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.