Can the Archimedean Property Prove 1/n < x for All Positive Reals?

[Samuelb88]

Homework Statement

Use the Archimedean property: For all x in the positive reals there exists an n in the naturals such that n > x.

to prove: For all x in the positive reals there exists an n in the naturals such that 1/n < x.

The Attempt at a Solution

Proof: Multiplying through the inequality by n (I am not including 0 in the set of all natural numbers) yields 1 < nx. Let x = a/b, where a,b are in the positive reals. Substituting into the inequality the subsequent value of x yields 1 < (an)/b. Dividing through the inequality by a/b yields b/a < n, where b/a is in the positive reals. By the Archimedean property, we may always find an n in the naturals such that n > b/a. This proves the first consequence of the Archimedean property. Q.E.D. ?

 

[tiny-tim]

Hi Samuelb8!

Yes, it's correct, but you won't get many marks for it, since it's too roundabout.

You talk about two new numbers, a and b (and you give no way of defining them), but all you use them for is defining b/a …

can you find a shorter version of your proof that doesn't need a or b?

 

[ZioX]

Hint: 1/x is a positive real too.