Question

Suppose \(A=\\{1,2,3,4\\}\) and \(B=\\{a, c\\}\) (a) \(A \times B\) (c) \(A \times A\) (e) \(\varnothing \times B\) (g) \(A \times(B \times B)\) (b) \(B \times A\) (d) \(B \times B\) (f) \((A \times B) \times B\) (h) \(B^{3}\)

Short answer

The answers are: (a) \(A \times B = \{(1,a), (1,c), (2,a), (2,c), (3,a), (3,c), (4,a), (4,c)\}\)(c) \(A \times A = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)\}\)(e) \(\varnothing \times B = \varnothing\)(g) \(A \times (B \times B) = \{(1,(a,a)), (1,(a,c)),(1,(c,a)), (1,(c,c)), (2,(a,a)), (2,(a,c)),(2,(c,a)), (2,(c,c)), (3,(a,a)), (3,(a,c)),(3,(c,a)), (3,(c,c)), (4,(a,a)), (4,(a,c)),(4,(c,a)), (4,(c,c))\}\)(b) \(B \times A = \{(a,1), (a,2), (a,3), (a,4), (c,1), (c,2), (c,3), (c,4)\}\)(d) \(B \times B = \{(a,a), (a,c), (c,a), (c,c)\}\)(f) \((A \times B) \times B = \{ ((1,a),a),((1,a),c), ((1,c),a),((1,c),c), ((2,a),a),((2,a),c), ((2,c),a),((2,c),c), ((3,a),a),((3,a),c), ((3,c),a),((3,c),c), ((4,a),a),((4,a),c), ((4,c),a),((4,c),c)\}\)(h) \(B^{3} = \{(a,a,a), (a,a,c), (a,c,a), (a,c,c), (c,a,a), (c,a,c), (c,c,a), (c,c,c)\}\)

Step-by-step solution

Calculate \(A \times B\)

This requires forming all possible combinations of elements from A and B, where an element from A comes first and an element from B comes second. So, \(A \times B = \{(1,a), (1,c), (2,a), (2,c), (3,a), (3,c), (4,a), (4,c)\}\)

Calculate \(A \times A\)

This is the combination of elements from A with itself. So, \(A \times A = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)\}\)

Calculate \(\varnothing \times B\)

Here, we have an empty set combined with B. Since there is no element in an empty set, we cannot form any tuple. Therefore, \(\varnothing \times B = \varnothing\)

Calculate \(A \times (B \times B)\)

This requires forming combinations between A and tuples from \(B \times B\), leading to \(A \times (B \times B) = \{(1,(a,a)), (1,(a,c)),(1,(c,a)), (1,(c,c)), (2,(a,a)), (2,(a,c)),(2,(c,a)), (2,(c,c)), (3,(a,a)), (3,(a,c)),(3,(c,a)), (3,(c,c)), (4,(a,a)), (4,(a,c)),(4,(c,a)), (4,(c,c))\}\)

Calculate \(B \times A\)

This time, an element from B comes first and an element from A second, so \(B \times A = \{(a,1), (a,2), (a,3), (a,4), (c,1), (c,2), (c,3), (c,4)\}\)

Calculate \(B \times B\)

The combination of B with itself would lead to the Cartesian Product \(B \times B = \{(a,a), (a,c), (c,a), (c,c)\}\)

Calculate \((A \times B) \times B\)

First, calculate \(A \times B\) as in step 1, then form combinations with B. We find that \((A \times B) \times B = \{ ((1,a),a),((1,a),c), ((1,c),a),((1,c),c), ((2,a),a),((2,a),c), ((2,c),a),((2,c),c), ((3,a),a),((3,a),c), ((3,c),a),((3,c),c), ((4,a),a),((4,a),c), ((4,c),a),((4,c),c)\}\)

Calculate \(B^{3}\)

This notation represents the Cartesian Product of B with itself three times. So, \(B^{3} = \{(a,a,a), (a,a,c), (a,c,a), (a,c,c), (c,a,a), (c,a,c), (c,c,a), (c,c,c)\}\)