# How is this sentence valid (with integers as domain of discourse): $\exists x >1 \forall y \exists z ~ ((y=xz) \lor (y=xz+1))$

This is a problem from the text An Introduction to Higher Mathematics (Exercises 1.4):

2 Using the integers as the universe of discourse, describe why the following are valid:

d) $$~~\exists x>1 \forall y \exists z ~ ((y=xz) \lor (y=xz+1))$$

I'm reading the sentence like: "There exists an $$x$$ greater than $$1$$ for all $$y$$, and for all $$y$$ there exists a $$z$$ such that $$y=xz$$ or $$y=xz+1$$."

In more normal English, I read: "y equals some number greater than 1 (x) times some relation of y (z), or that product plus 1."

As you can tell from my readings, I see that x is a fixed number and z can be a fixed number or some relation of y (like 2y). But I don't understand why the sentence is valid, and I can't understand what the sentence is trying to say in general, as I assume it's not just an arbitrary statement.

The only test that I can find that seems to work is $$Let~x=2~and~z=y/x$$, in which case the first operand expression always comes out true, and the latter seems trivial. Also, I'm not even sure if/how division is defined in the domain of integers.

Thanks for any help.

• If $y$ is not prime it is the product of two factors $xz$, if it is prime it is odd, so you can find an even number $2z$ or $x2$ (and many others) such that the successive number is $y$.
– N74
Commented Jun 24, 2017 at 6:17
• I would read this (in a non-native speaker's English): There exists an integer $x>1$ such that given any integer $y$ we can find an integer $z$ such that either $y=xz$ or $y=xz+1$ holds. The question about having division of integers defined does not arise at all. As Yngwie Malmsteen said, $x=2$ works, so the sentence is true. Commented Jun 24, 2017 at 6:52
• It's true, but I wouldn't call it valid ... in logic we typically say that a sentence is a valid sentence when it is true for all interpretations, i.e when it is a logical tautology. Commented Jun 24, 2017 at 15:55

Yes it is valid. If $x=2$, then $y$ being an integer implies that either $y$ is even or $y$ is odd, i.e. that either $y = xz$ for some integer $z$ or $y = xz+1$ for some integer $z$.