# Is the set {1, -1} closed under multiplication and/or addition?

$\left\{1 , - 1\right\}$ is closed with respect to multiplication, but not addition.
Here's a multiplication table for $\left\{1 , - 1\right\}$...
underline(color(white)(0+0$)|color(white)(0+)1color(white)(0)color(white)(|0)-1color(white)(0)) $\textcolor{w h i t e}{0 +} 1 \textcolor{w h i t e}{0} | \textcolor{w h i t e}{0 +} 1 \textcolor{w h i t e}{0} \textcolor{w h i t e}{| 0} - 1 \textcolor{w h i t e}{0}$$\textcolor{w h i t e}{0} - 1 \textcolor{w h i t e}{0} | \textcolor{w h i t e}{0} - 1 \textcolor{w h i t e}{0} \textcolor{w h i t e}{| 0 +} 1 \textcolor{w h i t e}{0}$Regardless of which element we multiply by which, we get an element of the set. So $\left\{1 , - 1\right\}$is closed under multiplication. The same cannot be said of addition, since $1 + \left(- 1\right) = 0$is not in the set $\left\{1 , - 1\right\}$$\textcolor{w h i t e}{}$Footnote Closure is one of the axioms of a group: A group is a set $S$equipped with a binary operation $\circ$satisfying the following properties: • Closure: If $a , b \in S$then $a \circ b \in S$• Identity: There is an element $I \in S$such that $a \circ I = I \circ a = a$for any $a \in S$• Inverse: For any $a \in S$there is an element $b \in S$such that $a \circ b = b \circ a = I$A commutative group also satisfies: • Commutativity: $a \circ b = b \circ a$for all $a , b \in S$The set $S = \left\{1 , - 1\right\}$with multiplication $\circ = \times$satisfies all of these axioms, so is a commutative group. The normal name of this particular group is ${C}_{2}$, the cyclic group of order $2\$.