Hi,

Can anyone help with this one:

Establish the following facts:

√p is irrational for any prime p.

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- Feb 4th 2007, 06:18 PMlylaProving irrationality
Hi,

Can anyone help with this one:

Establish the following facts:

√p is irrational for any prime p. - Feb 4th 2007, 06:52 PMThePerfectHacker
Euclid's proof.

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Assume,

$\displaystyle \sqrt{p}=a$ a positive integer.

By definition it is equivalent to say,

$\displaystyle p=a^2$.

Now the prime decomposition of $\displaystyle a^2$ has an even amount of prime factors because of the square. While $\displaystyle p$ does not. By uniquness this is impossible.

Pythagorus' Proof.

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Assume that $\displaystyle \sqrt{p}=\frac{n}{m}$ where $\displaystyle n,m$ is a reduced fraction, meaning no common factors.

Then,

$\displaystyle p=\frac{n^2}{m^2}$

$\displaystyle m^2p=n^2$.

Note, the right hand side is divisible by $\displaystyle p$ because the left hand side. Meaning $\displaystyle p$ divides $\displaystyle n^2$. But then $\displaystyle n$ itself is divisible by $\displaystyle p$ by properties of prime numbers. That is $\displaystyle n=pk$.

Subsitute,

$\displaystyle m^2p=k^2p^2$

$\displaystyle m^2=k^2p$

But then the left hand side is divisble by $\displaystyle p$, that is $\displaystyle o$ divides $\displaystyle m^2$. But then $\displaystyle p$ divides $\displaystyle m$. Hence $\displaystyle n$ and $\displaystyle n$ 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 $\displaystyle x^2-p\in \mathbb{Z}[x]$ fits the conditions of Eisenstein irreducibility criterion for $\displaystyle p$. Thus, there are no solution in $\displaystyle \mathbb{Z}$ and hence none in $\displaystyle \mathbb{Q}$.

Rational Roots Proof.

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