Proof that inf{ 1/n such that n is a natural number} = 0

Razack

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Oct 9, 2011
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Hi! I haven't had allot of exposure to writing proofs, and am a little lost.

I am to prove that the greatest lower bound of the natural numbers is zero. I think the archimidian property might have something to do with the proof. That is, for every real number x, there exists a natural number n such that n>x.

So for the statement:

inf{ 1/n such that n is a natural number} = 0

It is possible to pick a sufficiently large real number x, such that 1/n is essentially zero, via a sort of limiting process (although limits haven't been introduced in this proof class, so I can't use them).

How would I convey this in a "rigorous" proof?
 
That is, for every real number x, there exists a natural number n such that n>x.
So for the statement:
inf{ 1/n such that n is a natural number} = 0
First note that 0 is a lower bound for \(\displaystyle F = \left\{ {\frac{1}{n}:n \in \mathbb{Z}^ + } \right\}\).

Show that no positive number is a lower bound for \(\displaystyle F\).

If \(\displaystyle c>0\) then \(\displaystyle \left( {\exists k \in \mathbb{Z}^ + } \right)\left[ {\frac{1}{c} < k} \right]\)

Does that show that \(\displaystyle c\) is not a lower bound for \(\displaystyle F~?\)
 
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