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[parent] proof that the rationals are countable (Proof)

Suppose we have a rational number $ \alpha = p/q$ in lowest terms with $ q>0$. Define the “height” of this number as $ h(\alpha) = \vert p\vert + q$. For example, $ h(0) = h(\frac{0}{1}) = 1$, $ h(-1) = h(1) = 2$, and $ h(-2) = h(\frac{-1}{2}) = h(\frac{1}{2}) = h(2) = 3.$ Note that the set of numbers with a given height is finite. The rationals can now be partitioned into classes by height, and the numbers in each class can be ordered by way of increasing numerators. Thus it is possible to assign a natural number to each of the rationals by starting with $ 0, -1, 1, -2, \frac{-1}{2}, \frac{1}{2}, 2, -3, \ldots$ and progressing through classes of increasing heights. This assignment constitutes a bijection between $ \mathbb{N}$ and $ \mathbb{Q}$ and proves that $ \mathbb{Q}$ is countable.

A corollary is that the irrational numbers are uncountable, since the union of the irrationals and the rationals is $ \mathbb{R}$, which is uncountable.



"proof that the rationals are countable" is owned by alozano. [ owner history (1) ]
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See Also: rational number, irrational number, countable, irrational

Keywords:  countable, rational number

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Cross-references: union, uncountable, irrational numbers, countable, bijection, natural number, numerators, increasing, classes, rationals, finite, height, number, lowest terms, rational number
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This is version 5 of proof that the rationals are countable, born on 2001-11-16, modified 2002-02-25.
Object id is 927, canonical name is ProofThatTheRationalsAreCountable.
Accessed 8561 times total.

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AMS MSC: 03E10 (Mathematical logic and foundations :: Set theory :: Ordinal and cardinal numbers)
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