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I just wanted to check my answers for this because I'm still not that comfortable with it.

Which of the following statements are true?

(i) $(\forall x \in \mathbb R)$ $x+1>x$

(ii) $(\forall x \in \mathbb Z)$ $x^2>x$

(iii) $(\exists x \in \mathbb Z)(\forall y \in \mathbb Z)$ $x \le y$

(iv) $(\forall y \in \mathbb Z)(\exists x \in \mathbb Z)$ $x \le y$

(v) $(\forall \epsilon > 0)(\exists \delta >0)(\forall x \in \mathbb R)$ $ [0<\lvert x-1 \rvert < \delta] \implies [\lvert x^2-1\rvert<\epsilon] $

I have:

(i) true

(ii) false

(iii) true (not sure about this one)

(iv) true (not sure about this one)

(v) true (not sure about this one)

Basically, the book just says that the order of quantifiers matters and that one statement is true and the other is not; there's no real explanation of how to interpret it, though. Any input would be appreciated. Thanks!

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    $\begingroup$ How about if you explain the reasons for your answers? It will be much more useful to see if your logic is right, than just to see if you got the right answer. $\endgroup$ – Nate Eldredge Feb 18 '15 at 21:39
  • $\begingroup$ Well, the first two seem fairly straight forward to me. For (iii) I interpret it as "there exists an element x in Z for all elements y in Z for which x is less than y is true." So, sure, if you give me any integer, there'll always be an integer that's equal to or less than it. But, then I look at (iv) and don't know how to interpret that differently. My intuition tells me that one is true and one is false, but I couldn't tell you which one. $\endgroup$ – thisisourconcerndude Feb 18 '15 at 21:44
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    $\begingroup$ Perhaps it would be better to interpret (iii) as "There exists $x \in \mathbb{Z}$ such that, for all $y \in \mathbb{Z}$, $x \leq y$." So, does there exist such an $x$? $\endgroup$ – Ken Feb 18 '15 at 21:53
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    $\begingroup$ Hmm. I suppose not. That would mean that there'd have to be a "lowest" integer that is less than all other integers, which I assume doesn't exist. Right? $\endgroup$ – thisisourconcerndude Feb 18 '15 at 21:56
  • $\begingroup$ Yes! Now, we interpret (iv) as "For every integer $y$, there exists an integer $x$ such that $x \leq y$." What do you think here? $\endgroup$ – Ken Feb 18 '15 at 22:33

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