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5 February, 18:44

Prove inf (e) = 0 where e is the set of rationals greater than 0

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  1. 5 February, 21:51
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    Let u=inf (e).

    Suppose u<0

    Note - u/2 >0 and u + (-u/2) = u/2<0

    Hence u + (-u/2) is not in e

    Contradiction arises.

    Suppose u>0.

    Note 0
    Contradiction arises.

    Hence inf (e) = u=0.
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