I know that the open interval $(1,2)$ is uncountable. So there are an infinite amount of real numbers in the interval. Does there exist a ratio of rational numbers to irrational numbers?
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2The set of rational numbers is countable. – Count Iblis Jun 29 '17 at 22:00
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3There are as many irrationals as real numbers in any interval. The Lebesgue measure of the set of irrational numbers in $(a,b)$ is $b-a$. – Jun 29 '17 at 22:00
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2Yes, the ratio is $0$. – Bobson Dugnutt Jun 29 '17 at 22:00
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Put simply, the ratio of a countable set and an uncountable set would be 0 right? Since the uncountable set is infinite? Sort of like dividing by n as n goes to infinity? – Garrett Jun 29 '17 at 22:06
4 Answers
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We can put the set of rational numbers and natural numbers in one-to-one correspondence, and hence show that the number of natural numbers is equal to the number of rational numbers.
Hence, we have $\mathbb{R}$ (Real numbers) an uncountable infinity.
$\mathbb{Q}$ (Rational numbers) a countable infinity, and, $\mathbb{R}-\mathbb{Q}$ (Irrational numbers) an uncountable infinity.
Hence, the number of irrational numbers is strictly greater than irrational numbers.
Hence, the ratio of rational and irrational numbers is $0$.
Jaideep Khare
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The function
$$f(x):=\frac1{x-1},\quad x\in(1,2)$$
is a bijection from $(1,2)$ to $(1,\infty)$ with the property that map rational numbers to rational numbers and irrational numbers to irrational numbers.
If you knows that $(1,\infty)$ have the same number of rational and irrational than $\Bbb R$ you are done.
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The ratio is zero, the set of rationals is denumerable (countable) where as irrationals are uncountable. For a proof of cardinality of $\mathbb Q$ see Produce an explicit bijection between rationals and naturals?. Then note that $((1,2)\cap\mathbb Q)\cup((1,2)\cap(\mathbb R - \mathbb Q))=(1,2)$ where $\mathbb R - \mathbb Q$ denotes the irrationals. Since $(1,2)\cap\mathbb Q$ is countable and $(1,2)$ is uncountable it must be the case that $(1,2)\cap(\mathbb R - \mathbb Q)$ is uncountable as the finite union of countable sets is countable.
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In other words, you are asking what is the Lebesgue integral of the Dirichlet function (the characteristic function of the rationals): $$ D(x) =1_{\mathbb Q}(x)=\left\{\begin{matrix} 1 & \mbox{if } x\in\mathbb Q \\ 0 & \mbox{if } x \notin \mathbb Q \end{matrix}\right. $$ Because the rationals are countable, the Lebesgue measure of the rational numbers is $0$. Therefore the Dirichlet function equals $0$ almost everywhere and the answer is $\int D(x)=0$.
Dvir R
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