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Does there exists any form of Algebra where operators can be assumed as variables?

For example: $$ 1+2\times3=7 $$ can be considered as: $$ 1\:(\mathrm{\,X})\:2\:(\mathrm{Y})\:3=7 $$ ?

TheVal
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  • Let $G$ be a finite set of operations, then you can define a group structure on $G$ and if you want you can add this operation to $G$. If the set is infinite, I guess you can do something still but it is harder. – aeyalcinoglu Aug 11 '13 at 19:52
  • Usually, when we do represent them this way, we have to make sure we know what order things are in. You can have $1X2Y3$ mean $1+(2\times 3)$ when $X=+, Y=\times$ and $(1\times 2)+3$ when $X=\times$ and $Y=+$. $1X2Y3$ needs to be assumed to be one order of operations, $(1X2)Y3$ usually. – Thomas Andrews Aug 11 '13 at 19:58
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    And it is rare to write them this way. Usually, we write in functional notaion: $Y(X(1,2),3)$. Essentially, operations are just functions, and we use variable notation for functions all the time. – Thomas Andrews Aug 11 '13 at 20:01
  • @KonformistLiberal How do you define a group structure on the set ${+,\times}$? I think you've misunderstood the question. – Thomas Andrews Aug 11 '13 at 20:02
  • @ThomasAndrews: Our operation is $f$ and $f(+,\times) = \times , f(+,+) = +, f(\times,+) = \times, f(\times,\times) = +$. – aeyalcinoglu Aug 11 '13 at 20:06
  • And that gives you something useful? You can define group operations on any set, but what do those group operations mean? Why is $+$ the identity element and not $\times$, for example? – Thomas Andrews Aug 11 '13 at 20:08
  • @ThomasAndrews: Should it mean something? I did algebra with operations. – aeyalcinoglu Aug 11 '13 at 20:10
  • Well, the OP is asking if we ever study something. Your group structure doesn't "study" those operators, except as an arbitrary set. So, yeah, it matters if what you defined doesn't in any way relate to what the OP was asking about. – Thomas Andrews Aug 11 '13 at 20:13
  • @KonformistLiberal But the other reason it matters is that there are operator groups, such as the group of invertible square matrices, which are unary operators on a vector space. The risk is that you confuse the poor OP with something you made up and which no mathematician would possibly study. – Thomas Andrews Aug 11 '13 at 20:17
  • @ThomasAndrews: If I would think that my answer is sufficient, I wouldn't comment, I would answer. So, it was an approach, not a whole answer, not wrong. OP asked for existence, I indicate something. What is wrong with that? – aeyalcinoglu Aug 11 '13 at 20:25
  • It wasn't just insufficient, it was misleading, confusing, and unhelpful. @KonformistLiberal – Thomas Andrews Aug 11 '13 at 20:46

2 Answers2

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I would say the answer is most definitely yes, if you construe the operators to be functions. For example, we could represent the "operator" $+$ (on $\mathbb{Z}$, the integers, for example) to be the function $p: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ given by $p(z_1, z_2) = z_1 + z_2$ for $(z_1, z_2) \in \mathbb{Z} \times \mathbb{Z}$; likewise the "operator" * would be a function $t: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ such that $t(z_1, z_2) = z_1*z_2$; with these conventions, the equation $1 + 2*3 = 7$ would be written $p(1, t(2, 3)) = 7$. Now observe that $p$ and $t$ are "merely" elements of the set $\mathcal{F}$ of functions from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Z}$; letting $f_1, f_2$ denote variables ranging over the set $\mathcal{F}$, we see that our expression $p(1, t(2, 3))$ in fact corresponds to letting $f_1 = p$ and $f_2 = t$; clearly many other such choices are possible, though most of them would yield an $f_1(1, f_2(2, 3))$ which is more difficult to compute! But in fact a similar paradigm occurs over and over again in certain areas of mathematics.

One problem which might be encountered in this approach is that of parsing an expression like $1(X)2(Y)3$ into appropriate functional form, how to group the elements and so forth. But I'll leave that discussion for future research and questions!

Hope this helps! Cheers, Bob Lewis.

Robert Lewis
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Absolutely! In abstract algebra, you study general operations, besides $+$ and $\times$, that satisfy certain identities. An example of an identity is your equation $1\mathop X (2\mathop{Y}3)=7$. This equation constrains the variables $X$ and $Y$, subject to the constants 1, 2, 3, and 7. So you could say that $X={+}$ and $Y={\times}$ is a "solution" to the equation, whereas $X={\times}$ and $Y={+}$ is not a solution. However, unless you're working with differential equations, I don't think that terminology is very common. Usually you would give a name to a kind of operation that satisfies a given identity. For example, an operation $\circ$ is called commutative if $a\circ b = b\circ a$ for all $a$ and $b$. Then you would say that addition is commutative, but subtraction is not commutative.

In universal algebra, you go even further and treat the identities themselves as variables! From this viewpoint, you can consider all possible sets of identities and their interrelationships. I'm horribly rusty, so please don't ask me to expand on this idea. :)

Chris Culter
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