a.)If y>0, then there exists n in N such that n-1(less than or equal to) y<n.

Prove

b.)Also: Prove that the n in part (a) is unique...

Any help would be greatly appreciated!!!!

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- Feb 24th 2009, 07:38 AMtrojanlaxx223Real Analysis Homework Help!!!!!!
a.)If y>0, then there exists n in N such that n-1(less than or equal to) y<n.

Prove

b.)Also: Prove that the n in part (a) is unique...

Any help would be greatly appreciated!!!! - Feb 24th 2009, 07:05 PMbenes
Let A={nεN:n>y}.Now that set is not empty because the Archimedean principal tell us that:

for all y>0 there exists nεN SUCH THAT n>y

SINCE A is not empty and also a subset of the natural Nos N, then according to the well ordering principal of natural Nos :

there exists a kεA AND $\displaystyle \forall n. k\leq n$.

But since kεA ,THEN y<k. ALSO k-1<k.

If now y<k-1 ,then (k-1)εN which implies $\displaystyle k\leq k-1$ a contradiction ,hence : $\displaystyle k-1\leq y<k$

For part (b).

Assume that the set A HAS two minimums k,m both belonging to A

But since m is a minimum ,$\displaystyle k\leq m$...........(1)

Also since k a minimum ,$\displaystyle m\leq k$.............(2)

From (1) and(2) we conclude k=m ,therefor k is unique

Do not forget that a minimum of a set is an infemum that belongs to the set.

If you do not understand part (b) i can give a more detailed proof - Feb 24th 2009, 07:25 PMtrojanlaxx223Thanks!!!
Thanks for the help. This question was a part of my review sheet that I couldn't understand. The proof for uniqueness was perfect. Thanks again!!