Prove by contradiction that the interval (a,

b. has no minimum element.

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.