How many Irrational numbers?

Question

How many irrational numbers can exist between two rational numbers ? As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? Or there exist some specific irrational which could be told by some methods ? for eg. irrational numbers between 2 and 3 are 5^(1/2),7^(1/2) etc... If there exists only some specific irrationals then why so?

Answer 1

Let $r_1 < r_2$ be two rational numbers. There are countably many rational numbers $q \in (r_1, r_2)$, and there are uncountably many $x \in (r_1,r_2)$. So there are $uncountably$ many irrational numbers between any two rational numbers.

Answer 2

There are uncountably infinite number of irrational numbers and countably infinite number of rational numbers.

Suppose we have a irrational numbers $q_0<q_1$ then we can construct a bijective mapping from $\mathbb R$ to $(q_0,q_1)$ that preserves rationality. This would show that the number of rational numbers and irrational numbers respectively in the interval is the same as the number of rational and irrational numbers overall.

We construct these by using a bijective mapping $\theta:\mathbb R\to(0,1)$ with the same properties and then use $f(x) = q_0 + (q_1-q_0)\theta(x)$. The mapping $\theta$ can be given as:

$$\theta(x) = \begin{cases} {1\over 2-x} & \text{ if } x<0 \\ {1+x\over 2+x} & \text{ if } x\ge 0 \end{cases}$$

Answer 3

We know that the set of rationals, $\mathbb{Q}$ is dense in $\mathbb{R}$. Then, $\mathbb{Q} + i$ where $i$ is an irrational number (which is a subset of the set off all irrationals) is also dense in $\mathbb{R}$ (since $\mathbb{R}+i = \mathbb{R}$). Therefore, the set of irrationals is dense in $\mathbb{R}$ and there exists an irrational in every non-empty interval in $\mathbb{R}$. As there are infinite (unaccountably) such intervals, there are infinite irrationals between and two rational numbers.

Answer 4

Let $r_1,r_2$ be rational numbers such that $r_1<r_2$. And let $i_1$ be an irrational number. We can show that there is a irrational number $I_1$ in the form $I_1=r_1+i_1/k$ where $k$ is a non-zero real number such that $$r_1<r_1+i_1/k<r_2$$ Therefore, $$r_1+i_1/k<r_2$$ $$k>\sqrt {(i_1 )}/(r_2-r_1 )$$ So, when, $$k=⌊\sqrt{(i_1 )}/(r_2-r_1 )⌋+1$$

$$r_1<r_1+i_1/(⌊\sqrt {(i_1 )}/(r_2-r_1 )⌋+1)<r_2$$ There can be infinite number of irrational numbers satisfying this condition.