About determining whether the statements are true or false

Question

Determine whether the statements are true or false

$\mathscr R(x): x$ is odd

1. $\exists y \in \mathbb Z$ such that $\forall x \in \mathbb Z, \mathscr R(x + y)$
2. $\forall x \in \mathbb Z$, $\exists y \in \mathbb Z$ such that $\mathscr R(x + y)$

For 1., let $x = 2n, y = 2m+1$, thus $x+y= 2(n+m) + 1$. This is similar to $2k + 1$ and it is odd. At here, the result is the same as 1., so is this statement true? I am not sure.

Also for 2. I am still confused, thank you!

Answer 1

For 1., it's obvious that this statement is false. Suppose some $y\in\mathbb{Z}$ was chosen. It must be either even or odd. If odd, then for all odd $x\in\mathbb{Z}$, $x+y$ will be even. Likewise, if $y$ is even, then for all even $x\in\mathbb{Z}$, $x+y$ will be even. So regardless of the choice of $y$, this is false.

Now, the second part is true. Let $x\in\mathbb{Z}$ be arbitrary. It must be either even or odd. We can now pick $y\in\mathbb{Z}$ such that it is odd if $x$ is even or even if $x$ is odd. As such, $x+y$ will be odd regardless of the choice of $x$.

Answer 2

If $x+y$ is odd and $x'=x+1$ then $x'+y$ is not odd; it is even. So (1) is false.

If $x$ is odd and $y=0$ then $x+y$ is odd.

If $x$ is even and $y=1$ then $x+y$ is odd.

So (2) is true.

This is an example of what can happen if you interchange an "$\exists$" and a "$\forall$".