Irrational Square Roots
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From: Gc on 8 Aug 2010 08:48 On 5 elo, 00:17, bill <b92...(a)yahoo.com> wrote: > Conjectrure: If N is an integer and Sqrt(N) is a > decimal number; then Sqrt(N) is irrational. > > Let R be any real number with a square root. > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > cannot be an integer. Therefore, Sqrt(N) is an > infinite decimal. > > regards, Bill J You can easily generelize the original proof that squart(2) is irrational. Now let` s say that squart(N) can be expressed in a form a/b where g.c.d(a,b) = 1. Then a^2/b^2 = N, thus a^2 = N*b^2, thus either b=1 or g.c.d(a,b) > 1..
From: Bill Dubuque on 8 Aug 2010 11:42 Gerry Myerson <gerry(a)math.mq.edu.au> wrote: > > In any event, my feeling is this: if you had gone up to Euclid > and asked him, in terms that would have made sense to him, > whether the square root of 101 is rational, he would have been > able to answer the question without hesitation and he would have > been able to provide you with a proof of (what we would call) > the irrationality. I can't cite any documentation for this feeling, > so you would be justified in remarking that I am not being serious, > but let me ask you: do you think my feeling is wrong? I'm in no position to comment on your "feelings". As I said earlier, there are many subtleties involved when interpreting the Elements, not the least of which involves the various different conceptions of numbers throughout (arithmetical, geometrical, ratios, etc). Not being an expert on the mathematics of that period I would not be so naive to offer any opinion on such without serious study. Compare also what Franz Lemmermeyer wrote in [1], esp. his remark "while this is something Euclid might not have done", viz. In book VIII, Prop. 9, Euclid claims that if two coprime numbers a and b can be joined by a feometric sequence of length n+1, then the same is true for the pairs 1, a and 1, b. I'm wondering whether it has been noticed before how powerful this result is. For example, assume that a, c, b is such a sequence; then there exist numbers d, e such that 1, d, a and 1, e, b are geometric, i.e., a and b are perfect squares. Now a, c, b geometric implies b:c = c:a, i.e., ab = c^2. Similarly, Prop. VIII.6 implies the irrationality of square roots of nonsquares (or, more generally, the corresponding result for arbitrary powers): if a, b, ..., h is geometric and a divides h, then a divides b. Thus if n:1 = p^2:q^2 with p/q in lowest terms, then q^2:pq:bq^2 and q : p : nq are geometric, hence q divides p, and n is a perfect square. While this is something Euclid might not have done (in n:1 = p^2:q^2 we have ratios of numbers on the left and ratios of solid numbers on the right), it should have been ok for the contemporaries of Fermat and Euler. Was this noticed by anyone? --Bill Dubuque [1] Historia matematica, May 15, 2005, Euclid's VIII.9 http://mathforum.org/kb/thread.jspa?threadID=1150675
From: bill on 9 Aug 2010 16:16 On Aug 4, 2:17 pm, bill <b92...(a)yahoo.com> wrote: > Conjectrure: If N is an integer and Sqrt(N) is a > decimal number; then Sqrt(N) is irrational. > > Let R be any real number with a square root. > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > cannot be an integer. Therefore, Sqrt(N) is an > infinite decimal. > > regards, Bill J Some corrections: Conjecture: If N is an integer and Sqrt(N) > floor[( Sqrt(N)], then Sqrt(N) is not rational. Suppose that Sqrt(N) is rational. Then Sqrt(N) can be written as an integer + a fraction, say (I + F) Or (N) = I^2 + 2*I*F + F^2. Can (2*I*F + F^2) be an integer? Only if F = 0.9... . Therefore: The conjecture is true for all Sqrt(N) < (I + 0.9...). The conjecture is true for all N iff, 1) no integer N has a square root of (I + 0.9...), or 2) (I + 0.9...) is not rational BJ
From: Gerry Myerson on 9 Aug 2010 18:45 In article <7d557612-615b-4a98-9f40-ca165762da59(a)s24g2000pri.googlegroups.com>, bill <b92057(a)yahoo.com> wrote: > On Aug 4, 2:17�pm, bill <b92...(a)yahoo.com> wrote: > > Conjectrure: �If N is an integer and �Sqrt(N) is a > > decimal number; then Sqrt(N) is irrational. > > > > Let R be any real number with a square root. > > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > > cannot be an integer. Therefore, Sqrt(N) is an > > infinite decimal. > > > > regards, Bill J > > Some corrections: > > Conjecture: If N is an integer and > Sqrt(N) > floor[( Sqrt(N)], then Sqrt(N) is not rational. To you, it's a conjecture. To anyone whose mathematical education has gotten up to where Euclid was, it's a simple theorem, indeed, it's just a minor rewriting of the theorem we've been talking about since the beginning of this thread. > Suppose that Sqrt(N) is rational. Then Sqrt(N) can > be written as an integer + a fraction, say (I + F) Where, by "fraction", you mean, a rational number strictly between zero and one. > Or (N) = I^2 + 2*I*F + F^2. Can (2*I*F + F^2) > be an integer? Only if F = 0.9... . True, but I'd like to see you prove it! -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: spudnik on 10 Aug 2010 13:30
and, it's simpler to say that, it is rational, iff the decimal part (after a finite number of places) repeats, *including* a tail of zeroes or nines. thus: you don't give us any kind of an experimental verification, perhaps because you only have your toe in it. the surface of the sphere is pi*d*d, and it is four times the surface of the great circle -- a thing that Bucky apparently didn't know, oddly enough. it is pretty laughable, taht you'd think that you are dysproving F"L"T, because it is clear from the available stuff that it was the key to his method (along with the fact that he basically created numbertheorie, dood .-) --les ducs d'Enron! http://tarpley.net --Light, A History! http://wlym.com |