Question

1. Let A = {x,y,z} B = {100,28,39}, find

1. A×B

2. B×A

2. Find the truth set of the following predicate

1. P(x): x < x2, the domain is Z, the set of integers.

2. Q(x): x2 = 2, the domain is Z, the set of integers.

3. R(x):3x + 1 = 0, the domain is R, the set of real numbers.

Answer 1

Solution for the problem is provided below, please comment if any doubts:

A.

Here we need to find the Cartesian product of two sets. The Cartesian product of two sets is the ordered pair of elements of the two sets in order.

That is Cartesian product of two sets A and B is defined as

A×B = {(a, b), where a belongs to A and b is belongs to B}

Here, A = {x, y, z} B = {100, 28, 39}

1. A×B ={x, y, z} ×{100, 28, 39}

={(x, 100), (x, 28), (x, 39), (y, 100), (y, 28), (y, 39), (z, 100), (z, 28), (z, 39)}

2. B×A ={100, 28, 39}×{x, y, z}

={(100, x), (100, y), (100, z), (28, x), (28, y), (28, z), (39, x), (39, y), (39, z)}

B.

Truth set of a predicate is the set of values that satisfies the given predicate,

1.

Here the predicate the x<x² , we know that for all integers other than 0 and 1, the square of a number will be greater than the number.

Thus the truth set of the predicate = Z-{0, 1}

2.

Here the predicate is for some integer x, x² = 2.

We know that for no integer x, its square can be 1.

Thus the truth set of the predicate is NULL set, = {}

3.

Here the predicate is for some real number x, 3x+1=0.

For 3x+1=0, 3x=-1, x=-1/3

That is there is only one value for x.

Thus the truth set of the predicate = {-1/3}