From: Ken.Pledger@vuw.ac.nz (Ken Pledger)
Newsgroups: sci.math
Subject: Irrationals (was Re: please don't laugh...)
Date: Wed, 09 Dec 1998 15:05:55 +1200
In article <14055-366DB789-76@newsd-153.iap.bryant.webtv.net>,
ol3@webtv.net (Oscar Lanzi III) wrote:
> ....
> Q. Is sqrt(2) the first number ever proven irrational? And how many
> other irrationals were known as such in ancient times?
>
> --OL
Sqrt(2) was indeed the first, as far as we can tell. Its
irrationality was discovered perhaps between about 430 and 410 B.C. by a
Greek whose name is uncertain. Shortly after then, Theodorus of Cyrene
generalized the result by proving the irrationality of the square roots of
3, 5, ..., 15, but got stuck at sqrt(17) because of the method he was
using. His proof is lost, but historians of ancient mathematics have had
a great time trying to reconstruct it.
Theaetetus of Athens, a student of Theodorus, may have had a lot to
do with proving that sqrt(n) is irrational whenever the natural number
n is not a perfect square, and also with proving the irrationality of
more elaborate nests of square roots like sqrt(5 + 2 sqrt(5)) - sqrt(5 -
2 sqrt(5)). The theory of such things, without modern notation, has
survived in the notoriously difficult Book X of Euclid's "Elements" (c.300
B.C.).
So the simple answer to your second question is "infinitely many, but
of a rather special kind based on square roots".
Ken Pledger.
==============================================================================
From: Ken.Pledger@vuw.ac.nz (Ken Pledger)
Newsgroups: sci.math
Subject: Re: Irrationals (was Re: please don't laugh...)
Date: Mon, 14 Dec 1998 12:56:17 +1200
In article <74ps3l$f26$1@cantuc.canterbury.ac.nz>,
mathwft@math.canterbury.ac.nz (Bill Taylor) wrote:
> |> > Q. Is sqrt(2) the first number ever proven irrational?
>
> |> Sqrt(2) was indeed the first, as far as we can tell.
>
> I don't think this is quite accurate, in fact. The first such number
> was the golden ratio, (sqrt5 +1)/2, as expressed in the ratio of lengths
> in a regular pentacle. I'm pretty sure this was the one that was
> first discovered by the Pythagoreans, allegedly the unfortunate Hypasus,
> subsequently denavisated....
There's a lot of folklore about this sort of thing, especially as
most of the historical evidence is indirect, and also because many popular
writers pay scant attention to what historical evidence there is. In
particular, the story of Hippasus rests on very shaky evidence.
Our most reliable surviving account of the early history of
irrationality occurs in Plato's dialogue "Theaetetus," which mentions the
earlier work of Theodorus in proving the irrationality of the square roots
of 3, 5, ..., 15. The usual assumption is that he started with sqrt(3)
because the irrationality of sqrt(2) was already known and he was trying
to generalize it. A very detailed modern study of the whole business is
Wilbur R. Knorr, "The Evolution of the Euclidean Elements," 1975. An
alternative view of some aspects is in David H. Fowler, "The Mathematics
of Plato's Academy," whose second edition is about to appear.
Although such scholars may differ in various ways, they're usually
much too careful to say anything like your "The first such number was the
golden ratio".
Ken Pledger.
==============================================================================
From: Richard Carr
Newsgroups: sci.math
Subject: Re: how to prove: if x is not a perfect square, sqr (x) is irrational,
x positive integer
Date: Sat, 12 Dec 1998 23:12:34 -0500
On Sat, 12 Dec 1998, TS wrote:
:Date: Sat, 12 Dec 1998 20:40:37 GMT
:From: TS
:Newsgroups: sci.math
:Subject: how to prove: if x is not a perfect square, sqr (x) is irrational,
x positive integer
:
:I read this theorem somewhere. In class we proved sqr (2) and sqr (3)
:irrational by contradiction.
:
:How do you prove all non-perfect square integers have irrational
:square roots?
There was a post here recently showing that x=sqrt(2) is irratioonal by
looking at x+1 and x-1. Then (x+1)(x-1)=1 and x-1=a/b gives
a/b+2=b/a, so that (a+2b)/b=b/a. Without loss of generality (a,b)=1 and
then also (a+2b,b)=1 and the conclusion (Euclid VII.20) is that a+2b=b,
b=a, which gives a=b=0, a contradiction.
We can use a similar idea, although we won't be able to use reciprocals,
generally. Suppose sqrt(n) is rational.
Then for any a/b ((a,b)=1), sqrt(n)-a/b=c/d is rational, where w.l.o.g.
(c,d)=1, d>0.
We get c/d+2a/b=n-a^2/b^2.
So cb^2=d(nb^2-a^2-2ab). Now b^2|RHS, but b^2 is coprime to nb^2-2ab-a^2,
since any prime factor of b^2 is a prime factor of b and hence also of
nb^2-2ab. If it were also of nb^2-2ab-a^2, then it would divide a^2 and
hence a, contrary to (a,b)=1.
Thus we have b^2|d. Write d=b^2r.
Then c=r(nb^2-a^2-2ab). Now any prime factor of r divides c. Since r|d and
(c,d)=1, r is a unit. Since d>0 and b^2>0, r>0. So r=1.
Then d=b^2.
So sqrt(n)=a/b+c/b^2=(ab+c)/b^2. Choosing a/b=0/1, sqrt(n)=c, an integer,
so that n=c^2 is the square of an integer.
Thus, if n is not a square, then sqrt(n) is irrational.
This is not the standard way to do it, but it is sort of interesting.