# Proving irrationality

• Feb 4th 2007, 06:18 PM
lyla
Proving irrationality
Hi,

Can anyone help with this one:
Establish the following facts:
√p is irrational for any prime p.
• Feb 4th 2007, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by lyla
Hi,

Can anyone help with this one:
Establish the following facts:
√p is irrational for any prime p.

Euclid's proof.
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Assume,
$\sqrt{p}=a$ a positive integer.
By definition it is equivalent to say,
$p=a^2$.
Now the prime decomposition of $a^2$ has an even amount of prime factors because of the square. While $p$ does not. By uniquness this is impossible.

Pythagorus' Proof.
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Assume that $\sqrt{p}=\frac{n}{m}$ where $n,m$ is a reduced fraction, meaning no common factors.
Then,
$p=\frac{n^2}{m^2}$
$m^2p=n^2$.
Note, the right hand side is divisible by $p$ because the left hand side. Meaning $p$ divides $n^2$. But then $n$ itself is divisible by $p$ by properties of prime numbers. That is $n=pk$.
Subsitute,
$m^2p=k^2p^2$
$m^2=k^2p$
But then the left hand side is divisble by $p$, that is $o$ divides $m^2$. But then $p$ divides $m$. Hence $n$ and $n$ have common factors, contrary to assumption.
(This is also a similar approach to Fermat's principle of infinite descent.)

Eisenstein Proof.
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The polynomial $x^2-p\in \mathbb{Z}[x]$ fits the conditions of Eisenstein irreducibility criterion for $p$. Thus, there are no solution in $\mathbb{Z}$ and hence none in $\mathbb{Q}$.

Rational Roots Proof.
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The polynomial $x^2-p$ can only have zeros for $\pm 1,\pm p$. None of which work. Thus, it is irrational.
• Feb 5th 2007, 03:35 AM
lyla
I have to thank you very much for the help.