Adding and subtracting negative numbers

Question

Why is it such that adding and subtracting negative numbers is the way it is? Ex : $$- 8 - 8 = - 16$$ How is one supposed to write it ? $(-8) + (-8) =$ or $(-8) - (-8) =$ Yes, I am conscious of the fact that the answer is $(-8) + (-8) = -16$ but why? Why does the rule follow as such that when you have two negative numbers you add them and you put the sign of the largest number? ( absolute value ) I wish to understand the logic when you have:

$- x - x =$ ( ADD them and put the sing of biggest absolute value )

$+ x - x =$ ( subtract them and put the sing of biggest absolute value )

$- x + x =$ ( subtract them and put the sing of biggest absolute value )

Could someone kindly provide me with some intuition as to why this is the case?

Answer 1

Let us first concentrate on addition. To find the result of $a+b$, you start at $a$ on the number line and move to the right if $b$ is positive and move to the left if $b$ is negative.

For example $$(+5)+(+7) = +12$$ and $$(+5)+(-7) = -2$$and $$(-5)+(+7) = +2$$ and $$(-5)+(-7) = -12$$.

For subtraction use the definition, $$ a-b = a+(-b)$$ where $(-b)$ is the opposite of $b$.

For example $$ (-5)-(-2) = (-5)+(2)=-3$$ and $$ (-5)-(+7) = (-5)+(-7)=-12.$$

Answer 2

So let's suppose that you only know about the counting numbers $\mathbb N =\{0, 1, 2, 3, \dots\}$. And maybe you know how to add them and multiply them. We can use those to build a model for the integers, something that you can use to understand them. In my case I understand math with physics, and negative numbers are good to understand whether something is increasing our decreasing, like a bank balance or even toys in a toybox. And we can think about those changes with a pair of numbers: a flow in and a flow out.

So we think about pairs of counting numbers like $(0,1)$ or $(1,2)$ representing two units of flow out, one unit of flow in. More is going out than in, so the amount in a box will decrease by 1.

Except, we want to say that those are the same as $(2,3)$, they both represent one net unit of flow outward. The way we do that as mathematicians is to talk about the set of all the equivalent things, what we call an "equivalence class". Then two sets are equal when they contain the same elements. So we define zero change as all the flows that do not change what's in the box, $$ \bar 0=\{(0,0),(1,1),(2,2),\dots\},$$ while a positive change by 1 is the set$$ \bar 1=\{(1,0),(2,1),(3,2),\dots\},$$ and this equivalence rule does not need subtraction, it says that $(a,b)$ is the same change as $(c,d)$ when $a+d=b+c$. Another way to phrase this is that you can add the same number to both sides all the time, so $(a,b)$ is the same as $(a+m,b+m)$, which also doesn't require subtraction as long as your arguments also work for the smallest members of the class that look like either $(a,0)$ (positive integers) or $(0,b)$ (negative integers).

Addition of these flows is easy, take all the pairs of the two sets $(a,b)$ and $(c,d),$ and form the set of $(a+c,b+d)$, adding inflow and outflow separately. Negation turns outflow into inflow and vice versa so that is also easy, $(a,b)\mapsto(b,a)$.

We technically have to make sure that all operations are “well-defined,” so they don't generate any new sets or mess up any existing sets. So to do this just consider $(a+m, b+m) +(c,d)$ and you find that these rules give you $(a+c+m, b+d+m),$ which is indeed equivalent to $(a+c,b+d)$.

To get subtraction you combine addition and negation, $(a,b)-(c,d)=(a+d,b+c).$ So the inflow of the first is combined with the outflow of the second, and vice versa. And that is how we can use one flow to cancel out another.

So now you have negative eight minus eight, that is $(0,8)-(8,0)$ and the rules say that this is the same as $(0,8+8)=(0,16)$. We can also describe it as adding the negative, $x-y=x+(-y)$—that is just a pattern that we can see here.

The rules you have been given say that you can pretend that when the things to be added have opposite signs, you subtract the bigger (in absolute value) from the smaller, then you make it negative if that bigger one was negative. That is also true, and it is all about trying to go from some $(a,b)$ to the “simplest” representative, either $(a-b, 0)$ if $a>b$ or $(0,b-a)$ if $b>a$. So you first figure out whether the outflow or inflow is going to be bigger and then you subtract the other.

Appendix: multiplication

So multiplication of integers then works by extending the “distributive law” of multiplication, so that $(a, b)$ times $(c,d)$ works like $(ac+bd, ad+bc)$. This one we really should also prove is well-defined: imagine that we had used $(a+m, b+m)$ instead of $(a, b)$, we would get $(ac+bd+m(c+d), ad+bc+m(c+d))$, so this is well-defined. And you can prove that this does the distributive law right and does multiplication right between positive numbers $(a, 0)\times (c, 0)=(ac,0)$.

You didn't ask about this but I thought it might be helpful for understanding why a negative times a negative is positive or why negation is the same as multiplying by negative one.