Answer to Question #132726 in Discrete Mathematics for Promise Omiponle

Question

(1) Let Q(x, y) be the statement "x+y=x-y." If the domain for both variables consists of all integers, what are the truth values?
(a)Q(1,1)
(b)Q(2,0)
(c)∀yQ(1, y)
(d)∃xQ(x,2)
(e)∃x∃yQ(x, y)
(f)∀x∃yQ(x, y)
(g)∃y∀xQ(x, y)
(h)∀y∃xQ(x, y)
(i)∀x∀yQ(x, y)

Answer 1

Solution to a:

Q(1,1)

$x+y=1+1=2$ $x-y=1-1=0$

since x+y is not equal to x-y, the truth value of Q(x,y) is FALSE

Ans: FALSE

Solution to b:

Q(2,0)

$x+y=2+0=2$ $x-y=2-0=2$

since x+y=x-y, the truth value of Q(x,y) is TRUE.

Ans: TRUE

Solution to c:

$\forall y Q(1,y)$

let y=1;

$x+y=1+1=2$ $x-y=1-1=0$

since $\exists y$ for which x+y is not equal to x-y (e.g y=1), the truth value of Q(x,y) is FALSE.

Ans: FALSE

Solution to d:

$\exists xQ(x,2)$

$x+2=x-2$ $x-x=-2-2$ $0=-4$

since the above equation does not give a definite solution of x, the truth value of Q(x,y) is FALSE.

Ans: FALSE

Solution to e:

$\exists x\exists yQ(x,y)$

let x=2, y=0;

$x+y=2+0=2$ $x-y=2-0=2$

since $\exists x$ and $\exists y$ such that x+y=x-y (i.e x=2, y=0), the truth value of Q(x,y) is TRUE.

Ans: TRUE

Solution to f:

$\forall x \exists y Q(x,y)$

let x=1,y=1;

$x+y=1+1=2$ $x-y=1-1=0$

since x+y is not equal x-y when x=1 and y=1, the truth value of Q(x,y) is FALSE

Ans: FALSE

Solution to g:

$\exists y \forall xQ(x,y)$

when y=0;

$x+0=x-0$ $x=x$

since x+y=x-y for all x when y=0, the truth value of Q(x,y) is TRUE

Ans: TRUE

Solution to h:

$\forall y \exists x Q(x,y)$

let y=1;

$x+1=x-1$ $x-x=-1-1$ $0=-2$

since the above equation does not give a definite solution of x, the truth value of Q(x,y) is FALSE.

Ans: FALSE

Solution to i:

$\forall x\forall yQ(x,y)$

let x=1, y=1;

$x+y=1+1=2$ $x-y=1-1=0$

since x+y is not equal to x-y, the truth value of Q(x,y) is FALSE

Ans: FALSE