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I was going over counting and letters with my students by starting at a letter and counting how many moves it takes to get to another.

That got me thinking. Could addition between an integer, $x$, and a letter, $Γ$, (in the context of the English alphabet) be commutative? That is, does $Γ+x=x+Γ$, assuming that $|x|$ is never greater than the number of letters that precede or proceed $Γ$?

$A+1=B$, $F+3=I$, $S+(-7)=S-7=L$, and so on. It seems okay giving such equations because the first operand lets one know where to start and the second gives how many moves to make and in what direction. However, $1+A=B$ seems off.

I understand that commutativity has to do with the operation. Me using a number and a valueless letter as operands within the same expression certainly doesn't seem to allow for the commutativity of addition to apply. And if that's the case, it doesn't seem that $A+1$ is a valid expression. Maybe there's another way I could be looking at this?

Eugene Mason IV
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    Before we can ask anything about addition between a letter and a number we have to define what it means. And how we define what it means determines if it's commutative. If we define $n + \Gamma = \Gamma + n$ as the letter (if any) is $n$ above $\Gamma$ then of course it is commutative (but not actually complet). If we define $\Gamma +n$ as the letter that is $n$ above $\Gamma$ and define $n+\Gamma$ as the number that is the cardinality of $\Gamma$ above $n$ then it is not commutative ans $\Gamma + n$ is a letter and $n + \Gamma$ is a number. But... to be continued.... – fleablood Dec 05 '22 at 05:04
  • If we formalize your question in the obvious way (mapping ${1,2,3,\cdots}\to{A,B,C,\cdots}$), then the answer is simply yes, because this is essentially addition of integers, just with different symbols. – Graviton Dec 05 '22 at 05:05
  • But as it is must mathematicians would say addition between numbers and letters simply isn't defined at all. – fleablood Dec 05 '22 at 05:05
  • "Yes, because this is essentially addition of integers, just with different symbols." This depends on what "essentially" means. The OP is certainly think that the letters are NOT numbers but are actually letters. We could argue, that math such would induce would be essentially equivalent to letters just being numbers with different symbols but we could just as easily define the math as set operation on $\mathbb Z\cup {\text{Latin Alphabet}}$ where addition is defined as a non-commutative, non-closed binary operation. – fleablood Dec 05 '22 at 05:10
  • "However, 1+A=B seems off." WHy? It's your system and your definition. You can make it mean whatever you want and make it as off or on as you want. If it were me I'd so the whole thing is off and say that Letters are just a set that can be indexed by numbers and not numbers and so adding numbers to letters is not addition. However I would allow $f:LETTERS\times \mathbb N\to LETTERS$ via $f(\Gamma, n)=$ the letter indexed by the index of $\Gamma$ plus $n$. THis is not commutative (as most binary functions aren't). – fleablood Dec 05 '22 at 05:16
  • @fleablood Thanks for the input! Would it help if the number and letter were involved in some contrived operation? Say $1·A=B$, where the 'spot' operation $(·)$ takes the letter and number to produce the appropriate letter, regardless of what order the operands are in? That way I'm not cheapening addition. – Eugene Mason IV Dec 05 '22 at 05:40
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    You can define anything anyway you want. But what I think you are intuively doing is doing operations on the index of the letters. – fleablood Dec 05 '22 at 05:43
  • @fleablood I see what you're saying, but even with the $\mathbb{Z}\cup\text{Latin Alphabet}$ approach, everywhere $a+b$ is defined will still be isomorphic to $(\mathbb{Z},+)$. This is what I mean by essentially equivalent. Mind you, we are dropping properties like closure, of course. – Graviton Dec 05 '22 at 06:21

2 Answers2

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I'd say you aren't actually adding numbers and letters together when you do that.

I'd say you are treating the alphabet as an indexed set. Which is to say you have a bijection $a$ from $\mathbb J_{26} = \{1,2,3,.....,26\}$ to the set $\mathbb{ALPHABET}=\{A,B,C,....,Z\}$ where $a(1) =a_1 = A,$ and $a(2) = a_2 = B$ and so on.

You have defined a binary relation on $\mathbb{ALPHABET}\times \mathbb J_{26}$ where $f(\Gamma, n) = a_{a^{-1}(\Gamma) + n}$.

As such it is not commutative as not only is $f(\Gamma, n) \ne f(n, \Gamma)$ but $f(n,\Gamma)$ is not even defined as the relation is on the set $\mathbb{ALPHABET}\times \mathbb J_{26}$ which is a completely different set than the set $\mathbb J_{26}\times \mathbb{ALPHABET}$. $(\Gamma F, n) \in \mathbb{ALPHABET}\times \mathbb J_{26}$ but $(n,\Gamma)\not \in \mathbb{ALPHABET}\times \mathbb J_{26}$

But that would be my definition.

Another person could say what you are doing is creating a set $K=\mathbb Z \cup \mathbb{ALPHABET}$ and defining a binary operation on $K$ as $a \bigoplus b =\begin{cases} a+b& a,b \in \mathbb Z\\\text{the }b\text{th letter past }a& a\in \mathbb {ALPHABET}, b\in \mathbb Z\\\text{the }a\text{th letter past }b& a\in \mathbb Z, b\in \mathbb {ALPHABET}\\\text{the letter that is the }a\text{'s position past }b&a,b\in \mathbb{ALPHABET} \end{cases}$

If so the $\bigoplus$ is commutative and $1+A = A+1=A+A = B$.

But of course we could define things any way we want. Maybe we define

$a \bigoplus b =\begin{cases} a+b& a,b \in \mathbb Z\\\text{the }b\text{th letter past }a& a\in \mathbb {ALPHABET}, b\in \mathbb Z\\ a + \text{the position of }b& a\in \mathbb Z, b\in \mathbb {ALPHABET}\\\text{the letter that is the }a\text{'s position past }b&a,b\in \mathbb{ALPHABET} \end{cases}$

In which case $\bigoplus$ is not commutative because $\Gamma + n$ is a letter but $n + \Gamma$ is a number. We'd have $1+1=2$ $A+1=B$ and $1+A =2$ and $A+A= B$.

Or maybe we define things differently. But untill we decide just what is is that we are doing, the question doesn't yet make sense or has a definitive answer.

fleablood
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In the context you present, it looks like the sum of a number $n$ to a letter $\Gamma$ is defined as the letter $n$ positions to the left of $\Gamma$ in the sequence of letters we know as the English Alphabet $(A,B,C,..,Z)$.

A real-world analog to this can be seen in programming (which may be something nice to look into in a classroom environment). Under ASCII encoding (which can be seen as a bijective map $f:C \to \{0,...,127\}$, where $C$ is a set of often used characters), programmers can do things as you suggest, such as use 'A' + 1 and 1 + 'A' to denote the character B.

I'm curious as to how you define "rollover", though. What is Z + 1? Is it A? A programming language would not assume so.

Of course, as @fleablood writes in their answer, this isn't really adding numbers to letters, just adding numbers to numbers which can happen to be mapped to the alphabet.

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