Consider $$n = \sqrt a $$ where $a$ is any integer. Is there a rigorous, systematic method of figuring out if $n$ will or will not be a rational number?
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1Yes: prime factorisation! relevant – Myself Jan 12 '15 at 23:15
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It's not hard to show by writing $n$ as a rational in reduced form that it is rational iff it is integral. – Travis Willse Jan 12 '15 at 23:16
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If $a\in\mathbb{N}$, $\sqrt{a}$ is rational iff $a$ is a square. – Jack D'Aurizio Jan 12 '15 at 23:17
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1@Jack D'Aurizio, Thank you very kindly - perfect answer. – Michael Lee Jan 12 '15 at 23:19
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See the various proofs in this thread. – Bill Dubuque Jan 13 '15 at 00:12
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The square root of an integer is rational exactly if it is itself an integer.
Namely, suppose $p/q=\sqrt a$. Then $p^2/q^2=a$ which we assume to be an integer. But this means that $q^2$ divides $p^2$.
Now consider the prime factorizations of $p^2$ and $q^2$. Every prime that appears in $q^2$ must appear in $p^2$ with at least the same exponent. But that means that every prime that appears in the prime factorization of $q$ must appear in $p$ with at least the same exponent (each exponent in the prime factorization of $p^2$ is simply twice the corresponding exponent in the factorization of $p$). In other words, $q$ divides $p$, so $p/q$ is an integer.
hmakholm left over Monica
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The argument implicitly uses in several places the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations. It may fail for other numbers, e.g. in $,R = \Bbb Z[\sqrt{12}],$ we have $, 3 = (\sqrt{12}/2)^2$ is the square of a proper fraction over $R$. Thus any proof must necessarily employ some property special to the ring of integers, in particular, some property not shared by all quadratic number rings, e.g. unique factorization or equivalent properties such as Euclid's Lemma. – Bill Dubuque Jan 13 '15 at 00:06