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olasank [31]
3 years ago
12

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.

Mathematics
1 answer:
Levart [38]3 years ago
6 0

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)

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Answer:

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