Let x and y be distinct real numbers such that x < y, and that no real number z exists such that x < z < y. There must exist some real number a such that x + a = y. Since the real numbers are a field, this implies that a/2 is a real number, and since x + a/2 < x + a, there is a number z such that x < z < y. Contradiction, therefore x = y.

## Jeremy says:

Other proofs:

1/3 = 0.333…

2/3 = 2(1/3) = 2(0.333…) = 0.666…

1 = 3/3 = 3(1/3) = 3(0.333…) = 0.999…

Let x and y be distinct real numbers such that x < y, and that no real number z exists such that x < z < y. There must exist some real number a such that x + a = y. Since the real numbers are a field, this implies that a/2 is a real number, and since x + a/2 < x + a, there is a number z such that x < z < y. Contradiction, therefore x = y.

28 September, 2016 — 12:18 pm