# Is addition more fundamental than subtraction?

When I followed an introductory! course group theory and throughout all my Math courses as a physicist, subtraction was always defined in terms of the inverse element and addition.

Is this the only way? I.e.: can subtraction be defined without addition?

If this is too broad a question, perhaps focus on groups and other physically relevant concepts.

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Here's teh start of a bunch of C++ programmers discussing the question: chat.stackoverflow.com/transcript/message/4935773#4935773, where I held that I believe that the axioms of math are simpler if subtraction is defined in terms of addition, but I'm not a math guy, so we brought it here. –  Mooing Duck Aug 14 '12 at 18:00
Addition is more fundamental in the sense that the natural numbers (the most fundamental set of numbers) is closed under addition but not subtraction. –  Austin Mohr Aug 14 '12 at 18:02
And appending the negative numbers and zero to the natural numbers obviates subtraction: $a-b = a+(-b)$ –  alancalvitti Aug 14 '12 at 18:11
Fundamental to what? –  Qiaochu Yuan Aug 14 '12 at 22:50
We start with two statements "0", "a+b" and "a-b" which we understand and want to translate into one another. a+b=a-(0-b), a-b=a+(-b) addition is defined using only subtraction and subtraction isnt defined using anly addition, because to define -b, in terms of "a+b" and "a-b" statements we have to write -b=0-b, so writing a-b=a+(-b) isnt very fundamental ti should be a+(0-b). Both equation needs definition of 0.So subtraction is more fundamental –  Qbik Aug 14 '12 at 23:10

Well, since you know a little from group theory, maybe this helps.

We always require a group operation to be associatve. But - is not: \begin{align*} (a-b)-c &= a-b-c\\ a-(b-c) &= a-b+c \end{align*}

Moreover but less importantly, + is abelian, while - is not.

Hence from a group theory perspective, - does not make much sense. It is not even an operation, its a short way of writing "add the inverse of the next element".

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This doesn't strictly answer the question: can subtraction be defined without addition. –  DeadMG Aug 14 '12 at 18:09
Yes i know, i just wanted to provide some background. –  Joachim Aug 14 '12 at 18:19
You also have an issue with identities: while $x - 0 = x$, you typically do not have $0-x=x$ unless $x=0$ –  Henry Aug 15 '12 at 19:01
Nice comment @Henry, that is certainly true. –  Joachim Aug 15 '12 at 20:35

Tarski did this thing, where he defined "group" with a single operation called $x:y$ to be thought of as $x y^{-1}$, such that a single equation is satisfied identically. So, in fact, you can start with subtraction first and then define addition, zero, inverses from that.

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See cs.unm.edu/~mccune/projects/gtsax for some short axiom systems, including a 19 symbol axiom in 3 variables and the single operation of subtraction (division) that characterizes groups. –  Jack Schmidt Aug 14 '12 at 18:31
Thanks Jack. So, in fact Tarski did this only for "abelian group". I am old enough to have attended a lecture by Tarski where he mentioned this... –  GEdgar Aug 15 '12 at 2:58

Yes, you can define addition in terms of subtraction if you want to. But suppose I have a set $S$ which is non-empty and closed under subtraction as I would recognise it. Then:

• $S$ contains $0$, since it's just $x-x$ for some $x\in S$
• Hence, for all $x\in S$, $-x\in S$, since $-x = 0-x$.
• Hence, for all $x,y\in S$, $x+y\in S$, since $x+y=x-(-y)$.

Hence $S$ is a group under addition. The converse doesn't hold: there are structures, like the natural numbers, closed under $+$ but where $-$ doesn't make sense.

From this I conclude that $+$ is a strictly more general operation than $-$, hence more often applicable, and therefore (in addition[1] to it just generally being "nicer" in the sense of associativity and commutativity) more worthy of study and consideration. Wherever you find subtraction, it's because there's really some addition around, whereas addition need not come with a corresponding subtraction operation.

And one more thing: ask yourself what subtraction means without referring to addition. The simplest way to describe the concept of subtraction, rather than the implementation, is certainly to refer to addition, which itself is described in terms of the successor operation or something similar (the cardinality of a disjoint union, perhaps?)

[1]: Pun... sorta intended. Don't judge me.

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In the natural numbers, there are no inverses, so you simply can't define subtraction as addition of the inverse element there. One way to define $a - b$ is as the unique solution, if it exists, of the equation $a = b + x$. But you can also avoid addition entirely and define subtraction recursively as \begin{align} a - 0 &= a, \\ S(a) - S(b) &= a - b. \end{align} Then you declare $0 - S(a)$ to be undefined for all $a$, and that covers all the cases. So yes, in Peano arithmetic you can define subtraction without addition, but of course this may not generalize well to other number systems.

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I can at least think of a situation where subtraction intuitively comes up before addition: the context of a metric space. If you're interested in the metric properties of, say, the real line, then you certainly care more about subtraction than addition, since that's how you get the distance between two points.

If you started with $\mathbb{R}$ and the distance function $d(x,y)$ between any two points $x$ and $y$, then you could define subtraction as $$x - y = \begin{cases} d(x,y) &\mbox{if } x \geq y \\ -d(x,y) & \mbox{if } x < y \end{cases}$$ Of course, the distance function $d(x,y)$ is typically defined in terms of subtraction, and I think you'd be hard-pressed to get to this definition without defining addition first (defining the rational numbers, in particular, would be tough without multiplication on the integers, which seems hard to define without first defining addition).

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This sort of definition naturally comes first in axiomatic geometry: the starting point is the notion of a directed line segment. –  Hurkyl Aug 14 '12 at 19:42
now use triangle unequality... –  Qbik Aug 14 '12 at 23:21
@Qbik The triangle inequality? As in $d(x,z) - d(x,y) \leq d(y,z)$? No problem! –  MartianInvader Aug 14 '12 at 23:24
A subtraction like this does not conform to the natural definition subtraction has. A metric is positive definite, subtraction is not. –  rubenvb Aug 15 '12 at 9:18
@rubenvb: the context of the word "metric" here is not of a Riemannian metric where positive definite is relevant, but instead of a distance function $d(x,y)$. –  Lee Mosher Aug 16 '12 at 14:01

Addition can be defined in terms of subtraction just as easily as subtraction defined in terms of addition. The two definitions are completely equivalent, and they define the same relationship between addition and subtraction: one is the other, where the second operand has been inverted.

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Reversing the two concepts just leaves you with two isomorphic concepts that does not answer the question. –  rubenvb Aug 14 '12 at 18:00
completely equivalent I think not. More like completely equivalent inverses. –  Drise Aug 14 '12 at 18:01
@Drise: completely equivalent after the isomorphic transformation. So in Math speak: completely equivalent. –  rubenvb Aug 14 '12 at 18:02
@rubenvb I know you intend no offense, but many mathematicians find the the phrase "math speak" condenscending. –  Austin Mohr Aug 14 '12 at 18:06
@AustinMohr: I'll keep that in mind. Sorry about that. –  rubenvb Aug 14 '12 at 18:07

Actually it should not matter, the inverse of + is just going into the other direction.

Since we 'defined' + as going up and - as going down, there is no different at all.

The only complication is when starting at 0 going up means going into positive numbers were most people are more familiar with than negative numbers.

(I know this is not a formal explanation, I'm a software engineer, and purely seen from logic, there is no difference).

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