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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.