Why define addition with successor?

Question

I'm reading Russell's Introduction to Mathematical Philosophy Russell defines the sum of two numbers in terms of successors. I don't understand why:

Suppose we wish to define the sum of two numbers. Taking any number $m$, we define $m+0$ as $m$, and $m+(n+1)$ as the successor of $m+n$. In virtue of (5) this gives a definition of the sum of $m$ and $n$, whatever number $n$ may be.

Is this any different than doing $m+ (n+1) = (m+n) +1$ ?

What is the purpose of $+1$? We might as well write $m+n$.

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Edit:

Russell cites Peano's 5. Axiom as proof. How does Peano's 5. Axiom (mathematical induction axiom) prove addition here?

Answer 1

Note that he is defining addition. So to avoid circularity he uses recursion, defining $m+(n+1)$ in terms of two things already defined: $m+n$ and the successor operation. So his recursive definition of $m+k$ for any nonnegative integer $k$ is

1. $m+k=m$ when $k=0$, and
2. $m+k=\mbox{succ}(m+n)$ when $k>0$, in particular, when $k= n+1$ for any integer $n \geq 0$.

Answer 2

It would be beneficial for you to delve into the field of Set Theory, particularly focusing on Transfinite Induction and Transfinite Recursion. In Set Theory, we can define $\alpha+1=\alpha\cup\{\alpha\}$, and arithmetic operations are defined on a unique class of sets known as Ordinals. We only have the definition of $\alpha+1$, and addition is represented as a mapping from $\mathrm{Ord}^2$ to $\mathrm{Ord}$, or more simply, from $\mathbb{N}^2$ to $\mathbb{N}$. Transfinite Recursion provides a systematic and comprehensive way to define such mappings.

Transfinite Recursion is particularly tailored for Ordinals. If your focus is solely on arithmetic within $\mathbb{N}$, then Mathematical Induction is sufficient, and you might refer to it as Mathematical Recursion.

In summary, what we require is a rigorous and general method for defining a mapping through recursion. For instance, a strict definition of a function on $\mathbb{N}$ (or on $\mathrm{Ord}$) with the following holds: $$f(0)=a_0, \text{ and } f(n+1)=F(f(n)).$$ (For the $\mathrm{Ord}$ case, you can refer to any standard Set Theory textbook.)

We are so accustomed to arithmetic within $\mathbb{N}$ that we seldom question the existence of such mappings. However, their existence is guaranteed through certain forms of induction. This is the mechanism of recursion. To gain a better understanding of induction and recursion, in my opinion, a better way is to study ordinal arithmetic, a fundamental and essential concept in Set Theory. Because Ordinals are abstract enough to necessitate a departure from our intuition and prior experiences, compelling us to contemplate the underlying general principles.

Answer 3

The point here is that we don't have to assume that the numbers come to us already imbued with an additive structure. All we need is for the numbers to have the much simpler notion of “successor”:

0. There is a first number.
1. Each number has a following number.

From this tiny amount of structure, we can construct the whole idea of arithmetic addition. Any group of things with those two properties must include a thing that behaves like $2$, and a thing that behaves like $4$, and that do $2+2 =4$. It must also have $197 + 313 = 510$. You get all this just from those two properties.

Russell is confusing you by using the standard notation with “$0$” and “$n+1$” for these abstract ideas (“first number” and “number following $n$”) without showing that they really match up. He's combined two steps that should be separate.

• First, any system with a “first number” and “each number has a following number” can be shown to have a notion something like addition, based just on those two simpler ideas
• And if we do that with the ordinary numbers, taking the “first number” to be $0$, and “the number following $n$” to be $n+1$, then the addition-like thing we get is actually regular addition

Why do we do this at all? Because it allow us to understand addition as a result of simpler properties. For example, we expect that addition should the property that $n+m $ always equals $m+n$. We know this is true for ordinary addition of numbers. But is is a consequence of same the tiny bit of structure? Or does it come from somewhere else? We can show that actually no more properties are needed: any group of things with those two properties not only has something like addition, but that something is commutative.

In contrast, though: ordinary numbers have a “$<$” relation and it's always true that $a < a+b+1$. Can we get that from the two simple properties above? It turns out we can't, we need something extra! Isn't that interesting?

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Another way this turns out to be useful is that it lets us model addition in systems that are very unlike ordinary numbers. For example, suppose we are studying computation, and we have a simplified model of a computer. We would like to show that this model can perform all the computations that real computers can, such as additions. If we can show that the model can represent zero, and successors, and manipulate them in a few simple ways, Russell's construction gives a recipe for how the model can do addition.

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In the comment you asked:

All I see is that Russell assumes addition when he uses n+1. Calling n+1 “successor” or “number following n” does change the fact that addition is assumed. Can you explain how addition follows from your two initial assumption?

Let's undo the confusion by using different notation. Let's write $Z$ for the first number and $S(n)$ for the number after $n$. Notice that $S$ is not addition, it is simpler and more limited than addition. Addition combines any two numbers. $S$ just tells you the number that comes after any single number.

We define $a+b$ this way: Either $b$ is the first number, or it comes after some other number.

• If $b$ is the first number, $Z$, then $a+b = a$.
• If $b$ comes after some other number $p$, so that $b=S(p)$, then $a+b = S(a+p)$.

If we define “$+$” this way, we can prove that it has all the properties we expect addition to have. For example: $0+n = n+0$; $p+(q+r) = (p+q) +r$; $2+2=4$; $197+313=510$.

Notice that Russell has actually given this definition, but has confused you by using “$0$” to mean the first number and “$n+1$” to mean the number after $n$. By writing $n+1$ for the number after $n$, Russell has made it look like he is using addition to define addition. But he isn't, he is using the simpler and more limited operation of finding the next number after $n$.