Q) Let $P(x,y)$ be the predicate $y=2x$. Consider the statements
a)$\forall x \exists y P(x,y)$
b)$\forall y \exists x P(x,y)$
c)$\exists y \forall x P(x,y)$
where $x$ and $y$ range over the integers. Write whether each statement is true or false and give a very short explanation of why
For a) it is false, since for all the values of $x$ won't give the same value of a particular $y$.
For b) it is false, since all the values of $y$ cannot be formed by any particular $x$. In the case of $y=1$ it requires $x=0.5$, which is not a discrete number.
c) false, some values of $y$ cannot be obtained by all the values of $x$. It would require a $x=y/2$ value.
Are my answers correct? Does the justification make sense? Any help would be appreciated. Thanks!
$P(x, y)$ is the statement $y = 2x$, where $x$ and $y$ are integers.
Thus, the statement $\forall x \exists y P(x,y)$ means that for each integer $x$, there exists an integer $y$ such that $y = 2x$. Since the integers are closed under multiplication, this statement is true.
The statement $\forall y \exists x P(x, y)$ asserts that every integer is even. You correctly supplied a counterexample to this statement by choosing $y = 1$.
The statement $\exists y \forall x P(x, y)$ asserts that there exists a particular integer $y$ that is the double of every integer $x$. The statement is false since the function $y = 2x$ is strictly increasing. If $x_1 < x_2$, then $y_1 = 2x_1 < 2x_2 = y_2$. In particular, there does not exist an integer $y$ such that $y = 2 \cdot 0 = 0$ and $y = 2 \cdot 1 = 2$.