1. Discrete Math

Apparently this question is hard

Let $\displaystyle Q(a,b)$ be a property which involves a and b. Let $\displaystyle P(c)$ be a property which involves c. For this problem, the domain will be the reals. For each of the following below, find properties $\displaystyle P(c)$ and $\displaystyle Q(a,b)$ such that:

a.) $\displaystyle (\forall x)(\forall z)[Q(x,z) \implies Q(z,x)]$ is true.

b.) $\displaystyle (\forall x)(\forall z)[Q(x,z) \implies Q(z,x)]$ is false.

c.) $\displaystyle (\forall x)[P(x) \implies (\exists z) Q(x,z)]$ is true.

d.) $\displaystyle (\forall x)[P(x) \implies (\exists z) Q(x,z)]$ is false.

2. a) x=z
b) x<z
c) x is rational then for some y, xy is irrational.
d) x is an even integer then for some integer xy is odd.

3. Originally Posted by Plato
a) x=z
b) x<z
c) x is rational then for some y, xy is irrational.
d) x is an even integer then for some integer xy is odd.
Two questions:

where does y come from, and for c do you mean "if x is rational, then for some z, xz is rational" ?

4. Originally Posted by DiscreteW
Two questions:

where does y come from, and for c do you mean "if x is rational, then for some z, xz is rational" ?
NO!
P(x)="x is rational"
Q(x,y)="xy is not rational"
These predicates work for both (c)

For (d) Q(x,y)="x+y is not rational"

5. Originally Posted by Plato
NO!
P(x)="x is rational"
Q(x,y)="xy is not rational"
These predicates work for both (c)

For (d) Q(x,y)="x+y is not rational"
Why y? Why not z?

Oh I see I guess it doesn't matter. So, my answer can be:

a) x=z
b) x<z
c.) If $\displaystyle P(x) = x \in \mathbb{Q}$, then $\displaystyle Q(x,z) = xz \notin \mathbb{Q}$
d.) If $\displaystyle P(x) = x \in \mathbb{Q}$, then $\displaystyle Q(x,z) = x + z \notin \mathbb{Q}$

Right? Although not sure how c and d are both not in the rationals (that is irrational), since one is true and the other is false, and we know that rational + rational = rational and similarly for rational*rational.

c is true, but isn't $\displaystyle xz \notin \mathbb{Q}$ false?