Is there a math equivalent of the ternary conditional operator as used in programming?

a = b + (c > 0 ? 1 : 2)

The above means that if $c$ is greater than $0$ then $a = b + 1$, otherwise $a = b + 2$.

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    $\begingroup$ @Alex $a = b + 2 - u(c)$ $\endgroup$ – eyeballfrog Mar 5 '19 at 5:45
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    $\begingroup$ A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer). $\endgroup$ – Winther Mar 5 '19 at 9:04
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    $\begingroup$ In C-derived languages at least, you have made a grievous error! Your expression parses as a = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary. $\endgroup$ – TonyK Mar 5 '19 at 11:38
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    $\begingroup$ Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics? $\endgroup$ – Konrad Rudolph Mar 5 '19 at 15:21
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    $\begingroup$ What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator. $\endgroup$ – user Mar 5 '19 at 15:44

14 Answers 14


From physics, I'm used to seeing the Kronecker delta,$$ {\delta}_{ij} \equiv \left\{ \begin{array}{lll} 1 &\text{if} & i=j \\ 0 &\text{else} \end{array} \right. _{,} $$and I think people who work with it find the slightly generalized notation$$ {\delta}_{\left[\text{condition}\right]} \equiv \left\{ \begin{array}{lll} 1 &\text{if} & \left[\text{condition}\right] \\ 0 &\text{else} \end{array} \right. $$to be pretty natural to them.

So, I tend to use $\delta_{\left[\text{condition}\right]}$ for a lot of things. Just seems so simple and well-understood.


  1. Basic Kronecker delta:
    To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$ \delta_{ij} \Rightarrow \delta_{i=j} \,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.

  2. Conditional operator:
    The "conditional operator" or "ternary operator" for the simple case of ?1:0:$$ \begin{array}{ccccc} \boxed{ \begin{array}{l} \texttt{if}~\left(\texttt{condition}\right) \\ \{ \\ ~~~~\texttt{return 1;} \\ \} \\ \texttt{else} \\ \{ \\ ~~~~\texttt{return 0;} \\ \} \end{array} ~} & \Rightarrow & \boxed{~ \texttt{condition ? 1 : 0} ~} & \Rightarrow & \delta_{\left[\text{condition}\right]} \end{array} _{.} $$Then if you want a non-zero value for the false-case, you'd just add another Kronecker delta, $\delta_{\operatorname{NOT}\left(\left[\text{condition}\right]\right)} ,$ e.g. $\delta_{i \neq j} .$

  3. Indicator function:
    @SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$ \begin{array}{ccccc} \underbrace{a=b+1+\mathbb{1}_{(-\infty, 0]}(c)} _{\text{their example}} & \Rightarrow & \underbrace{a=b+1+ \delta_{c \in \left(-\infty, 0\right]}} _{\text{direct translation}} & \Rightarrow & \underbrace{a=b+1+ \delta_{c \, {\small{\leq}} \, 0}} _{\text{cleaner form}} \end{array} \,. $$

  4. Iverson bracket:
    Iverson bracket notation, as suggested in @FredH's answer, is very similar to the generalized Kronekcer delta. For example:$$ \delta_{i=j} ~~ \Rightarrow ~~ \left[i = j \right] \,.$$Dropping the $`` \delta "$ reduces backwards-compatibility withe the basic Kroncker delta, plus it weakens the signal about what the notation means, so it's probably not as good in general contexts right now. However, Iverson bracket notation should be easier to read and write, so when reinforcing the meaning of the notation isn't a big issue, it could be preferable.

Note: "Conditional operator" rather than "ternary operator".

The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).

Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.

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    $\begingroup$ wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need. $\endgroup$ – dataphile Mar 5 '19 at 15:05
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    $\begingroup$ The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands. $\endgroup$ – Konrad Rudolph Mar 5 '19 at 15:22
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    $\begingroup$ The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain $\{0,\,1\}$, wraping the Iverson bracket. $\endgroup$ – J.G. Mar 5 '19 at 17:17
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    $\begingroup$ Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective. $\endgroup$ – Nat Mar 6 '19 at 21:32
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    $\begingroup$ Finally someone who does't call ?: "the ternary operator". Conditional operator doesn't describe it properly though. It's not an operation that's coditional, but one that selects a value based on a condition. For that reason, I always call it the "selection operator", as that's what it does: based on the condition, select either option A or option B. $\endgroup$ – Clearer Mar 8 '19 at 13:32

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.

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    $\begingroup$ Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2. $\endgroup$ – Džuris Mar 5 '19 at 9:40
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    $\begingroup$ @Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator. $\endgroup$ – frabala Mar 5 '19 at 10:08
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    $\begingroup$ Once again, clarity beats compactness when it comes to mathematics. $\endgroup$ – Asaf Karagila Mar 5 '19 at 12:25
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    $\begingroup$ Well-chosen compact notations can contribute to clarity rather than detract from it, though. $\endgroup$ – hmakholm left over Monica Mar 5 '19 at 12:34
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    $\begingroup$ @Džuris No it isn’t. The link you cite says, in its very first sentence, that “?: is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow). $\endgroup$ – Konrad Rudolph Mar 5 '19 at 15:25

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 $$ a = b + 2 - [c \gt 0]. $$

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    $\begingroup$ aka en.wikipedia.org/wiki/Iverson_bracket $\endgroup$ – qwr Mar 5 '19 at 6:50
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    $\begingroup$ I would find it clearer to write $a = b + 1[c>0] + 2[c\le0]$. $\endgroup$ – user856 Mar 5 '19 at 7:00
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    $\begingroup$ I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation. $\endgroup$ – J. M. isn't a mathematician Mar 5 '19 at 12:05
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    $\begingroup$ Surely this is a unary operator, which works much like for example ! in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression. $\endgroup$ – pipe Mar 5 '19 at 14:46
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    $\begingroup$ Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket". $\endgroup$ – safkan Mar 7 '19 at 6:49

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

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    $\begingroup$ Indicator is definitely the way to go, since the conditional can define an arbitrary set. $\endgroup$ – eyeballfrog Mar 5 '19 at 5:47
  • $\begingroup$ Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios. $\endgroup$ – dataphile Mar 5 '19 at 5:52
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    $\begingroup$ I often use - and have seen others use - $\mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$ $\endgroup$ – Christopher Mar 5 '19 at 13:26
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    $\begingroup$ The indicator function isn’t a ternary operator, it only has two operands. $\endgroup$ – Konrad Rudolph Mar 5 '19 at 15:23
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    $\begingroup$ I used the wikipedia definition, $\mathbb{1}_A(x)$ where $A$ is a set. $\endgroup$ – Siong Thye Goh Mar 7 '19 at 11:53

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.


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 + \[c\not>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 $b+(c>0\mid 1\mid 2)$ 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 $$ a_n=(n\bmod 6+1\mid 0,1,1,0,-1,-1) $$ (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 $$ ( m \mid 31,(y\bmod 4=0\land y\bmod 100\neq0\lor y\bmod400=0\mid 29\mid 28) ,31,30,31,30,31,31,30,31,30,31). $$

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    $\begingroup$ So counting months starting from 1 is a mistake? Using matrix indiices starting from 1 (much de rigeur in mathematics) is a mistake? I think the choice made in Algol 68 is based on the very basic concept of counting and cannot be dismissed in such a cavalier way. $\endgroup$ – Albert van der Horst Aug 17 '19 at 0:47
  • $\begingroup$ @AlbertvanderHorst It is of course just an opinion, but yes I do think the using any other number than $0$ (and in particular using $1$) as default starting value for counting is a mistake (I never mentioned counting months in particular, but the general judgement applies). In the sense that it has unnecessary disadvantages (which outweigh and advantages it might have). I believe this is true both in programming and in mathematics, though it is nowadays more accepted in the former. Having $1$-based indexing costs machine cycles, either at runtime, or (with a smart compiler) at compile time. $\endgroup$ – Marc van Leeuwen Aug 18 '19 at 8:51
  • $\begingroup$ ... As for mathematics, there are numerous advantages for indexing from $0$ as a general rule (with exceptions, like for the sequence $(\frac1n)_{n\geq1}$), even though this is not the current convention. I know this from experience as a combinatorist, where I practice indexing from $0$ everywhere, in spite of the drawback of deviating from literature. As an example. consider a Vandermonde matrix; indexing matrices starting from $1$ causes awkwardness in defining them that just does not happen when indexing from $0$. But I do not want to embark on this discussion in these comments. $\endgroup$ – Marc van Leeuwen Aug 18 '19 at 8:59
  • $\begingroup$ Good luck with introducing that the general public."I play first fiddle in the London Philharmonic Orchestra." "You main there is some one playing the zeroth fiddle who is more senior". "Why the hell do you thing that???? " "We mathematicians think that if you play first fiddle you actually mean second fiddle." $\endgroup$ – Albert van der Horst Sep 19 '19 at 18:23

The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.

Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=\begin {cases} y&x\neq 0\\ z & x=0 \end {cases}$$

Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.

About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.


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.


Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or $$a=\begin{cases}b+1 &\text{if }c>0\\b+2 &\text{otherwise}\end{cases}$$ is much harder to read than either of those two things.

  • $\begingroup$ The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion. $\endgroup$ – dataphile Mar 7 '19 at 13:59
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    $\begingroup$ Most of any mathematics paper is natural language. There's a reason for that. $\endgroup$ – David Richerby Mar 7 '19 at 15:05

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.

  • $\begingroup$ "in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 or INT_MAX when that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used (x>1) as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard. $\endgroup$ – user Mar 5 '19 at 15:55
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    $\begingroup$ @aCVn The C11 standard states, “Each of the operators < (less than), > (greater than), <= (less than or equal to), and >= (greater than or equal to) shall yield 1 if the specified relation is true and 0 if it is false.” Similarly for equality and inequality, “Each of the operators yields 1 if the specified relation is true and 0 if it is false.” $\endgroup$ – Davislor Mar 5 '19 at 16:31
  • $\begingroup$ @aCVn What you wrote is true, but the relational operators in C will only return the values 1 or 0. $\endgroup$ – Davislor Mar 5 '19 at 16:44
  • $\begingroup$ @aCvn There is nothing wrong with counting how many objects are greater than 10 using n += a[i]>10; . If you don't understand this idiom, you're just a poor c-programmer. $\endgroup$ – Albert van der Horst Aug 17 '19 at 1:02
  • $\begingroup$ @AlbertvanderHorst Let’s please be kind. $\endgroup$ – Davislor Aug 17 '19 at 1:40

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:

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


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

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:

$$d\lambda + e(1-\lambda)$$

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

  • $\begingroup$ And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um... $\endgroup$ – David Richerby Mar 7 '19 at 11:48

The best idea is probably to split the world into different cases above the context in which your expression lives, so you consider the cases where $c$ is positive and the cases where $c$ is nonpositive separately. Or impose conditions on $c$ like $|c| = 1$ .

Another alternative is to create some new ad hoc indicator-like functions whose algebraic properties are as strong as possible. I'm partial to signum because it's odd and multiplicative.

b + (c > 0 ? 1 : 2)

Can be written nicely using two new functions $S$ for signum and $Z$ the zero indicator function:

$$ S(x) \stackrel{\small{\text{def}}}{=} \begin{cases} \;\;1 & x > 0 \\ \;\;0 & x = 0 \\ -1 & x < 0 \end{cases}$$

And $Z$ is $1$ when its argument is $0$ and $0$ otherwise.

So we can write the expression as

$$ b + \frac{3}{2} + \frac{1}{2} S(c) - \frac{1}{2} Z(c) $$


$$ b + 1 + S(c) + \frac{1}{2} S^2(c) $$

with $S^2(c)$ denoting $S(c^2)$ or $S(c)^2$ since they're equivalent.

This expression isn't that great in this case, but $S$ and $Z$ have some nice properties:

$$ S(a)S(b) = S(ab) $$ $$ S(a)Z(a) = 0 $$ $$ Z(a) = 1 - S(a)^2 $$ $$ S(S(a)) = S(a) $$ $$ S(a)^n = S(a^n) \;\;\;\;\forall n \in \mathbb{N} $$ $$ S(a)^m = S(a)^m \cdot S(a)^{(2n)} \;\;\;\;\forall m, n \in \mathbb{N} $$


The concepts from lambda calculus and combinatory logic is worth mentioning here.

We define lambda calculus by two constructions: - you can make a function using abstraction, e.g. to define the function $f(x)=x+1$ (assuming that numbers and addition is defined), we write $\lambda x.x+1$. - you can use a function using application, e.g. given a function $M$, we can evaluate it at $x$ by $M\ x$.

Surprisingly, these two constructions are enough to express all of mathematics, without anything else assumed! For natural numbers, there is Church numerals to use. To define boolean values, we can write $$\mathbf{True} \equiv \lambda x. \lambda y. x$$ and $$\mathbf{False} \equiv \lambda x. \lambda y. y.$$ From this we can define all of Boolean algebra. Specifically, for the tenary operator $b?x:y$, we can construct the expression $b\ x\ y.$

  • $\begingroup$ Very interesting. Being Turing-complete this where math and computers meet. I'll look into this further. $\endgroup$ – dataphile Mar 13 '19 at 5:03

For mathematical logic, let's say that the ternary operator A ? B : C represents "for the predicates $A$, $B$, $C$, if $B$ is true iff $A$ is true, and $C$ is true iff $A$ is false". We can then represent your problem as $A := (c > 0)$, $B := (a = b + 1)$, and $C := (a = b + 2)$. We can write this in predicate logic as the true statement

$$(\neg A \iff C) \land (A \iff B)$$

We can verify the correctness of this with a truth table:

$$\begin{array} {|c c c | c |} \hline A & B & C & (\neg A \iff C) \land (A \iff B) \\ \hline F & F & F & F \\ \hline F & F & T & \color{blue}T \\ \hline F & T & F & F \\ \hline F & T & T & F \\ \hline T & F & F & F \\ \hline T & F & T & F \\ \hline T & T & F & \color{blue}T \\ \hline T & T & T & F \\ \hline \end{array} $$


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