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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)
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