# Naive Set Theory: Functions

## Binary Operations

A **binary operation** $f$ on a set $A$ is a function $f : A^- \rightarrow A$, where $A^- \subseteq A \times A$. Intuitively, a binary operation takes in two elements of a set and spits out another element of the same set.

A set $S$ is **closed under** a binary operation $f$ if $f$ is of the form $f : A \times A \rightarrow A$. That is, if $A^- = A \times A$. In English, a set is closed under a binary operation if the binary operation is defined for all pairs of elements within the set and always outputs another element of the set.

Binary operations usually have special symbols associated with them. For example, the addition operation is denoted with the $+$ symbol. Likewise, binary operations are also written in a special way. A binary operation $*$ is not written as $*(a, b)$ but instead as $a * b$. We thankfully retain our ability to write $a + b$ with this accommodating helping of syntactic sugar.

## Problems

Little Timmy believes that the integers are closed under division. Prove Little Timmy wrong.

Little Timmy believes that $\div$ is defined for $\div : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$. However, what value in $\mathbb{Z}$ is equal to $1 \div 2$? Or what about $-20 \div 7$? There are many such ordered pairs for which division is not defined. So Little Timmy is wrong.

Little Timmy believes that the rational numbers are closed under division. Show why Little Timmy wrong again.

$\mathbb{Q}$ is

*almost*closed under division, but almost isn't good enough. We see that $0 \in \mathbb{Q}$, and there is no value $a \in \mathbb{Q}$ such that $a \div 0$ is defined.Consider the set $A = \{a, b, c \}$. Let the binary operation $f : A^- \rightarrow A$ be defined by $f(a,b) = c, f(a,c) = b, f(b,a) = c, f(b,c) = a, f(c,a) = b, f(c,b) = a$. Is $A$ closed under $f$? If so, explain why. If not, modify $f$ to make $A$ closed under it.

**No**. $A$ is not closed under $f$ because the pairs $(a,a)$, $(b,b)$, and $(c,c)$ are not defined for $f$. However, if we define outputs for these inputs such that, for example, $f(a,a) = c$, $f(b,b) = a$, and $f(c,c) = b$, then $A$ will be closed under $f$.