How to find what a definition defines?

Question

Are these definitions of the "+" and "mod" operators?

$m+0=m$........(1)

$0 \;\text{mod} \;2 = 0$.....(2)

To me, (1) defines the identity property of zero and (2) defines zero as an even number.

In my previous question Is Russell's proof of addition with Peano's 5. Axiom valid? several commentors read (1) as the definition of the "+" operator. Is there a rigorous mathematical mechanism to decide if these are definitions and if they are, what they define?

To me a definition of "+​" should be something like this:

Definition: "+" is a binary operator that takes two bundles of units and combines them into one bundle.

Since, (1) or (2) are not in this definition format, they do not define operators but use them.

Are there formal rules to decide a question like this?

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Russell's original quote:

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.

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Edit: I got this message: Your question has been identified as a possible duplicate of another question. If the answers there do not address your problem, please edit to explain in detail the parts of your question that are unique.

I already linked above to the question that was suggested as duplicate. It is not duplicate because I'm asking a different question here. All I want to understand is: If $m+0=m$ is a definition of addition. To me this expression uses the "+" sign so it cannot be a definition of addition. Definition with the equality sign must have this form: [What is to be defined] = [The definition of what is to be defined]. $m+0=m$ does not have this form. As explained in the original question, $m+0=m$ is simply the first step in Russell's use of mathematical induction.

Answer 1

$+$ is overloaded and has many different definitions depending on context. For example addition is defined differently for the naturals, integers, rational, real and complex numbers although we rarely draw that distinction when working with them. They're typically associative and mostly commutative binary operators with string addition being a common non-commutative exception. In algebraic settings addition typically signifies an Abelian group.

However modulo is not a binary operation, rather modulo $2$ is a unary operation. The $2$ in this case is not an input, rather a property of the function. So it's more accurate to think of modulo as as collection of functions indexed by the natural numbers which return the remainder when divided by their index. In the usual function notation it may look like $m_n(x)= x \mod n$ in the same way we might write $p_n(x)=x^n$ to index monomials.

The reason for this is we get a rich algebraic structure if we keep the modulo fixed most of our intuition about how numbers works carries over since they'll form a ring. When studying rings the Chinese remainder theorem gives us a way to examine how different modulo act in concert. So there are some nice results you can work to using this method which don't require you to mix modulo explicitly. If you're still unsure why we don't consider modulo a binary operation examine some of the properties it might have such as commutativity and associativity if it was. It should provide some insight into why we do it this way.

Answer 2

Short answer

I think the OP might be confused by issues related to the semantics of English words and the pragmatics of how the English language is used, rather than any genuine mathematical issue.

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Long answer

Are these definitions?

Are these definitions of the "+" and "mod" operators?

Of course not. No one would say so*. For instance, $m+0=m$ doesn't say anything about additions where both numbers are nonzero. And $0\mod 2=0$ doesn't say anything about what happens when you take nonzero numbers modulo $2$.

*For instance, when you thought that Henry was saying so in the comment thread on that other question and you said "I guess by 'this definition' you mean $m+0=m$", Henry was correcting you when saying "$m+0=m$ is part of the definition you quote for $+$".

What would a definition look like?

Definition with the equality sign must have this form: [What is to be defined] = [The definition of what is to be defined]

For this format (which isn't always followed), a definition of addition on the nonnegative integers could be written as something similar to the following:

"addition $=$ [the binary operation written with the symbol '$+$' that satisfies $\forall m:\,m+0=m$ and $\forall n,m:\,n+s(m)=s(n+m)$]."

Then why do people say these things?

The main sentence/phrasing of contention

we define $m+0$ as $m$, and $m+s(n)$ as the successor of $m+n$.

"We define X as Y" is used in mathematical English not just for general cases like "we define addition as the binary operation that...", but also for partial/special cases, just as in this quote.

In context, if you prefer, you could think of "We define $m+0$ as $m$" to mean something like "we define $+$ to be a binary operation satisfying certain properties, and for expressions of the form $m+0$, it outputs $m$ (where the universal quantifier is implied as it often is in mathematics)".

Using the symbol

To me this expression uses the "+" sign so it cannot be a definition of addition.

That simply does not match how English is used in mathematics.

For another example, the following is completely typical mathematical English:

We define the (unnormalized) sinc function by $\operatorname{sinc}x = \dfrac{\sin x}{x}$ for nonzero $x$ and $\operatorname{sinc}0=1$.

The meaning is essentially the same as:

$\operatorname{sinc}=\left(x\mapsto\begin{cases} \dfrac{ \sin x } x, & x \ne 0 \\[2px] 1, & x = 0\end{cases}\right)$

Answer 3

I decided to answer my own question combining all I learned from comments and answers.

Russell uses mathematical induction

1. To prove by mathematical induction that a property $P(n)$ is true for $\mathbb{N}$,

   (1) Prove the base case, $P(0)$,

   (2) Assume that $P(n)$ is true,

   (3) Assuming that $P(n)$ is true prove that $P(n+1)$ is true, then,

   $P(n)$ will be true for all numbers in $\mathbb{N}$.

2. In Russell's case (original quote is in the question), his statement (1) $m+0=m$ is his definition for the base case, the first step in mathematical induction.

3. Russell then states his definition for part two of mathematical induction as (2), $m+n+1 = m+n+1$.

4. So, (1) is the base case, that is, it is a statement about a property of zero, it is $P(0) = m+0=m$.

Russell never defines addition

5. Russell is stating his base case for mathematical induction. He is stating a property which is true for zero as required by induction. He is not defining addition. It makes no sense to define addition as the base case as a property of zero.

What is Russell's property $P(n)$ ?

6. It is possible to read $m+0=m$ two ways: (1) as the identity property of zero: When you add a number to zero you get that number unchanged. But this cannot be Russell's intention because in this case, $P(n)$ would not hold because only zero has the identity property on addition, other numbers don't.

7. Therefore, the correct reading is that (2) zero is a number that has the property to enter into addition operation.

8. Therefore, Russell assumes addition and uses addition to state that addition works with zero. Russell never defines addition. He only proves by mathematical induction that addition is valid for $\mathbb{N}$.

We already have a proper definition of addition

9. We already have an established definition of addition: $m+n=m+n$.

10. Addition takes two bundles of units $m$, $n$ and combines them into one bundle. We express this fact by writing $2+3 = 2+3$ as $2+3=5$.

11. By writing $5$ on the RHS instead of $2+3$, we indicate that we combined two bundles of units $m=\mid \mid$ and $n=\mid \mid \mid$ into one bundle $\mid \mid \mid \mid \mid$.

12. So Russell's definition $m+0=m$ cannot be a definition of addition for two reason: (1) Russell is using mathematical induction and he must state a property of zero, (2) $m+0=m$ does not conform to the definition of addition. If we want to add zero to a number $m$ we need to write $m+0=m+0$, this is correct. If define addition as $m+0=m$, addition will not work, e.g., $2+3 = 2$ is wrong.

13. Also, Russell's second step in his induction proof is just the definition of addition: $m+n=m+n$.

14. It makes no sense to write $m+n=m+n$ as $(m+n)+1 = m +(n+1)$ because addition is associative $(m+n)+1 = m +(n+1)$ and $m+n+1 = m +n+1$ are identical statements: $[(m+n)+1 = m +(n+1)] \equiv [m+n+1 = m +n+1]$.

15. Writing "+1"s on both sides of the definition is silly because it contributes no new information. Eliminating "+1"s from both sides we end up with $m+n=m+n$ which is the proper definition of addition.

16. If we have such a simple and established definition of addition as $m+n=m+n$ what is the point of another, more complex definition of addition in terms zero and successor? I think the problem here is that Russell rejects Euclid's definition of number as multitude of units, instead he defines number without units as a one-to-one relation between sets. But since Russell cannot use his unitless abstract numbers in arithmetical operations such as addition, he secretly continues to use Euclid's numbers made of units.

17. So I think Russell's attempt to prove that addition is valid for $\mathbb{N}$ is to give an example of mathematical induction.

18. There are several fundamental questions here. We can ask, for instance, if addition needs to be defined. Euclid never defined addition.