-1)\times 0 =0$. So, when we get to $(-1)\times (-1)$ we have to be one more than $0$ since $-1$ is one less than $0$.

It may also be good just to look at the table - it's very symmetrical. We see, for instance, in the second and fourth columns (multiplication by $1$ and $-1$) a very clear reversal of the ordering, which more or less tells us what multiplication by $-1$ is actually doing.

We know that $\mathbb{R}\subset \mathbb{C}$.

Let $x,y\in \mathbb{R}$ and $x<0, y<0$

Any number $n$ can be written in polar form as $n=|n|e^{i\theta}$ where $\theta$ is the angle made by the line joining origin and $n$ with positive $x$ axis.

Therefore $x=|x|e^{i\pi}$ and $y=|y|e^{i\pi}$

$xy=x=|x|e^{i\pi}|y|e^{i\pi}=|x||y|e^{i2\pi}=|x||y|$

$x$ and $y$ being negative but there product is $|x||y|$ which is positive!

NOTE

$e^{im}=\cos m +\iota \sin m$

• -1: This is an incredibly over-complicated argument about real numbers that actually holds in far more generality anyway. – Zev Chonoles Jun 12 '15 at 7:42
• @ZevChonoles Sir though it may feel complicated it is technically correct and provides and alternative approach to the given problem. I request you to kindly elaborate your comment. – Singh Jun 12 '15 at 7:56
• @Zev I agree ,but not only this, proving it's true in the complex plane does NOT necessarily imply it's true for real numbers only. To give a simple counterexample, the square roots of negative numbers can be multiplied in C, which does not carry over by inclusion to R. – Mathemagician1234 Jun 13 '15 at 3:17

If he already accepts that $1\cdot -1=-1$, then we can use some hand-waved algebra to show that $1=-1\cdot -1$ using symmetry, just like addition and subtraction.

$2-1=¿what?$ , This is the question.

$1+¿what?=2$ , But this is what we're really asking, in simpler terms.

If 1 times -1 is -1, then -1 times what is 1? Because 1/1 is 1 we don't have to worry about reciprocal fractions, so even though it's a little misleading to use multiplication as the inverse of multiplication, I think it works well enough for now.

$1\cdot -1=-1$

$¿what?\cdot -1=1$

If we say ¿what? is 1, then we have a contradiction, because then $1\cdot -1=-1$ in the first equation, then $1\cdot -1 = 1$ in the second equation. If we say ¿what? is -1, then there's no contradiction.

I think I have a pretty simple proof to why a negative times negative is a positive. We know that every $x$ has an inverse $-x$ which when added together equals $0$. Now as $x + (-x)=0$ we can multiply this by some arbitrary $-y$ to get $-(y)x + (-y)(-x) = 0$. But we also know that $xy + (-xy) = 0$ so since the additive inverse is unique this means that $xy = (-x)(-y)$. Of course this assumes $x(-y) = -(xy)$ but that can be proved by multiplying the first equation with $y$ instead of $-y$. I am open to any suggestions and feedback.

I reformat the most upvoted answer (also my favorite) with MathJax, from Reddit:

### Zerotan 42.2k points 7 months ago

Repost from 2 years back:

I give you three \$20 notes: $$+3 × +20 =$$ you gain$60

I give you three \$20 debts: $$+3 × -20 =$$ you lose$60

I take three \$20 notes from you: $$-3 × +20 =$$ you lose$60

I take three \$20 debts from you: $$-3 × -20 =$$ you gain$60

Why a negative times a negative can be reduced to the question of why -1 x -1 = 1. The reason for that is because it is forced upon you by the other rules of arithmetic.

1 + (-1) = 0 because of the definition of -1 as the additive inverse of 1

Now multiple both sides by -1 to get

-1(1+(-1)) = 0 because 0 times anything is 0

Use distributive law to get:

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

Now -1 * 1 = -1 because 1 is multiplicative identity.

So we have -1 + (-1)x (-1) = 0

Put -1 on the other side by adding 1 to both sides to get

(-1) x (-1) = 1

So -1 x -1 = 1.

Now for any other negative numbers x, y we have

x = (-1) |x| and y= (-1) |y|

So x * y = (-1) |x| * (-1) |y| = (-1) *(-1) * |x| * |y| = |x * y| is positive.

Now that you know the reason it really doesn't make much difference in understanding. This question is not really that important. It's like asking why is 1 raised to the 0 power equal to 1? Because that's forced upon you by other rules of exponents,etc.

A lot of time is wasted on this. This is not the kind of problem kids should be thinking about.

• I don't think this would be that accessible to a child who is only starting to learn how to multiply, so it does not really answer the question. – Dylan Nov 9 '15 at 6:44
• The short answer is forget about it kid. Just accept it as a rule of arithmetic. If a child doesn't understand this. then NO answer is going to satisfy him. There is no physical reason why it is so. Math is a man made. The rules can be anything. For instance why 1+1 = 2 and not 0? Well in other number systems it can be. 1+1 =0 . No reason to ask why. it just happens to be true. There is no universal truth in math. It is all in the axioms. Or why is parallel postulate true? It isn't? You can use other postulates. – larry Nov 9 '15 at 6:50
• I'm not planning to have children any time soon, but if I were, I certainly wouldn't want them to grow up just accepting things without questioning them or asking "why?". Also, that still doesn't change the fact this this is not an answer to the question which was asked. You could post it as a comment, if you like, but it is not an answer in my opinion. – Dylan Nov 9 '15 at 6:54
• But a lot of the time, we made the rules to model certain phenomenon or to capture certain intuitions, and it is perfectly valid to ask why the rules are the way that they are. The idea of the integers was around long before we started taking an axiomatic approach to them, and historically there were certainly other reasons to consider the product of two negative integers to be positive than just "them's the rules, kid". – Dylan Nov 9 '15 at 7:10
• Perhaps it is difficult to explain to a layperson, but that is precisely why this question is being asked: to gather input on what some approaches might be to explain the concept to a layperson, or--ideally--an eight-year-old. – Dylan Nov 9 '15 at 7:13

This is a sketch of an explanation that can easily be made more or less formal.

My job now is to explain to you (who have a clue) my way to explain to somebody else (who has almost no clue). That's why I use formal language (like define) and notation (like $f(x)$) in order to keep my answer neat. At the same time I stick to integers, as if this "somebody else" was a kid. Adjust the form and scope to your interlocutor's level.

Let's define

$$f_a(x) \equiv ax$$

The explanation goes as follows:

1. On $XY$ plane plot $y=f_0(x), \space y=f_1(x), \space y=f_2(x), \space … \space$ for $x \in \{0,1,2,3,…\}$; use different symbols/colors to distinguish the functions.
2. Notice (or better let the interlocutor notice) that the points of $f_0$ lay along a straight line, the points of $f_1$ lay along another straight line etc.
3. There is no reason this rule shouldn't apply when we consider $x \in \{…,-2,-1,0,1,2,…\}$; expand the plot.
4. Notice $\forall a \space f_a(0)=0$. Why?
5. Notice $\forall a \space f_a(1)=a$. Why?
6. There is no reason the three rules shouldn't apply for negative $a$; plot $y=f_{-1}(x), \space y=f_{-2}(x), \space … \space$ according to the rules.

It will look like this:

Only the blue points were obtained by actual multiplication. The rest of them were placed thanks to the rules found.

And here you go. The result of $(-1) \cdot (-1)$ is there among few others.

You can replace "there is no reason the rules shouldn't apply" with some formal proofs if your interlocutor can understand them. While explaining to a kid it should be enough to point out this is the way mathematics works – coherence, no unnecessary exceptions, rules as broad as possible. I think it can be quite reassuring on an early stage of education.

The following is based on this Reddit post that I edited for grammar and readability.

As you've probably heard, multiplication is just a fancy word for repeated addition. If you wanted to repeatedly add $5$ three times ($5 + 5 + 5$), multiplication allows you to rewrite that as $3 \times 5$.

What if you wanted to repeatedly add $-5$? You would write $(-5)+(-5)+(-5)$ as $3 x (-5)$.

Now what if I asked you to subtract $(-5)$ three times? Well, you can write that as $- (-5)- (-5)- (-5)$, or $(-3) \times (-5)$. This is just $5 + 5 + 5$: a positive number.

So why is subtracting a negative the same as adding a positive? Because as explained in the previous paragraph, when you subtract debt, you get money.

There is an abundance of intuitive answers to this question so it came as a surprise that a somewhat trivial (by the time someone finishes highschool he should have learned it..) but also fundamental example from physics is absent.

Long story short:

Opposites attract, Same repel

When two particles with the same electromagentic charge interact, either positive or negative, they produce the same result-they repel each other.

While if one has a negative charge and the other a positive, they attract each other.

Let's symbolize the interaction with I, a positively charged particle with $$(+)$$, a negatively charged one with $$(-)$$, attraction with $$A$$ and repulsion with $$R$$.

So $$(+)I(+)=R\\(+)I(-)=A\\(-)I(+)=A\\(-)I(-)=R$$

For me this looks like the best "natural" example, suitable to students relatively quickly.

• Interesting viewpoint, but how does the story translate to magnetism and magnetic poles? – Frenzy Li Nov 12 '18 at 22:56
• @FrenzyLi Well, the "hard math" answer involves the uniqueness of the inverse element in a ring, which obviously cannot be taught to highschool students the same time they learn about negatives. So instructors tend to look for examples from the physical world but end up talking about artificial creations like economics, running competitions etc. Electromagnetism is an integral part of the universe-one of the fundamental forces in particular..What could be more natural? – MathematicianByMistake Nov 12 '18 at 23:00

I like the most upvoted answer from Reddit much more than the 2nd most upvoted answer, but I reformat it with MathJax beneath.

### sjets3 14.1k points 7 months ago

Imagine you are watching a movie. The first number is how the person in the movie is moving. The second number is how you are watching the film (normal or in reverse).

$$1 \times 1$$ is a person walking forward, you watch it normal. Answer is you see a person walking forward, which is 1.

$$1 \times -1$$ is a person walking forward, you watch it in reverse. You see a person walking backwards. -1

$$-1 \times 1$$ is a person walking backward, you watch it normal. You see a person walking backwards. -1

$$-1 \times -1$$ is a person walking backwards, but you watch it in reverse. What you will see is a person that looks like they are walking forward. 1

As a funny example of the last para. overhead, compare https://gfycat.com/PopularFrighteningCormorant (original video) with https://i.imgur.com/ZCw2C81.gifv (film person walking backward then play backward).