PROOF:

(i) Let f be a function f: EVEN -> R0,1 that satisfies:

f(x) = 1/x for x ≠0 and f(0) = 0

Because f(x) is total and 1-to-1, cardinality(EVEN) <= cardinality(R01)

(ii) Suppose that there exists a function f that is total, 1-to-1 and onto, f: EVEN -> R0,1.

Each f(k), k ∈EVEN, can be represented by its infinite decimal expansions f(e sub k) = 0. ak1 ak2 ak3 ...

Construct number r = 0. r1 r2 r3... so that:

+ ri is any non-zero digit but aii (the ith digit after the period of f(i)) when i is even and

+ ri = 0 when i is odd.

It is easy to see that r ≠ f(0), r ≠ f(1), ...r ≠ f(i) for any i > 0. Therefore, function f is not onto.

==> there not exists a function f that is total, 1-to-1 and onto

==> cardinality(EVEN) ≠ cardinality(R0,1)

Combine (i) and (ii) we have: cardinality(EVEN) < cardinality(R0,1) Not regular, use the pumping lemma for RE.

Assume the language is regular. It must satisfy the pumping lemma. Let p = 2n be the

pumping length in pumping lemma.

Consider a string w = a^n b^n.

Since |w| ≥ p, a^n b^n can be written as xyz such that:

- xy^i z ∈ L for all i ≥ 0

- |xy| ≤ 2n

- |y| ≥ 1

• Since |y| ≥ 1, there are only 3 cases: y contains only a, only b or both a and b (y = a^s b^t, s > 0, t > 0).

• If y contains only a (or only b), xy2

z will contains more a than b (or more b than a). Thus xy^2z is not in L.

• If y contains both a and b (y = a^s b^t, s > 0, t > 0), xy^2z is not in the form of a^n b^n and thus is

not in L.