Explaining to a kid why a negative × negative = positive? [duplicate]

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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

Solution 5:

$\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).