Prove that $\sqrt{47}$ is irrational number.
I know that a rational number is written as $\frac{p}{q}$ where $p$ & $q$ are co-prime numbers. But I do not have any idea to prove it irrational number.
Thank you very much!
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2Look at the proof that $\sqrt{2}$ is irrational and emulate it. – Yuval Filmus Jul 24 '15 at 15:37
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See e.g. http://math.stackexchange.com/questions/64643/how-to-prove-that-sqrt-3-is-an-irrational-number – sjb Jul 24 '15 at 15:38
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See also http://math.stackexchange.com/questions/162134/is-this-proof-that-sqrt-2-is-irrational-correct – Dietrich Burde Jul 24 '15 at 15:38
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1if we assume that $\sqrt{47}=\frac{p}{q}$ and $(p,q)=1$ what can you say about the equation $47\cdot q^2=p^2$? – Dr. Sonnhard Graubner Jul 24 '15 at 15:38
2 Answers
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If $\dfrac p q =\sqrt{47}$ then $p^2=q^2\cdot 47$. That means the number of $47$s in the prime factorization of $p^2$ is one more than the number of $47$s in the prime factorization of $q^2$. But that cannot happen because $p^2$ and $q^2$, both being squares, must both have an even number of $47$s in their factorizations.
(This relies on the theorem that tells us that an integer cannot have more than one prime factorization.)
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$47$ is a prime, hence you may just mimic the proof of the irrationality of $\sqrt{2}$. Let:
$$\nu_{47}(m) = \max\{n\in\mathbb{N}: 47^n\mid m\}.$$ Assuming $\sqrt{47}=\frac{p}{q}$ with $\gcd(p,q)=1$, it follows that: $$ p^2 = 47\, q^2 $$ but $\nu_{47}$ of the LHS is even while $\nu_{47}$ of the RHS is odd, contradiction.
Jack D'Aurizio
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