Why is negative times negative = positive?

Question

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 this:

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?

Answer 1

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 two 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 to imply that 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 first 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.

Answer 2

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.

Answer 3

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!

Answer 4

$\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{(}\! +\,\color{#0a0}{xy}}\, $ in $\:\!2\:\!$ ways (note over/underlined terms $ = 0)$

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

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

Said equivalently, $ $ evaluate $\rm\,\ \overline{-a\, +\!\!} \overline{\phantom{+}\! \underline {\color{#c00}{a}}}\underline{\ + \color{#0a0}{-a}}\ $ in $\,2\,$ ways (note 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 other "number" systems, then the Law of Signs is a logical consequence of these basic laws (abstracted from those 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

$$\begin{align}\rm f(g)\ =\ (-f)\ (-g)\ =\:\! -(f(-g)) \iff\,&\rm \ f(-g)\ = -(f(g))\\ \rm \overset{ \large g(x)\,=\,x}\iff&\rm \ f(-x)\ = -f(x),\ \ \text{ie. $\rm\:f\:$ is odd} \end{align}\qquad$$

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).

Answer 5

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.

Answer 6

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 $(-a)(-b)=ab$.

Answer 7

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.

Answer 8

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.

Answer 9

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.

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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.

Answer 10

Here's a simple proof that relies on the distributive property, the transitive property of equality, and cancellation.

Theorem: If $x,y \in \mathbb{N}$, then $(-x)(-y) = xy$.

Proof. First, notice that $$(-x)(-y) + x(-y)= ((-x) + x)(-y) = 0(-y) = 0.$$ Similarly, $$xy + x(-y)= x(y + (-y)) = x(0) = 0.$$ By the transitive property of equality, $(-x)(-y) + x(-y) = xy + x(-y)$, and cancellation yields $(-x)(-y) = xy$, as desired.