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!!!!
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