Answer
a. False
b. True
c. False
d. True
e. False
f. False
g. True
h. True
Work Step by Step
a. $\mathbb{Z^+}$ has a minimum value which is 1. Therefore, there is no positive integer $x$ such that $1=x+1$.
b. $\mathbb{Z}$ is an unbounded set which implies that it does not have a minimum. Therefore, for every integer, a predecessor can be produced.
c. The statement is saying: there is a single value of $x\in \mathbb{R}$ such that $\forall y\in \mathbb{R}, x=y+1$. This statement is true only for subsets of $\mathbb{R}$, of the form $\{x,y\}$, such that $x=y+1$. There is no such value in $\mathbb{R}$. Note that if the quantifiers were reversed as in $\forall y\in \mathbb{R}, \exists x\in \mathbb{R}$ such that $x=y+1$, then that statement would have been true; but that is not the case here.
d. All real numbers except 0 have a multiplicative inverse. Since 0 is not a positive real number, this statement is true.
e. 0 is a real number that does not have a multiplicative inverse.
f. Counterexamples: If $x=y$ then $z=x-y=0\notin \mathbb{Z^+}$.
g. This statement is true because it is also valid for the counterexamples identified in the (f).
h. For $u=\frac{1}{2},\forall v\in \mathbb{R^+}, \frac{v}{2}
