Dick said:No, that doesn't work. Try formulating it as a proof by contradiction.
kmikias said:ok! so i tried it this way...
Suppose a > b+ε for every ε>0,then a≤b
gb7nash said:Unfortunately, no. To do a proof by contradiction, you need to assume the hypothesis and the negation of the conclusion.
kmikias said:ok..another try...i won't stop until i get this ...
suppose a>b and let ε = 1/2 (a-b) ...(i think you know where i get that)
so a < b+ε → a< b+(1/2a-1/2b)
then a-(1/2)a < b-(1/2)b → 1/2a<1/2b→ a<b CONTRADICTION ∅
The statement "a < b + ε" means that a is less than b plus any positive value of ε. In other words, a is always smaller than b plus a small amount ε that can be chosen arbitrarily small.
The symbol "≤" represents the "less than or equal to" relation, meaning that a is either smaller than b or equal to b.
Since the statement says that a is always smaller than b plus any positive value of ε, it means that a can never be greater than b. This implies that a must be either smaller or equal to b, satisfying the "less than or equal to" relation.
Yes, the statement can be rephrased as "no matter how small ε is chosen, a is always less than or equal to b".
This statement is useful in mathematical proofs because it allows us to show that a certain value is always smaller than or equal to another value, without having to consider all possible values of ε. It simplifies the proof and makes it easier to follow.