Is every integer a unary operation? I have been thinking about mathematical operations, and I was trying to ponder what an operation simpler than addition would be. I started thinking about what each operation did (exponents: repeated multiplication, multiplication: repeated addition). I figured that addition was repeated something, and I came up with a unary operation, 4 = 1 + 1 + 1 + 1, where each number is the input, and its value is the output. That would mean that our number system's numerals are just shorthand for repeated addition of one. (To extend this to negative numbers, we could define this unary function's opposite: -4 = -1 - 1 - 1 - 1.) I am an eighth grader taking high school geometry, and this sort of shocked me. Am I totally wrong, or is every integer just the value of a unary operation: itself?
 A: No, an integer is not a unary operation, it is a $0$-ary operation. A unary operation is essentially a function, it requires one input, for example, negation $x\to-x$ is a unary operation. The operation "5" requires zero operands, it has a constant value.
On the other hand, as pointed out in the other answers, the successor operation $x\to x+1$ is unary, and addition is repeated succession.
A: You've discovered Church numerals. Almost.
In Church encoding, numbers are in fact defined as operations. Actually, a bit more involved: the $n$-th Church numeral is a function that takes a function, and returns the $n$-fold composition of that function with itself. Written in lambda calculus, the first few numbers look thus:
$$
  0 = \backslash f\ x\mapsto x
$$
$$
  1 = \backslash f\ x\mapsto f\,x
$$
$$
  2 = \backslash f\ x\mapsto f(f\,x)
$$
$$
  3 = \backslash f\ x\mapsto f\bigl(f(f\,x)\bigr)
$$
etc. In other branches of maths, these would be written rather something like
$$
  0(f,x) = x
$$
$$
  1(f,x) = f(x)
$$
$$
  2(f,x) = f\bigl(f(x))
$$
$$
  \ldots
$$
Now you might ask, are these the natural numbers, or is it just some particular encoding for the naturals? But that's a philosophical question. The church numerals are a model satisfying the Peano axioms, so basically whenever you see numbers used it's possible to put in these definitions, it won't change the meaning.
A: Yes, you may see this in some more advanced course.  Say $S$ is the unary operation "add 1".  Then addition may be defined recursively in terms of $S$.
$$
x+0 := x;
\\
x+Sy := S(x+y)
$$  
Why not define multiplication next?
$$
x*0 := 0
\\
x*Sy := (x*y)+x
$$
Can you do powers $x^y$?
Look for "Peano's axioms" for more.
A: While is is true that to every integer $k$ one can associate a unary operation $n\mapsto n+k$ of adding$~k$ (which you can see as a function $\Bbb Z\to\Bbb Z$, or in case $k\in\Bbb N$ as a function $\Bbb N\to\Bbb N$) and this correspondence is injective (different integers give different unary operations), it would be wrong to define integers as such operations, and fairly confusing even to identify integers with such operations.
The problem is, these operations have to act on something, namely on integers, but if you want to define the integers as those operations, the integers that they must act upon have not been defined yet. So you end up with a circular definition.
You might want to define integers first, and then the operations on the integers, and then identify $k$ with the operation $n\mapsto n+k$. This would be similar to how one defines the integers, then the rational numbers in therms of them (a somewhat subtle affair), after which (for the purposes of arithmetic) the integer $k$ is identified with the rational number$~\frac k1$. This is not logically inconsistent, and would allow you to write $1(1)=2$ or even $1\circ 1=2$ instead of $1+1=2$, but there is not too much to be gained by this, and it would be less clear than what is usually done. Also it is fairly arbitrary: why identify $k$ with $n\mapsto n+k$ rather than with $n\mapsto nk$ o something else yet?
A: No, it is incorrect to say "an integer is a unary operation". What the other answers so far are actually saying instead is that any integer is the result of repeatedly applying a unary operation, which is quite different.
The true statement I think you are after is that any given integer is (or can be identified with) a nullary operation, specifically the operation taking no arguments and returning itself. See this math.SE thread for example.
A: The Peano axioms start with a unary operation called $S$ and intended as successor.  In fact we define $1=S(0), 2=S(S(0)),$ etc.  as $1,2,$ etc. are not part of the language.  Then there are axioms that define addition and multiplication to make them work the way they should.
