Is $\forall x\exists y((y<x)\land (x<y+5))$ true if $x,y \in \mathbb N$ and $x,y\in \mathbb Z$? $\mathbb N$ includes $0$.

I think that if $x,y \in \mathbb N$ then it's not true if $x=0$.

But for $x,y \in \mathbb Z$ I think it is true. Is it correct?

  • $\begingroup$ Over $\Bbb Z$ can you describe, in terms of $x$, a suitable $y$? $\endgroup$ – Lord Shark the Unknown Jun 27 '18 at 6:24
  • $\begingroup$ I think it's $x-5<y<x$ $\endgroup$ – user123429842 Jun 27 '18 at 6:29
  • $\begingroup$ The concept of predecessor may be useful. $\endgroup$ – AnyAD Jun 27 '18 at 6:35

Over ${\mathbb N}$ this is indeed false. In fact, already $\forall x \in {\mathbb N} \; \exists y \in {\mathbb N}:y<x$ is false; $x = 0$ is a counterexample.

Over ${\mathbb Z}$ this is indeed true. Given $x \in {\mathbb Z}$, take $y = x - 1$ (or $x - 2$, $x - 3$, $x - 4$, because, as you noticed, the condition is equivalent to $x - 5 < y < x$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.