elementary-number-theoryradicalsrationality-testing
I am trying to review some old algebra, and in particular I wanna show that $\sqrt2$ is irrational
Since integers are the only integral elements of $\mathbb Q$ over $\mathbb Z$, assume $r=\sqrt 2$ is rational, then $ r^2-2=0$ is a polynomial in wich $r$ is a root, so $r$ must be integer, but since $1<r<2$, this is a contradiction.
Am I doing this right? Can I from there generalize it?
The statement $x|a^2\Rightarrow x|a$ is not generally true (Counter-example : $8|4^2$ but we have not $8|4$).
The proof works for $x$ prime, because, then, this statement is true by Euclid's lemma.
That's a perceptive observation. In fact the proof works in domains more general than UFDs, e.g. it follows from the monic case of the Rational Root Test, i.e. it works in all integrally-closed domains, which are far from being UFDs. But let's looks closer at the specific proof you give. It employs $\rm\:(m,n)=1\:\Rightarrow\:(m^2,n^2)=1.\:$ This follows from a special case of Euclid's Lemma, viz. $\rm\:(a,b) = 1 = (a,c)\:$ $\Rightarrow$ $\rm\:(a,bc)=1.\:$ Namely,$\rm\: a=m,\,b=n=c\:$ yields $\rm\:(m,n^2)=1.\:$ Finally $\rm\:a=n^2,\,b = m = c\:$ yields $\rm\:(m^2,n^2) = 1.$
This special case of Euclid's Lemma holds true in any GCD domain, in particular in any UFD or any Bezout domain. But it is weaker, i.e. there are domains satisfying this identity which are not GCD domains, i.e. where some elements have no gcd. In fact this is equivalent to Gauss' Lemma, that the product of primitive polynomials is primitive (or the special case for degree $1$ polynomials). Indeed if $\rm\:(m,n)=1\:$ then $\rm\:m\,x\pm n\:$ is primitive, thus so is $\rm\:(m\,x-n)\,(m\,x+n) = m^2\, x - n^2,\:$ hence $\rm\:(m^2,n^2) = 1.\:$ Below is an excerpt of my sci.math post on 2003/5/3 (see here or here) which shows the relationships between various domains closely related to GCD domains. If you wish to understand precisely the class of rings satisfying such properties then see the links in said post, and also google "root-closed" domains.
There has been much study of domains related to GCD domains. Below are some of them, in increasing order of generality.
PID: $\ \ $ every ideal is principal
Bezout: $\ \ $ every ideal (a,b) is principal
GCD: $\ \ $ (x,y) := gcd(x,y) exists for all x,y
SCH: $\ \ $ Schreier = pre-Schreier & integrally closed
SCH0: $\ \ $ pre-Schreier: a|bc $\, \Rightarrow\, $ a = BC, B|b, C|c
D: $\ \ $ (a,b) = 1 & a|bc $\,\Rightarrow\,$ a|c
PP: $\ \ $ (a,b) = (a,c) = 1 $\,\Rightarrow\,$ (a,bc) = 1
GL: $\ \ $ Gauss Lemma: product of primitive polys is primitive
GL2: $\ \ $ Gauss Lemma holds for all polys of degree 1
AP: $\ \ $ atoms are prime [AP = PP restricted to atomic a]
Since atomic & AP $\,\Rightarrow\,$ UFD, reversing the above UFD $\,\Rightarrow\,$ AP path shows that in atomic domains all these properties (except PID, Bezout) collapse, becoming all equivalent to UFD.
There are also many properties known equivalent to D, e.g.
[a] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ a|bc $\,\Rightarrow\,$ a|c
[b] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ a,b|c $\,\Rightarrow\,$ ab|c
[c] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ (a)/\(b) = (ab)
[d] $\ \ $ (a,b) exists $\,\Rightarrow\,$ lcm(a,b) exists
[e] $\ \ $ a + b X irreducible $\,\Rightarrow\,$ prime for b $\ne$ 0 (deg = 1)
Best Answer
Sure you can try. Let $n$ be a square-free natural number, then you show $\sqrt{n}$ is irrational. You can use theory of polynomial to do this. To this end, if $x = \sqrt{n} \to x^2-n = 0$. Then if $\dfrac{p}{q}$ is a rational root,then $p \mid n, q \mid 1 \to q = \pm 1$. Thus any rational root of this equation must be of the form $x = p, p \mid n \to n = p^2$, contradiction to $n$ being square-free. Thus there is no rational root, it means $\sqrt{n}$ is irrational.