(Unconstructive) Proof that irrational numbers does exist can be following:Any real number between 0 and 1 in binary notation can be assigned (maped) to exactly one subset of set of natural numbers and vice versa. (for example if on n-th place [in expression like 0.01110...] is 0 then number "n" is not in the assigned subset; if there is "1", it is). Therefore there are as many (concept of "larginess" of infinite sets is nontrivial and I will assume you are familiar with it, because my language abilities are not good enough to explain it more; and I will refer to it simply as "larginess" since I don't know english expression for it) real numbers as there are subsets of natural numbers. Now - there are as many rational number as there are natural numbers, while any set of all subsets of some set is larger than original set itself. This proves the thing. (sry for uncompletness, i could prove both things but it would be needed to define "larginess" of infinite set which I'm not willing to do right now)Edit: "larginness" is called cardinality, read about it here if interested http://en.wikipedia.org/wiki/Cardinality