Bijection between reals and irrationals

[georgios]

Homework Statement

Prove that |R \ |Q ~ |R ... the irrationals are equinumerous to the reals

Homework Equations

The Attempt at a Solution

I can prove it using the Cantor Schroeder Bernstein theorem, but i was wondering if there
is a clever way of constructing the bijection explicitely ...
thanks,
g.

 

[mighty2000]

Dear g.,

Yes, there is a clever way to construct the bijection explicitly. First, let's define a set S as follows: S = {x + √2 | x ∈ Q}. In other words, S is the set of all numbers that can be obtained by adding the square root of 2 to any rational number. Next, let's define a function f: R → S as follows: f(x) = x + √2. This function is injective, since each real number maps to a unique number in S. Now, we need to show that f is also surjective.

To do this, let's consider an arbitrary number y ∈ S. We know that y can be written as y = x + √2, where x ∈ Q. But since x ∈ Q, we can also write it as x = a/b, where a and b are integers. Therefore, y = (a/b) + √2 = (a + b√2)/b. This shows that every number in S can be written as a rational number divided by an integer. Now, we can define a function g: S → R as follows: g(y) = a/b. This function is also injective, since each number in S maps to a unique rational number divided by an integer. Therefore, we have shown that there exists a bijection between S and R, which means that |R| = |S|. But since S is a subset of the irrationals, we can also say that |S| = |R \ |Q|. Therefore, we have shown that |R \ |Q ~ |R, which means that the irrationals are equinumerous to the reals.