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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.
a) When x = −2 and y = 1, is Q(x, y).
a. true
b. false
The hypothesis of Q(−2, 1) is__, which is___. The conclusion is___, which is____. Thus Q(−2, 1) is a conditional statement with a____hypothesis and a_____conclusion. So Q(−2, 1) is____.
b) Give values different from those in part (a) for which Q(x, y) has the same truth value as in part.
c) Give values different from those in part (c) for which Q(x, y) has the same truth values as in part.


Sagot :

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 [tex]Q(x,y)[/tex] is predicate that [tex]x<y[/tex] then [tex]x^{2} <y^{2}[/tex]. where [tex]x,y[/tex] are rational numbers.

a)

when [tex]x=-2, y=1[/tex]

Here [tex]-2<1[/tex] that is [tex]x<y[/tex]  satisfied. Then

[tex](-2)^{2}<1^{2}[/tex]

[tex]4<1[/tex] this is wrong. since [tex]4>1[/tex]

That is [tex]x^{2}[/tex][tex]>y^{2}[/tex] Thus [tex]Q(x,y)[/tex] [tex]=Q(-2,1)[/tex]is false.

b)

Assume [tex]Q(x,y)=Q(-5,2)[/tex].

That is [tex]x=-5, y=2[/tex]

Here [tex]-5<2[/tex] that is [tex]x<y[/tex] this condition is satisfied.

Then

[tex](-5)^{2}<2^{2}[/tex]

[tex]25<4[/tex] this is not true. since [tex]25>4[/tex].

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

Since in both [tex]x<y[/tex] satisfied and [tex]x^{2} >y^{2}[/tex] for both the points.

c)

if [tex]Q(x,y)=Q(3,8)[/tex] that is [tex]x=3[/tex] and [tex]y=8[/tex]

Here [tex]3<8[/tex] this satisfies the condition [tex]x<y[/tex].

Then [tex]3^{2} <8^{2}[/tex]

[tex]9<64[/tex] This also satisfies the condition [tex]x^{2} <y^{2}[/tex].

Hence [tex]Q(3,8)[/tex] exists and it is true.

d)

Assume [tex]Q(x,y)=Q(9,10)[/tex]

Here [tex]9<10[/tex] satisfies the condition [tex]x<y[/tex]

Then [tex]9^{2}<10^{2}[/tex]

[tex]81<100[/tex] satisfies the condition [tex]x^{2} <y^{2}[/tex].

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