# How to prove that $\sqrt 3$ is an irrational number? [duplicate]

Possible Duplicate:
$a^{1/2}$ is either an integer or an irrational number

I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?

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## marked as duplicate by JavaMan, Zev ChonolesSep 15 '11 at 6:28

How about showing your proof for $\sqrt 2$ so you can be guided on how to modify that proof for $\sqrt 3$? –  Guess who it is. Sep 15 '11 at 1:29
Nobody seems to have posted it, so: consider applying the rational root theorem to the polynomial $x^2-3$... –  Guess who it is. Sep 15 '11 at 3:12

If you follow through the usual proof for $\sqrt{2}$ substituting $3$ for $2$, it goes through just fine. Let $\sqrt{3}=\frac{p}{q}, p,q$ relatively prime. $3=\frac{p^2}{q^2}$, so $3$ divides $p$ and so on.

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The proof is very similar to the irrationality of square root of two.

Let $\sqrt{3} = \frac a b$, where a and b have no common factors besides $1$

As $3b^2 = a^2$ so $a^2$ is a multiple of $3$, and hence $a$ should be a multiple of $3$. Let $a = 3k$, then $b^2 = 3k^2$, and $b$ must also be a multiple of three. You will arrive at a contradiction to the earlier assumption that $a$ and $b$ have no common factors.

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Only we don't know which $\sqrt2$ proof OP knows. –  Gerry Myerson Sep 15 '11 at 1:36
That's how you do it in Discrete Mathematics. –  darksky Sep 15 '11 at 2:12

Another variation on a theme:

If $\sqrt 3 = m/n$, where $n$ is as small as possible, then $$\frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1} = \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$ and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).

This can be used to show (IIRC) that $\sqrt k$ is irrational for any non-square k by multiplying $\sqrt k$ by $\frac{\sqrt k - j}{\sqrt k - j}$ where $j = \lfloor \sqrt k \rfloor$.

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A continued fraction proof of the irrationality of $x = \sqrt{3} - 1$, from which the irrationality of $\sqrt{3}$ follows. (A continued fraction proof of $\sqrt{2}$ can be found here: How can you prove that the square root of two is irrational? )

Notice that $x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$

This can be re-written as

$$x(3+x) = 2+x$$

$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$

And thus

$$x = [1,2,1,2,\dots]$$

and so is irrational.

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How do you prove that an infinite continued fraction is irrational? Seems to me that you have to prove that a rational number has a finite continued fraction, which is not trivial, but not extremely difficult. In other words, this requires more machinery, which many of the other proofs here do not need as they are self contained. –  marty cohen Oct 19 '13 at 23:40
@martycohen: Most introductory textbooks on number theory should have that (for eg: Hardy & Wright, Ivan Niven & Zuckerman..., of course my sample space is small.). I don't understand your point, though. The more different proofs, the better, isn't it? This is math, and not stackoverflow... :-) –  Aryabhata Oct 20 '13 at 18:49

Suppose $\sqrt{3} = a/b$ where $a$ and $b$ have no common factor (and note $b\neq 1$). Then $3 = a^2/b^2$, but $a^2$ and $b^2$ no common factors to cancel to produce an integer, so we have a contradiction.

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A well-known variant of the usual proof may be clearer. If $\sqrt{3}=\frac{a}{b}$, then $a^2=3b^2$. Recall the Fundamental Theorem of Arithmetic and consider the exponent of $3$ in the prime factorization of both sides. On the left you have an even exponent. On the right you have an odd exponent, contradiction. This approach goes one to prove that $\sqrt m$ is rational iff $m$ is a square.

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Alternatively, you can use the rational root test on the polynomial equation $x^2-3=0$ (whose solutions are $\pm \sqrt{3}$). If $\frac{a}{b}$ is a solution to the equation (with $a,b\in \mathbb{Z}$ and $b\not=0$), then $b \vert 1$ and $a \vert 3$, hence $\frac{a}{b}\in \{\pm 1, \pm 3\}$. However, it is straightforward to check that none of $\pm 1, \pm 3$ are solutions to $x^2-3=0$. Therefore there are no such rational solutions and $\sqrt{3}$ is irrational.

In fact, in the above argument, if we replace 3 with an arbitrary prime $p\in \mathbb{N}$ and 2 with an arbitrary $m\in \mathbb{N}$, $m\geq 2$, the same argument shows that $\sqrt[m]{p}$ is irrational.

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