0
$\begingroup$

The Argument

Why is the second part where the quantifiers are interchanged false? Is there a concrete example for this?

$\endgroup$
  • $\begingroup$ The statement $(\exists m \in N)(\forall n \in N) m > n$ means that there is an upper bound on $N$, within $N$. If this statement were true, then it's true for $n = m$; but $m > m$ is certainly false. $\endgroup$ – user61527 Mar 14 '14 at 1:02
  • $\begingroup$ Thank you! I will reiterate my understanding: The highlighted sentence says "for every natural number 'n' (individual cases), there exists atleast one natural number 'm' such that m>n" while the second part states that "there exists atleast one 'm', for every (all) natural numbers 'n' such that m>n" which is definitely false. I think I was getting confused between 'every' and 'all' in the sense that 'every' & 'all' could mean individual cases or a collective group. Please let me know if you think my understanding is not correct. $\endgroup$ – va4az Mar 14 '14 at 2:20
  • $\begingroup$ For a lot of practice with this sort of thing, spend some time looking over the math StackExchange question A game with $\delta$, $\epsilon$ and uniform continuity. $\endgroup$ – Dave L. Renfro Mar 14 '14 at 15:00
1
$\begingroup$

Yes. The second statement reads "there exists a natural number $m$ such that for all other natural numbers $n$, $m >n$". Clearly this is false.

$\endgroup$

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.