You must have the completeness axiom or some equivalent to prove the first.
If is bounded above then there is a least upper bound .
But so .
That is a clear contradiction. Thus .
Hey all,
(this is my first post so bear with me haha)
I have to use the Archimedean principle (For every real number x, there is an integer n such that n > x) to prove this:
For every positive real number q, there is a positive integer N such that 1/n < q for every integer n >= N.
Now, my response to this was:
Because of the Archimedean Principle, there will always be an n greater than q. Thus when n >= N, 1/n < q because as n gets larger, it will reach infinity, and approach 0. Since q > 0, and n > q, 1/n < q.
Did I do this right? If not, how can I prove this with the AP? Thanks for the help!
I don't understand what you mean by there is a least upper-bound K. My teacher explained that:
1/q > 0
N > 1/q
Now, need to show that if n>= N, then 1/n < q
Bear in mind that this is meant to be a simpler homework assignment. Could you maybe explain the contradiction a bit more? And what you meant by least upper-bound K? Thanks a lot!
NOTE:
To clear something up in my original post, N is an integer, not the realm of Natural numbers.
If your class material does not include any discussion of least upper bounds (completeness of the real numbers) then I would say that the class is not ready for this problem. Actually we have a very sad situation here. Your teacher is sincere in wanting the class to understand the ideas involved here. BUT without be basics there is no way to fully understand this concept. I fear that you teacher, no matter good intentions, may not understand the fundaments. A classroom teacher, like a physicians, have an obligation to do no harm.
On the other hand, if your teacher gave you an axiom that says that the natural numbers are not bounded above then the proof I gave you above still works.
In any case, you can discuss the with the educational administrator in your area.