Results from set theory tell us that the set of all real numbers \mathbb{R} and the set of all irrational numbers \mathbb{R} \setminus \mathbb{Q} are equinumerous (have the same cardinality). So, how can we construct a bijection between them? This is a question that quite tests the reader’s imagination. Of course, once the answer is provided, it may no longer seem so mysterious. Below, I provide an example of a construction I developed, from which readers can see how such a mapping is built.
To construct such a bijection, a natural idea is to let all rational numbers and a portion of the irrational numbers map to each other within their own combined set, while the remaining irrational numbers are mapped via the identity map. To construct such a bijection, we first need to find a function whose values are always irrational. It is not difficult to find such functions; for example, we know:
1. The equation x^4 + 1 = y^2 has no rational solutions other than x=0, y=\pm 1; otherwise, it would contradict the result of Fermat’s Last Theorem for n=4.
2. The square root of an irrational number is still an irrational number.
Based on this information, it is sufficient to construct a bijection from the set of positive real numbers \mathbb{R}^+ to the set of positive irrational numbers \mathbb{R}^+ \setminus \mathbb{Q}^+. Then, with a slight modification, we can obtain a bijection from \mathbb{R} to \mathbb{R} \setminus \mathbb{Q}.
A bijection from \mathbb{R}^+ to \mathbb{R}^+ \setminus \mathbb{Q}^+ is as follows: f(x) = \begin{cases} \sqrt{x^4+1}, & x \in \mathbb{Q}^+ \\ \sqrt{x}, & x \in \mathbb{A} \\ x, & x \in \mathbb{R}^+ \setminus \mathbb{Q}^+ \setminus \mathbb{A} \end{cases}
Where the set \mathbb{A} is the set of a_i \, (i \geq 1) defined below (with initial values traversing all positive rational numbers): a_{n+1} = \sqrt{a_n} \ (n \geq 1), \quad a_1 = \sqrt{a_0^4+1}, \quad a_0 \in \mathbb{Q}^+ In other words, starting from any positive rational number q, let A_q be the set of all irrational numbers constructed from the above expression. Then, by letting q traverse all positive rational numbers and taking the union of all A_q, we obtain \mathbb{A}. Although it might seem that such a set has many more elements than the set of rational numbers, it is actually equinumerous to the set of rational numbers.
It is almost obvious that f(x) is a surjection (irrational numbers are covered by these categories). Next, we only need to prove it is an injection. If it were not an injection, the "accident" could only occur within the set \mathbb{A}; that is, there would exist positive rational numbers x, y and positive integers m, n such that: (x^4+1)^{\left(\frac{1}{2}\right)^m} = (y^4+1)^{\left(\frac{1}{2}\right)^n} Which is: (x^4+1)^{(2^n)} = (y^4+1)^{(2^m)} Assuming without loss of generality that n \geq m, we have: (x^4+1)^{(2^{n-m})} = y^4+1 If n-m is not equal to 0, then a=y, b=(x^4+1)^{(2^{n-m-1})} would be a rational solution to a^4+1=b^2, which is a contradiction. Therefore, n=m, which implies x=y. Thus, f(x) is an injection.
With a slight modification, we obtain the mapping from real numbers to irrational numbers: f(x) = \begin{cases} \sqrt{x^4+1}, & x \in \mathbb{Q}^+ \\ \sqrt{x}, & x \in \mathbb{A} \\ -\sqrt{(x-1)^4+1}, & x \in \mathbb{Q}^- \text{ or } x=0 \\ -\sqrt{-x}, & x \in \mathbb{B} \\ x, & x \in \mathbb{R} \setminus \mathbb{Q} \setminus \mathbb{A} \setminus \mathbb{B} \end{cases} Where the set \mathbb{A} is the set of a_i \, (i \geq 1) defined as: a_{n+1} = \sqrt{a_n} \ (n \geq 1), \quad a_1 = \sqrt{a_0^4+1}, \quad a_0 \in \mathbb{Q}^+ And the set \mathbb{B} is the set of b_i \, (i \geq 1) defined as: b_{n+1} = -\sqrt{-b_n} \ (n \geq 1), \quad b_1 = -\sqrt{(b_0-1)^4+1}, \quad b_0 \in \mathbb{Q}^- \text{ or } b_0=0
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