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?
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$\begingroup$ The sum of a rational and an irrational is irrational. The number $\sqrt{2}/n$ is irrational for any positive integer $n$ and can be made as small as desired by choosing $n$ large enough.Therefore, given two rationals $r_1 < r_2$, the irrational number $r_1 + \sqrt{2}/n$ will be between $r_1$ and $r_2$ for all sufficiently large $n$. This shows that there are infinitely many irrationals between any two rationals. (In fact, a much stronger statement is true: there are uncountably many irrationals between any two rationals.) $\endgroup$– user169852Sep 19, 2017 at 7:08
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4 Answers
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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:
$$\theta(x) = \begin{cases} {1\over 2-x} & \text{ if } x<0 \\ {1+x\over 2+x} & \text{ if } x\ge 0 \end{cases}$$
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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.
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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.
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$\begingroup$ Will there be a range of irrational numbers between two numbers or there will be some specific points for that ? $\endgroup$ Sep 19, 2017 at 7:33
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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.
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$\begingroup$ Please use MathJax to render the math. Thank you. $\endgroup$ Sep 13, 2020 at 10:34