If A={ -1,1}, then find $A\times A\times A$ .
Answer
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Hint: To find $A\times A\times A$ , we will use the property of the Cartesian product of a non-empty set A. This property states that the Cartesian product of the set A will be the ordered pairs of the elements of A or we can write it as $A\times A\times A=\left\{ \left( a,b,c \right):a,b,c,\in A \right\}$ .
Complete step by step answer:
We have to find $A\times A\times A$ . We are given that $A=\left\{ -1,1 \right\}$ . We know that for a non-empty set A, the Cartesian product of the set A will be the ordered pairs of the elements of A, that is,
$A\times A\times A=\left\{ \left( a,b,c \right):a,b,c,\in A \right\}$
We have to write the set of various combinations of $\left\{ -1,1 \right\}$ with the number of elements in each combination as 3.
$\Rightarrow A\times A\times A=\left\{ \left( -1,-1,-1 \right),\left( -1,-1,1 \right),\left( -1,1,-1 \right),\left( -1,1,1 \right),\left( 1,-1,-1 \right),\left( 1,-1,1 \right),\left( 1,1,-1 \right)\left( 1,1,1 \right) \right\}$
Note: Students must understand what ordered pair means. Sets of ordered pairs are the pair of elements that occur in particular order and are enclosed in brackets. If a set P contains x elements and another set Q contains y elements, then the number of elements in the cartesian product of the sets P and Q will be the product of x and y (that is, xy). Therefore, we can verify the result by confirming the number of elements in $A\times A\times A$ using this rule. Here, we are given that $A=\left\{ -1,1 \right\}$ , that is, the number of elements is 2. Then, the number of elements in $A\times A\times A$ can be found by multiplying 2 thrice.
$2\times 2\times 2=8$
Therefore, elements are present.
We can also find $A\times A\times A$ by an alternate way.
We will first find $A\times A$ using the property $A\times A=\left\{ \left( a,b \right):a,b\in A \right\}$ .
$\begin{align}
& \Rightarrow A\times A=\left\{ -1,1 \right\}\times \left\{ -1,1 \right\} \\
& \Rightarrow A\times A=\left\{ \left( -1,-1 \right),\left( -1,1 \right).\left( 1,-1 \right),\left( 1,1 \right) \right\} \\
\end{align}$
Now, we have to find $A\times A\times A$ .
$\begin{align}
& \Rightarrow A\times A\times A=\left\{ \left( -1,-1 \right),\left( -1,1 \right).\left( 1,-1 \right),\left( 1,1 \right) \right\}\times \left\{ -1,1 \right\} \\
& \Rightarrow A\times A\times A=\left\{ \left( -1,-1,-1 \right),\left( -1,-1,1 \right),\left( -1,1,-1 \right),\left( -1,1,1 \right),\left( 1,-1,-1 \right),\left( 1,-1,1 \right),\left( 1,1,-1 \right)\left( 1,1,1 \right) \right\} \\
\end{align}$
Complete step by step answer:
We have to find $A\times A\times A$ . We are given that $A=\left\{ -1,1 \right\}$ . We know that for a non-empty set A, the Cartesian product of the set A will be the ordered pairs of the elements of A, that is,
$A\times A\times A=\left\{ \left( a,b,c \right):a,b,c,\in A \right\}$
We have to write the set of various combinations of $\left\{ -1,1 \right\}$ with the number of elements in each combination as 3.
$\Rightarrow A\times A\times A=\left\{ \left( -1,-1,-1 \right),\left( -1,-1,1 \right),\left( -1,1,-1 \right),\left( -1,1,1 \right),\left( 1,-1,-1 \right),\left( 1,-1,1 \right),\left( 1,1,-1 \right)\left( 1,1,1 \right) \right\}$
Note: Students must understand what ordered pair means. Sets of ordered pairs are the pair of elements that occur in particular order and are enclosed in brackets. If a set P contains x elements and another set Q contains y elements, then the number of elements in the cartesian product of the sets P and Q will be the product of x and y (that is, xy). Therefore, we can verify the result by confirming the number of elements in $A\times A\times A$ using this rule. Here, we are given that $A=\left\{ -1,1 \right\}$ , that is, the number of elements is 2. Then, the number of elements in $A\times A\times A$ can be found by multiplying 2 thrice.
$2\times 2\times 2=8$
Therefore, elements are present.
We can also find $A\times A\times A$ by an alternate way.
We will first find $A\times A$ using the property $A\times A=\left\{ \left( a,b \right):a,b\in A \right\}$ .
$\begin{align}
& \Rightarrow A\times A=\left\{ -1,1 \right\}\times \left\{ -1,1 \right\} \\
& \Rightarrow A\times A=\left\{ \left( -1,-1 \right),\left( -1,1 \right).\left( 1,-1 \right),\left( 1,1 \right) \right\} \\
\end{align}$
Now, we have to find $A\times A\times A$ .
$\begin{align}
& \Rightarrow A\times A\times A=\left\{ \left( -1,-1 \right),\left( -1,1 \right).\left( 1,-1 \right),\left( 1,1 \right) \right\}\times \left\{ -1,1 \right\} \\
& \Rightarrow A\times A\times A=\left\{ \left( -1,-1,-1 \right),\left( -1,-1,1 \right),\left( -1,1,-1 \right),\left( -1,1,1 \right),\left( 1,-1,-1 \right),\left( 1,-1,1 \right),\left( 1,1,-1 \right)\left( 1,1,1 \right) \right\} \\
\end{align}$
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