Question

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Answer 1

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$. where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$  satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$$>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$is false.

b)

Assume $Q(x,y)=Q(-5,2)$.

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$.

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$.

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$.

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$.

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)