To give the context, I've been trying to look at different ways to convince myself how $-\times - = +$

Additive inverse of $a$ is written as $-a$

As an example the additive inverse of $-3$ is written as $-(-3)$
Also $-1$ times $-3$ is written as $(-1)\times (-3)$

Both above expressions evaluate to the same quantity $3$.
I guess it is easy to see why the additive inverse of $-3$ equals $3$ simply by staring at the equation $3+(-3) = 0$

However it must be very difficult to convince oneself why the second expression $(-1)\times (-3)$ evaluates to $3$ too. Both these operations seem related. I'm trying to figure out connection/intuition behind taking additive inverses and multiplying by $-1$. Help is appreciated. Thanks!

  • $\begingroup$ Duplicate of math.stackexchange.com/questions/304422/…? $\endgroup$ – Martín-Blas Pérez Pinilla Mar 15 '18 at 9:41
  • $\begingroup$ @Martín-BlasPérezPinilla Ty :) I've been to that page before and went through few answers. My question is not really about why -x- = +. I feel my question is specifically about the link between additive inverses and multiplication.. I haven't got full clarity yet I'm still working $\endgroup$ – AgentS Mar 15 '18 at 9:46
  • $\begingroup$ Mathologer did a nice video on this youtu.be/ij-EK-MZv2Q $\endgroup$ – rrauenza Mar 15 '18 at 15:04

The standard proof goes along the lines of $$ 0\times(- 3)=0\\ (1+(-1))\times(-3)=0\\ 1\times (-3)+(-1)\times(-3)=0\\ (-3)+(-1)\times(-3)=0 $$ and we see that $(-1)\times(-3)$ is an (the) additive inverse of $(-3)$.

  • $\begingroup$ Clever use of distribution law! Awesome! Ty :) Since $(-3) + 3$ also equals $0$ and the additive inverses are unique, I think it may also be reasonable to conclude that $(-1)\times ( -3 )= 3$ $\endgroup$ – AgentS Mar 15 '18 at 8:58
  • $\begingroup$ Wow! Two birds with one stone! $\endgroup$ – user535339 Mar 15 '18 at 13:43

A way to provide insight that additive inverses add up to $0$ is:

You are doing a certain distance in a direction in a coordinate plane. Let that distance be $x$ units. You are going the same distance in the opposite direction, or going to where you initially were. We moved the same distance, $x$ both times, but you were moving "backwards" $x$ the second time, which can be denoted by $-(x)$. Since you end up at your original location, you can claim that $x+(-x)=0$.

Now, we know additive inverses add up to $0$. Now for the multiplication of two negatives.

Using the information that additive inverses add up to $0$, define $x=ab+(-a)(b)+(-a)(-b)$, where $a$ and $b$ are real numbers.

$$x=ab+(-a)(b)+(-a)(-b)\implies x=ab+(-a)[(b)+(-b)]\implies x=ab+(-a)[0]$$

Which means $x=ab$. Also:

$$x=ab+(-a)(b)+(-a)(-b)\implies x=b[(a)+(-a)]+(-a)(-b)\implies x=b[0]+(-a)(-b)$$

Which means $x=(-a)(-b)$. Since $x=ab$ and $x=(-a)(-b)$, what can you tell?

  • $\begingroup$ Be careful using the apple analogy. It goes wrong pretty fast and especially in a lot of people that struggle with maths, they bring up the apples into situations where they are not at all applicable. On top of that, an analogy doesn't prove anything, it just gives insight into it. On the other hand, that $ab=(-a)(-b)$ proof is quite nice! $\endgroup$ – vrugtehagel Mar 15 '18 at 13:22
  • $\begingroup$ Yes, thank you for your suggestion. I will try to improve my answer! $\endgroup$ – user535339 Mar 15 '18 at 13:41
  • $\begingroup$ It's a nice proof, but since the whole point was to prove that $(-1)(-3)=-(-3)$, you're assuming what we want to prove when you say "Note that [...] $(-a)(b)=-ab$" Also, if we've already proven that $(-a)(b)=-ab$, then applying it twice together with commutativity of multiplication is, in my opinion, an easier way to prove $(-a)(-b)=ab$. $\endgroup$ – Arthur Mar 15 '18 at 13:57
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    $\begingroup$ @Arthur Actually, I can delete the part that $(-a)(b)=-ab$, it is quite irrelevant, multiplication by $0$ makes the term $0$. $\endgroup$ – user535339 Mar 15 '18 at 13:59
  • $\begingroup$ @idk Right, I missed that. It doesn't matter what $(-a)(b)$ is, since you're never evaluating it, you're just factoring out either the $(-a)$ or the $b$. $\endgroup$ – Arthur Mar 15 '18 at 14:03

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