(Advanced Calculus I) Problem to follow.

- zepdrix

(Advanced Calculus I) Problem to follow.

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- zepdrix

Suppose that the real number \(\large\rm a\) has the property that for every natural number \(\large\rm n\),
\(\large\rm a\le\dfrac{1}{n}\).
Prove that \(\large\rm a\le0\).

- zepdrix

I'm thinking I need to use the Archimedean Property in some way here...
struggling with it though... hmm

- sohailiftikhar

how it can be true ok take n=5 so 1/5 is not smaller than zero right ?

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## More answers

- zepdrix

If you look at all the fractions on the right side of the inequality,
the larger and larger than n gets, the smaller and smaller the fraction gets overall.
So it's getting really really close to zero.
So a, which is smaller than those set of values has to be zero or below since the fractions take up all of the positives.
Conceptually I think I understand.
I'm just having trouble proving it XD

- sohailiftikhar

oh! yes you can use your word to prove it use them into language of math . just make two equations and draw the conclusion .

- zepdrix

Do I need to do something with the Completeness Axiom? :o
Like if I can show that 0 is the infimum of the set of these (1/n)'s,
does that get me on the right track?
@zzr0ck3r @ganeshie8 Gotta get the big guns up in hereee XD

- zzr0ck3r

This is Archimedes all the way

- anonymous

lol @zzr0ck3r yeah in fact this is the property itself :)

- sohailiftikhar

oh yes just suppose a>0 is correct and you will find that this is not correct then a<0 will be correct that's all.

- zzr0ck3r

\(a\le 0\)

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