Can any one tell me how to prove that the set of rational numbers are countable? Prove give me a prove?
Thanks.
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4The standard argument is the visual argument found here: http://aminsaied.wordpress.com/2012/05/21/diagonal-arguments/ – Lepidopterist Mar 17 '13 at 22:41
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It's not countable. There are an infinite number of rational numbers just in between 0 and 1. – Brian Silva Mar 17 '13 at 22:42
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2@BrianSilva: They are countable – MITjanitor Mar 17 '13 at 22:43
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1@BrianSilva: So what? "Countable" means "has the same cardinality as the natural numbers", which is infinite too. (Or sometimes "has the same cardinality as the natural numbers, or less than that", but that's not relevant here). – hmakholm left over Monica Mar 17 '13 at 22:43
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1@BrianSilva You're confused about the definition of countable. Countable means finite or equinumerous to $\Bbb N$ – Git Gud Mar 17 '13 at 22:44
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Of the 399 results of searching for rational numbers countab;le on this site's built-in search, the first three ask exactly this question. Two of them even have accepted answers. – hmakholm left over Monica Mar 17 '13 at 22:45
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8I can't make head or tails of the sentence "Prove give me a prove?" – Pedro Mar 17 '13 at 22:45
3 Answers
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Map each rational $\frac{a}{b}$ into the integer $2^a 3^b$. This shows that the number of rationals is at most the number of integers.
If you want to handle the negative rationals, map the sign ($-1$, $0$, or $+1$) to $5^{\mathrm{sign}+1}$ and stick it on the end, so the mapping is $\mathrm{sign} \times \frac{a}{b} \to 2^a \, 3^b \,5^{\mathrm{sign}+1}$.
If you find this troubling, that's OK. You are not the only one.
wchargin
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marty cohen
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How would you go about proving that $\mathbb{N}\leftrightarrow{\text{integers in the form }2^a3^b5^{\text{sign} + 1}}$. Proving this is not trivial (at least for me.) – Nairit Jul 02 '17 at 04:08
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hello, the function doesn’t have to be a bijection. We know that any infinite subset of the naturals (a sequence) has the same cardinality as the naturals. So you just have to find an injective function of Q into N, since Q is infinite. This one is injective because the decomposition of an integer in primes is unique. So each a/b (Irreducible) goes to a unique element in N and there you have it. – Sahdo Jul 07 '20 at 23:23
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Here is another argument:
Consider the map $\varphi:\mathbb{Q}\rightarrow \mathbb{Z}\times\mathbb{N}$ which sends the rational number $\frac{a}{b}$ in lowest terms to the ordered pair $(a,b)$ where we take negative signs to always be in the numerator of the fraction. This map is an injection into a countably infinite set (the cartesian product of countable sets is countable), so therefore $\mathbb{Q}$ is at most countable. Since $\mathbb{Q}$ is not finite, it must be countably infinite.
Jared
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Well, I became confused. So what does mean "has the same cardinality as the natural numbers"? Does countable mean that we count them or mean anything else? – LoveMath Mar 17 '13 at 23:01
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1Two sets have the same cardinality if we can exhibit a bijection between them, i.e., a map that is one-to-one on the elements. Any set that can be put in one-to-one correspondence in this way with the natural numbers is called countable. In some sense, this means there is a way to label each element of the set with a distinct natural number, and all natural numbers label some element of the set. – Jared Mar 18 '13 at 00:33
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For an explicit enumeration of positive rationals, you can use the Calkin-Wilf sequence: $$q_{i+1}= \frac{1}{ \lfloor q_i \rfloor +1 - \{q_i\} }, \ q_0=1.$$
More details can be found in Proofs from THE BOOK.
Seirios
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