Show that $\sqrt{13}$ is an irrational number.
How to direct proof that number is irrational number. So what is the first step.....
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6Perhaps you should clarify what do you mean by "direct" proof - do you mean "not by contradiction"? – Luka Mikec Dec 26 '13 at 11:22
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possible duplicate of use contradiction to prove that the square root of $p$ is irrational – Doktoro Reichard Dec 26 '13 at 11:26
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1@DoktoroReichard That question specifically says to provide a proof by contradiction while this specifically says not to. – Git Gud Dec 26 '13 at 11:27
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@DoktoroReichard The english is a bit dodgy, but the OP wrote 'How to direft proof...' – Git Gud Dec 26 '13 at 11:29
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@Git sorry, misread. Still, it might be relevant. – Doktoro Reichard Dec 26 '13 at 11:29
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@DoktoroReichard Related, yes. Relevant? Maybe. In any case just as you feel the need to keep the possible duplicate comment there, I feel the need to leave my comment warning it isn't so too. – Git Gud Dec 26 '13 at 11:30
5 Answers
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Consider the polynomial $x^2-13\color{grey}{\in \mathbb Z[x]}$. The rational root theorem guarantees its roots aren't rational and since $\sqrt {13}$ is a root of the polynomial, it is irrational.
Git Gud
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3how good that would be to use Rational root theorem for an elementary number theory problem... Keeping that aside, I really love this way :D – Dec 26 '13 at 11:44
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3@ulead86 The rational root theorem isn't deep at all, it has a simple elementary proof. – Git Gud Dec 26 '13 at 11:51
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If by "direct" the question meant "not by contradiction", then actually this proof is not valid, if you'll forgive me the pun. The rational root theorem uses either Gauss's lemma or Euclid's lemma, both of which uses proof by contradiction. ;-) – Balarka Sen Jan 13 '14 at 16:41
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The standard proof that $\sqrt{p}$ is irrational for any prime $p$ is as follows
Let $\sqrt{p} = \frac{m}{n}$ where $m,n\in\mathbb N.$ and $m$ and $n$ have no factors in common.
Now $\frac{m^2}{n^2} = p \Rightarrow m^2 = p \cdot n^2$
Since $p$ is prime and $m^2$ is a multiple of $p$ then $m$ is multiple of $p$
So substitute $m = p \cdot k$
Now $\frac{(p \cdot k)^2}{n^2} = \frac{p^2 \cdot k^2}{n^2} = p\Rightarrow n^2 = p \cdot k^2$
Since $p$ is prime and $n^2$ is a multiple of $p$ then $n$ is multiple of $p$
We now have a contradiction since $m$ and $n$ must have no common factors (except 1) but we have proved that if $\frac{m}{n}$ exits then $m$ and $n$ must have common factor $p$
So $\frac{m}{n}$ can not exist and the square root of any prime is irrational.
Warren Hill
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You can try it this way:
A number is irrational, if you can not find a finite continued fraction.
Now try to write $\sqrt{13}$ as continued fraction and you'll see, its periodic:
$$\sqrt{13}=[3;\overline{1,1,1,1,6}]$$
Hope that is what your you looking for.
ulead86
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The equation $m^2=13n^2$ is a direct contradiction to the Uniqueness part of the Fundamental Theorem of Arithmetic, since the left side has evenly many $13$’s, while the right side has oddly many.
Lubin
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As mentioned, one can quickly prove the irrationality of square roots using the Rational Root Test, or uniqueness of prime factorizations, or other closely related propeerties such as Euclid's Lemma or Bezout's gcd identity. Below is a simple proof using Bezout that I discovered as a teenager.
Theorem $\quad \rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \ $ Note that $\rm\,\ \color{#0a0}{r = a/b},\ \ \gcd(a,b) = 1\ \Rightarrow\ \color{#C00}{ad\!-\!bc \,=\, \bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\,\ $ by Bezout.
$\rm\color{#C00}{That\,}$ and $\rm\: r^2\! = \color{orange}{\bf n}\:\Rightarrow\ \color{#0a0}{0\, =\, (a\!-\!br)}\, (c\!+\!dr) \ =\ ac\!-\!bd\color{orange}{\bf n} \:+\: \color{#c00}{\bf 1}\cdot r \ \Rightarrow\ r \in \mathbb{Z}\ \ \ $ QED
The proof easily generalizes to roots of monic quadratic polynomials (and to higher degrees).
Bill Dubuque
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