I know that it works, but I'm seeking a way to think about it so it makes sense.

It seems like a weird question to me. I might not be able to articulate it, and it might not even have an answer (other than it is what it is; it's just defined that way so it works). Let me try to demarcate what I don't understand by saying what I already do understand:

On a number line, for $0<a<b$, the subtraction $b-a$ is the difference (and the distance) between the two points. If you translate both $a$ and $b$ by the same amount left or right, the relative position of the two won't change - even if shifted so far left that $a<0<b$ or $a<b<0$.

A similiar view is "zeroing" at point $a$ (marking point $a$ as "$0$"). The distance from point $0$ to point $b$ is now $b-a$, which we can just read off.

In terms of arithmetic,

  1. for $0<a<b$, their difference $b-a$ is simply subtraction.

  2. For $a<0<b$, the distance between them is the sum of their distances from $0$, or $|a|+|b|$. Since $|a|=-a$ and $|b|=b$, the sum $|b|+|a|=b-a$.

  3. For $a<b<0$, the distance is $|a|-|b|$ (because both magnitudes are reflected, $|a|>|b|$). This is $(-a)-(-b)=-[(-b)-(-a)]=b-a$.

Why does subtraction work out so neatly, so the difference is always $b-a$? What is the property of the rules of arthmetic that make this work? Proofs seem to be about showing $what$ is true, but not $why$. Biology and engineering often have a story about design decisions: what is the underlying design here?

I'm not asking about $|b-a| = |a-b|$; but the simpler, basic subtraction. (To show I understand that equality: by definition $|C|=|-C|$. If $b-a=C$ then $|b-a|=|-(b-a)|$, which is $|a-b|$.)

Background: I've just refreshed my arithmetic by doing all problems in the Khan Academy arithmetic unit (and all material in the negative numbers subsection).

EDIT related question: Subtraction of a negative number

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    $\begingroup$ It seems like you understand quite a bit, but it's hard for me personally to get a sense of what exactly you'd like to know; what you're unsure about. $\endgroup$ – pjs36 Jan 28 '18 at 2:46
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    $\begingroup$ What is it that you don't understand? Everything you say seems to make perfect sense, unless I'm missing something. $\endgroup$ – saulspatz Jan 28 '18 at 2:48
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    $\begingroup$ Remember that subtraction is define as $x-y = x + (-y)$. $\endgroup$ – steven gregory Jan 28 '18 at 2:52
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    $\begingroup$ You sound like you'd really enjoy abstract algebra, OP. $\endgroup$ – G Tony Jacobs Jan 28 '18 at 2:58
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    $\begingroup$ The facts that the structure $[\mathbb Z, +]$ is an abelian group and $x-y \equiv x+(^-y)$ should explain a lot. $\endgroup$ – steven gregory Jan 28 '18 at 3:50

One way to dig deeper into this is to look at a set-theoretic definition of integers in terms of the natural numbers.

For positive $n$, define the integer $n$ as the collection of ordered pairs of natural numbers $(a,b)$, such that $a=b+n$. Then define $-n$ as the collection of ordered pairs $(a,b)$, such that $a+n=b$. This gives us a way of finding the opposite of an integer: To find $-5$ for example, just pick a representative out of the collection representing $5$, and reverse its entries. If you pick $(7,2)$ as a representative of $5$, then $(2,7)$ is a representative of $-5$.

Now, we can define addition: To add the integer $n$ represented by $(a,b)$ to the integer $m$ represented by $(c,d)$, define $m+n$ as the integer represented by $(a+c,b+d)$. Define subtraction by $n-m$ means $n+(-m)$. This gives you a framework for seeing why addition and subtraction don't really "care" if the numbers they're working on are positive or negative.

Does that help at all?

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    $\begingroup$ Thanks! It's a little complicated yet promising, but I fried my brains just articulating my question! I'll need a break before I can visualize it intuitively enough to tell if it helps. $\endgroup$ – hyperpallium Jan 28 '18 at 3:57
  • $\begingroup$ I've gone through it, and I'm afraid it is a little too complicated to give me intuition. The first definition encodes integers as differences (between positives) - but (I now see) differences are part of my difficulty! I saw $n$ represented by all (directed) lengths $n$, on a positive-only number line, with negtive $n$'s points swapped (so in the opposite direction, like vectors). I algebraically rederived addition $(a+c, b+d)$, yet don't really see it... I note that you, like others, defined subtraction in terms of addition of the inverse (negated/opposite). This might be important for me. $\endgroup$ – hyperpallium Jan 29 '18 at 3:13
  • $\begingroup$ My seeing it as a set of directed lengths, $(a,b)$ starts at $a$, ends at $b$ seemed nice, but doesn't fit with your addition. A second idea is to see $b$ as the size of a container (say a rectangular area), and $a$ is a quantity (how full it is, say, shaded area on left) - the space left is the integer represented. Your addition now makes sense, for positives, because summing the containers, and summing the quantities, will also sum the unfilled spaces. Not sure about summing negatives... a negative is overfilled comtainer (perhaps better vice versa: + is overfilled, - is underfilled?) $\endgroup$ – hyperpallium Jan 31 '18 at 1:10
  • $\begingroup$ Can also see the pair $(a,b)$ as $a-b=n$. So swapping them negates their value. Like directed lengths. $\endgroup$ – hyperpallium Feb 1 '18 at 3:47

It seems to me you are mystified about the notion of negative numbers or you're wondering how the operation of subtraction can be extended to the negative integers. (I might not have gotten exactly what you want, but so you mightn't too; I suggest you re-analyse what the problem is, carefully. Perhaps you might be able to articulate your points more accurately.)

Nevertheless, the negative integers were accepted in order to have a set of integers closed under subtraction. Thus $3-5$ reduces to $0-2$, which is usually abbreviated as $-2$, where one way to understand the symbol before the numeral is as indicating that $-2$ is to the left of the point $0$ on the integer number line.

We may use this number line to understand also why subtraction "works" for the negative integers. Recall that subtraction $b-a$ of positive integers $a$ and $b$ so that $a\le b$ is geometrically equivalent to starting from $b$ and moving $a$ units to the left. The positive integer $c$ where we land is such that $c=b-a$. Indeed when $b\lt a$ the definition extends nicely, only now the answer is a negative integer.

Now the question is, how does this extend to the case when we have two negative integers? First however let us settle the case when we have one negative and one nonnegative integer, say $n$ and $m$ respectively. No, pick concrete integers $-2$ and $3$ for example. We may immediately interpret $-2-3$ as before, namely starting from $-2$ and moving $3$ units to the left, landing us on $-5$. Now to understand $3-(-2)$ we must start from $3$ and move $2$ units to the right, which by walking on the integer line gives us $5$. Lastly if we have two negative integers $p$ and $q$, say $-5$ and $-1$, then $-5-(-1)$ is the place we land by starting from $-5$ and moving a unit to the right, which is $-4$ and similarly, $-1-(-5)$ gives $4$.

These geometric intuitions can lead us to notice why negative integer subtraction is consistent with, and thus a natural generalisation, of positive integer subtraction. There are other arguments as to why this is the case (for example purely algebraic arguments) but I have used the geometric one to give a motivation in as intuitive a way as one can get.

Hope this helped; otherwise you may need to edit your question after a little more introspection as to exactly what the problem is.

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  • $\begingroup$ Thanks; I'd edited my title a few times, trying to sum it up; of course they varied in accuracy. A big part of my question is why subtraction gives "difference" - which is a confusing statement, because to me they mean the same thing. It's really geometric difference, or distance. BTW it reads like something of a gap, when you said "can lead us to notice why...[it's consistent]", but don't go on to explain that "why" (although there's several why's before then). Laying out the consistency would be helpful to me - perhaps a text would be helpful? (Though they never seem to explain, only define) $\endgroup$ – hyperpallium Jan 29 '18 at 3:42
  • $\begingroup$ @hyperpallium After having read your question again after the edit, I think I might be closer to getting your meaning. However I still have a few questions. You want to know why $m-n$ always gives the "distance" for arbitrary real numbers $m$ and $n$, but how exactly do you use the word distance here? Specifically, which distance are you talking about? Take two positive numbers, say $3$ and $8$, then how is either of $3-8$ or $8-3$ a distance in the sense you mean? -- for using the interpretation in my explanation above, both subtraction results are (to use integers now) the numbers where... $\endgroup$ – Allawonder Jan 30 '18 at 0:33
  • $\begingroup$ ...you land, so to say, if you start at $m$ and move forwards or backwards $n$ units depending on the combination of signs before $n$. Thus this subtraction always gives you a point as answer, not a distance. I'd like you to consider this, because calling the answer a distance either uses a different interpretation or gets something wrong. If the former, please explain why the answers are distances and not points. Even for arbitrary reals $r$ and $s$, the subtraction $r-s$ is the point you get to if you start from $r$ and move a distance $s$ either forwards or backwards accordingly. $\endgroup$ – Allawonder Jan 30 '18 at 0:40
  • $\begingroup$ I've been trying to define my terms (on paper) after reading your comment, and noticed that a point isn't a distance, too. I haven't come to grips with this yet, but... so far, the subtraction gives a point equal to the distance; and $|b-a|$ is widely described as the "distance" between $a$ and $b$. Understanding that is my end goal; the simpler "$b-a$, where $a<b$" is a step along the way. Still working on it. $\endgroup$ – hyperpallium Jan 30 '18 at 2:01
  • $\begingroup$ @hyperpallium Note that there's a difference between $|a-b|$ and $a-b$. Whereas the first is a distance between two points, the latter is a point related to the point $a$ by some operation. As you say you're trying to analyse what the problem may be further, I shall see what you come up with. Sometimes, understanding what the problem is is the hardest part; so stick with it. Good luck! :) $\endgroup$ – Allawonder Jan 30 '18 at 2:09

Subtraction is defined as the operation that undoes addition:

If $a+b = c$ then $c-b=a$.

Similarly, Division is defined as the operation that undoes multiplication.

If $a\times b = c$ then $c\div b = a$.

The fact that $a \times 0 = 0$ for all $a$ is why division by zero is not allowed.

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  • $\begingroup$ Thanks; I think this undoes/inverse insight might be important, but I don't yet see how it relates. $\endgroup$ – hyperpallium Jan 29 '18 at 3:44

How do we generalize subtraction to negative integers?

One way uses a number line. For points $a<b$, if we move both one unit left, or both one unit right, the number of units in between stays the same (i.e. the difference or distance between them). So, if we shift both until $a$ is at $0$, the shifted $b$ will be the same distance away, and we can read that number off the number line.

We move $a$ to $0$, and move $b$ by the same amount:

  • if $a$ is left of $0$, move $a,b$ right by $|a|$ units.
  • if $a$ is right of $0$, move $a,b$ left by $|a|$ units.

Translating this to arithmetic, move left = subtract, move right = add:

  • if $a<0$, add $|a|$ to $a,b$: $a'=a+|a|=0, b'=b+|a|$
  • if $0<a$, subtract $|a|$ from $a,b$: $a'=a-|a|=0, b'=b-|a|$

Now as it happens,

  • if $a<0$, then $|a|=-a$ (neg flipped to pos), so $b+|a|=b-a$.
  • if $0<a$, then $|a|=a$ (already pos), so $b-|a|=b-a$.

That is, both cases can use the one expression: $b-a$.

That is nice and simple... but is it just a coincidence, or is there some sense to it? Is it like bit twiddling, where clever boolean algebra gives a surprising result, or it is natural that it works out?

Another simplication involves the additive inverse $-a$, such that $a+(-a)=0$. We don't know which of $a$ and $-a$ is positive, and which is negative, just that there's one of each. This gives a direct way of moving $a$ to $0$ which doesn't require knowing left or right, or negative or positive, because adding its inverse to itself always gives $0$, by definition. Now we can say:

$$ \begin{align} a' &=a+(-a)=0\\ b' &=b+(-a) \end{align} $$

Or using the textual convention that defines subtraction as the addition of a negative:

$$ \begin{align} a' &=a-a=0 \\ b' &=b-a \end{align} $$

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Another approach is to define positive numbers as saying "next" for each next one. e.g. so $3$ would be "next, next, next". We can also reverse this with "back". Once we run out of "next"s to undo, we start just having "negative" numbers e.g. $-2$ is "back, back". Similarly, we can reverse these with "next"s. These negatives behave just the same as positives, in that "back" and "next" still work, just as if we were using numbers composed purely of "next"s (e.g. if we were at a higher number than the lowest negative we needed). This generalizes positive numbers to negative numbers.

For $a<b$, the difference between $a$ and $b$ is how many "next"s it takes to get from $a$ to $b$. Or

\begin{align} a+X=b \end{align}

where $X$ is the difference.

Algebra says if we do the same thing to both sdes of an equation, it remains true. So, let's get rid of the $a$ by doing the reverse of it. If $a$ is "next,next", then do "back, back" to both sides. If $a$ is "back,back,back", do "next,next,next" to both sides.

\begin{align} a+(-a)+X=b+(-a) \end{align}

Because $a+(-a)=0$, this gives:

\begin{align} X=b+(-a) \end{align}

I think the difficulty is that the natural meaning of subtraction, of "taking away", that works for a set of physical objects (pebbles, fingers, people) i.e. natural numbers, doesn't work for negative numbers, because you can't have a negative number of physical objects.

Negative numbers, like rationals, surds/radicals, and imaginary numbers are pretend. Their notation suggests this, by instead of stating them, shows a way to calculate them. e.g.

\begin{align} -2, 1/3, \sqrt{2}, \sqrt{-1} \end{align}

The text "$-2$" isn't a number, but the unary operation of negation ($-$) applied to 2. At least Euclid could point to a figure to show a negative, rational or irrational number.... but to be fair, his constructions were also calculations.

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