Hey. I have homework due in analysis and I am stuck on what I am doing in this problem. I'm not sure if my approach is right on this problem, so please help me out if I'm not doing it right.
Let a, b ? R. Prove: a ? b, i? a ? b + ? for all ? ? (0, ?).
Assume that for every ? > 0, a ? b + ? , then a ? b .
Let p ? for every ? > 0, a ? b + ? and q ? a ? b.
Therefore p -> q
Let ~p ? there exists ? > 0, a > b + ? and ~q ? a > b
If, a>b, then there exists ? > 0, a > b + ?
Suppose that a > b then let ? = (a-b)/2 > 0
So a > b + ?
a > b + (a-b)/2 > 0
a – b > (a-b)/2 > 0
a – b > (a-b)/2 > 0
Let a=4, b =2
4 – 2 > (4-2)/2 > 0
2 > 2/2 > 0
2 > 1 > 0
Therefore, if a > b, then there exists ? > 0, a > b + ?. Thus, if, for every ? > 0, a ? b + ? , then a ? b . So, a ? b, i? a ? b + ? for all ? ? (0, ?).
Let a, b ? R. Prove: a ? b, i? a ? b + ? for all ? ? (0, ?).
Assume that for every ? > 0, a ? b + ? , then a ? b .
Let p ? for every ? > 0, a ? b + ? and q ? a ? b.
Therefore p -> q
Let ~p ? there exists ? > 0, a > b + ? and ~q ? a > b
If, a>b, then there exists ? > 0, a > b + ?
Suppose that a > b then let ? = (a-b)/2 > 0
So a > b + ?
a > b + (a-b)/2 > 0
a – b > (a-b)/2 > 0
a – b > (a-b)/2 > 0
Let a=4, b =2
4 – 2 > (4-2)/2 > 0
2 > 2/2 > 0
2 > 1 > 0
Therefore, if a > b, then there exists ? > 0, a > b + ?. Thus, if, for every ? > 0, a ? b + ? , then a ? b . So, a ? b, i? a ? b + ? for all ? ? (0, ?).