Question

Find the truth set of each of these predicates where the domain is the set of integers.

(a) \(P\left( x \right):{x^2} < 3\)

(b) \(Q\left( x \right):{x^2} > x\)

(c) \(R\left( x \right):2x + 1 = 0\)

Short answer

(a) \(\left\{ { - 1,0,1} \right\}\)

(b) \(\left\{ {..... - 3, - 2, - 1,2,3,4......} \right\}\)

(c) \(\left\{ \phi \right\}\)

Step-by-step solution

Truth set

The truth set contains all integers for which the given equation is true.

To find the truth set of the given predicate (a) 

The truth set of \(P\left( x \right):{x^3} < 3\)consist of all integers x for which the inequality \({x^2} < 3\) is true.

\({x^2} \ge {2^2} = 4\)if \(x \ge 2\)

if \(x \ge 2\)

Now, the integers for which \({x^2} < 3\) hold true are -1, 0 and 1.

Hence, the truth set of \(P\left( x \right):{x^3} < 3\) is the set containing 0, 1 and -1.

Thus, truth set for \(P\left( x \right):{x^3} < 3\) is \(\left\{ { - 1,0,1} \right\}\).

To find the truth set of the given predicate (b) 

The truth set of \(Q\left( x \right):{x^2} > x\)consist of all integers x for which the equality \({x^2} > x\) is true.

If x is a negative integer \({x^2} > x\) as \({x^2}\) is positive and all positive integers are greater than negative integer.

Thus, the truth set must contain all negative integers.

Now, since \({0^2} = 0\) , 0 is not an element of the truth set.

And since \({1^2} = 1\) , 1 is not an element of the truth set.

If \(x \ge 2\), then \(x < {x^2}\) and thus all positive integers of 2 are in the truth set.

Thus, truth set for \(R\left( x \right):{x^2} > x\) is \(\left\{ {..... - 3, - 2, - 1,2,3,4......} \right\}\)

To find the truth set of the given predicate (c) 

The truth set of \(R\left( x \right):2x + 1 = 0\)consist of all integers x for which the equality \(2x + 1 = 0\) is true.

\(2x + 1 = 0\)

Solve the above equation and find value of x

\(\begin{array}{c}2x = - 1\\x = - \frac{1}{2}\end{array}\)

Hence, the equality \(2x + 1 = 0\) is only true for \(x = - \frac{1}{2}\)

But \( - \frac{1}{2}\) is not an integer and thus the truth set is empty.

Thus, truth set for \(R\left( x \right):2x + 1 = 0\) is \(\left\{ \phi \right\}\).