Solution to (2) Determine the truth value of each of these statements if the domain of each … - Sikademy
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Archangel Macsika

(2) Determine the truth value of each of these statements if the domain of each variable consists of all real numbers. (a)∀x∃y(x^2=y) (b)∀x∃y(x=y^2) (c)∃x∀y(xy= 0) (d)∃x∃y(x+y=/=y+x) (e)∀x(x=/= 0 => ∃y(xy= 1)) (f)∃x∀y(y=/= 0 => xy= 1) (g)∀x∃y(x+y= 1) (h)∃x∃y(x+ 2y= 2^2x+ 4y= 5) (i)∀x∃y(x+y= 2^2xy= 1) (j)∀x∀y∃z(z= (x+y)=2)

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Here's the Solution to this Question

Here given,

The domain of each variable consists of all real numbers.

(a)∀x∃y(x^2=y)


For every real value of x function x=y exist means there exist value y which also a

real number.So truth value = TRUE.



(b)∀x∃y(x=y^2)


function is x=ywhich is not true for every x . if x=(-1) then not exist y which square value is (-1) . So, truth value = FALSE.


(c)∃x∀y(xy= 0)


function is xy = 0. here for every value of y there exist x that is x=0. so, xy=0.

So truth value = TRUE.


(d)∃x∃y(x+y {=}\mathllap{/\,} y+x)

function is x+y {=}\mathllap{/\,} y+x is false because addition of real number is commutative.

So, truth value = FALSE.


(e)∀x(x{=}\mathllap{/\,}0\implies ∃y(xy= 1))

for any nonzero x, the assignment y = 1/x has the property that is xy = x.(1/x) = 1.

So truth value = TRUE.


(f)∃x∀y(y{=}\mathllap{/\,} 0 \implies xy= 1)


This is false. If it were true, then there would be an x for which the proposition is true for

y = 2 and 3 simultaneously, i.e., 2x = 1 and 3x = 1. Taking the difference of these

two equations, we find x = 3x − 2x = 1 − 1 = 0,


but x · y = 0 · y = 0 for any y, making our premise absurd.


(g)∀x∃y(x+y= 1)


function is x+y =1 is true when we take y = 1-x . so there exist value of y which make

x+y =1.


So, truth value TRUE.



(h)∃x∃y(x+ 2y= 2 \land 2x+ 4y= 5)


here 2x+4y = 5 divide this eqution by 2 then we get the eqution x+2y = 5/2;

compare with line eqution y = mx+c

The line it determines is of the same slope as the first equation, but they have distinct

y-intercepts, which means that they are parallel and so, never intersect. So, they

cannot have a common solution (x, y). 

So, truth value TRUE.


(i)∀x∃y(x+y= 2 \land xy= 1)


if x= 1 and y=1 then this is solution that satifies given condition of eqution .

So, truth value TRUE.


(j)∀x∀y∃z(z= (x+y)=2)


here x+y = 2 means that we can take y = 2-x; so real of x and y are exist that gives z=2;


So, truth value TRUE.


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Question ID: mtid-5-stid-8-sqid-3747-qpid-2446