Kdtyjb

The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$

You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.

In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$

Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$

In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".

There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.

One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.

Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.

With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.

So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.

Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.

Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:

$$a = b + 1lambda + 2(1-lambda)$$

where

$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$

You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.

Knowing that

$$frac{|c|}{c} in {1,-1}$$

we can calculate the $lambda$ as follows:

$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$

Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:

$$dlambda + e(1-lambda)$$

which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.

There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).

It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.

Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.

I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.

As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$

It can be viewed as inline function:

$$f(c)=begin {cases} 1&c gt 0\
2& c le 0 end {cases}$$

so it comes as:

$$a = b + f(c)$$
used in everyday mathematics.