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28 November, 08:52

Prove by contradiction that the interval (a,

b. has no minimum element.

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  1. 28 November, 10:16
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    Assume (a, b) has a minimum element m.

    m is in the interval so a < m < b.

    a < m

    Adding a to both sides,

    2a < a + m

    Adding m to both sides of the first inequality,

    a + m < 2m

    So

    2a < a+m < 2m

    a < (a+m) / 2 < m < b

    Since the average (a+m) / 2 is in the range (a, b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.
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