Prove that the set of irrational numbers in 01 is not counta

Prove that the set of irrational numbers in [0,1] is not countable.

Solution

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Theorem :

Prove that the set of irrational numbers is not countable.

Proof :

Let us consider A and B be countable sets. Suppose that the both A and B are finite. Then this conveys that the A B is finite, and hence could be countable.

Now, suppose one of the A and B sets are finite in the order and the other is considered as the countably infinite. Now that A is finite and since B is countably infinite, then there exists a function
named f : B which have integer or the whole numbers and which is a one-to-one correspondence.

Now, let C = A B. Thus A B = B C.

If C = , then A B = B which is countably infinite. Thus assume that C != . And suppose that C = {c1, c2, ...cn}.

Let f\' : B C for all integers and be defined as : f0(cj) = j and f0(bi) = n + i.

So that clearly, the function f\' is a one-to-one and onto related. Thus B C is countable.

But it is to be noted that B C = A B, thus making the A B is countable.

If that both A and B are infinite and that A B = . But given that the functions A and B are both countable, and there exists functions f : A and g : B that are one-to-one correspondences.

So let us consider the function h : A B where h(n) = f(n/2), if n is even and h(n) = g((n + 1)/2), if n is odd.

Since both f and g are one-to-one, then so is h. Parallely, as f is onto, every element in A is grabbed and since g is onto, every element in B is covered, so
h is also a onto function. Therefore, h is a one-to-one correspondence, and hence h is countable.

Suppose that both A and B are infinite and that A B != . Let C = B A.

Then AB = AC and AC = .

If C is countably infinite, then AB = AC is countable by the previous case.

And if C is finite, then A B A C is countable by the above second case.

So that in any of the cases, A B is countable.

Now that assume by the way of contradiction that the set of all irrational numbers is countable. And we know that the rationals are countable. Since the real numbers are the union of the irrationals and rationals, as stated above proofs, thus the real numbers must be countable. Therefore the irrationals are uncountable.

Hence Proved.

Hope this is helpful.