- #1

- 1,462

- 44

Assume for contradiction that there does exist a function ##f : \mathbb{N} \rightarrow (0,1)## that is a bijection. ##\forall m \in \mathbb{N}##, ##f(m)## is a real number between 0 and 1, and we represent it using the decimal notation ##f(m) = .a_{m1}a_{m2}a_{m3}...##. We assume that every real number is in the set ##\{ f(1), f(2), f(3),...\}##. Now, define ##x \in (0,1)## to be ##x=.b_1b_1b_3...## where ##b_n = 2## if ##a_{nn} \ne 2## and ##b_n = 3## if ##a_{nn} =2##. Then ##\forall n \in \mathbb{N} ~~x \ne f(n)##. This means that ##f## is not onto, and hence not a bijection, which is a contradiction.

My only question about this proof is that do we need to take into account the fact that, for example, ##0.59999... = 0.6##? Do we need to state at the beginning of the proof that don't allow strings of 9's? Is this the only way to resolve this issue?

My only question about this proof is that do we need to take into account the fact that, for example, ##0.59999... = 0.6##? Do we need to state at the beginning of the proof that don't allow strings of 9's? Is this the only way to resolve this issue?

Last edited: