Irrational Square Roots
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From: Bill Dubuque on 7 Aug 2010 12:17 Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: > Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> wrote: >> Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: >>> Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> wrote: >>>> Butch Malahide <fred.galvin(a)gmail.com> wrote: >>>>> erry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>> wrote: >>>>>> bill <b92...(a)yahoo.com> wrote: >>>>>>> >>>>>>> Conjectrure: If N is an integer and Sqrt(N) is a >>>>>>> decimal number; then Sqrt(N) is irrational. >>>>>> >>>>>> If you manage to give a precise definition of decimal number >>>>>> and do it in such a way that integers don't count as decimal >>>>>> numbers, you will find that your "conjectrure [sic]" was proved >>>>>> 2000 years ago. >>>>> >>>>> Are you sure? 2000 years ago seems early for such a general result. >>>>> And yet, I recall reading that Theodorus is supposed to have proved >>>>> the irrationality of the square roots of nonsquare integers up to and >>>>> including 17, sometime--let me look it up--in the 5th century BC. >>>>> Yeah, I guess another 4 or 5 centuries would be enough time for >>>>> somebody to work out the generalization. Do you know who it was? >>>> >>>> No. But with guys like Archimedes and Diophantus and Eudoxus >>>> and Euclid all hanging around, I'm confident that someone figured >>>> it out back then. >>> >>> Hopefully the above remark was not meant to be taken seriously. >>> >>> Euclid did in fact prove some results about geometric sequences that, >>> to us, easily implies such results about the irrationality of sqrts. >>> But no one appears to have noticed such applications in Euclid's time. >>> Indeed, only rarely are such applications noticed even nowadays. >> >> Perhaps Bill is referring to Proposition 22 of Book VII of the Elements. >> Here is a paragraph from Roger Cooke's essay, Life on the mathematical >> frontier, Notices of the AMS 57 (April 2010) page 473: >> >> It thus appears that in Plato¹s time a general proof that the square >> root of a nonsquare integer is irrational had not been found. However, >> the Greeks did eventually find a proof. Proposition 22 of Book VII of >> the Elements asserts that if three integers are in proportion and the >> first is a square, then the third is, also. In our language, if a, b, >> and c are integers, a/b = b/c, and a is the square of an integer, then >> c is the square of an integer. (By writing this paragraph, I have >> deliberately produced a pseudo-syllogism of type 2. Euclid would not >> have recognized the proposition in the first sentence, since the phrase >> irrational number would have been an oxymoron to him.) > > One has to be very, very careful when interpreting ancient mathematics. > In particular I would not trust any such claim unless it was made by > someone who was an expert in the mathematics of that period. No such > expert has ever made such a claim - to the best of my knowledge. I regret to report that - after perusing some of Roger Cooke's works on historical matters - the above cautionary remark is not strong enough. Mr. Cooke's writings not only contain historical inaccuracies but also serious mathematical errors. For example, the first few places that I looked in his book "Classical Algebra - It's Nature, Origins, and Uses" had errors that would be obvious even to a beginning algebra student: On p.156 he incorrectly states the definition of a UFD - by completely omitting any mention of some necessary finiteness condition (e.g. ACCP) "In any ring, a prime is irreducible. If the converse is true, the ring is called a unique factorization domain or a Gaussian domain" Note also the above confusion between rings and domains. Even with an appropriate finiteness condition, it is not necessarily true that a unique factorization *ring* is an *integral domain*. In fact there isn't even a unique definition of unique factorization rings. In a couple other places he makes remarks that seem to imply that he believes a polynomial is irreducible if it has no rational roots, e.g. on p.114 he writes "x^6 + [...] + 31 is irreducible [over Q] since any solution of it would have to be an integer that divides 31" Later he reinforces this misconception when he writes on p.161 "This principle also converts the finding of rational roots of an equation with rational coefficients into a finite search. Hence that problem can be solved algorithmically, and it is always possible to tell when a polynomial with rational coefficients is irreducible over the rational numbers." I stopped perusing the book after finding such serious errors in the first few topics that I randomly chose to peruse. --Bill Dubuque
From: Bill Dubuque on 7 Aug 2010 13:35 Gerry <gerry(a)math.mq.edu.au> wrote: > On Aug 7, 8:02 am, bill <b92...(a)yahoo.com> wrote: >> erry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>> wrote: > >>> Ah. How do you get from b^2 divides a^2 to b divides a? >>> That's the crux of the matter - it's not hard, but it does >>> require proof. > >> Given If B^2/A^2 is an integer, then B/A is an >> integer. >> >> If B is prime, B^2 is divisible only 1, B or B^2. >> Then A is 1, or B >> >> If B is not prime, then B is the product of several >> prime factors. >> >> The prime factors of B^2 consist solely of the >> prime factors of B. >> >> If B^2 is divisible by A^2, then all of the prime factors of A^2 are >> prime factors of B^2. >> >> The prime factors of B^2 are prime factors of B. >> The prime factors of A^2 are prime factors of A. >> Then, the prime factors of A are prime factors of B. >> Therefore, A is a factor of B. > > You had me until that last line. > > The prime factors of 18 are prime factors of 12. > Therefore, 18 is a factor of 12. > Back to the drawing board, Bill. He's overlooking multiplicity, e.g. p^n & p have same prime factors but not p^n|p if n>1. It fails precisely when multiple factors occur: LEMMA TFAE for naturals M, N 1) M|N if prime p|M => p|N 2) M is squarefree, i.e. p|M => not pp|M PROOF (1 => 2) if pp|M then N = M/p is contra 1) (2 => 1) e.g. p,q,r|N => pqr|N for nonassoc primes p,q,r --Bill Dubuque
From: bill on 7 Aug 2010 18:19 On Aug 6, 3:42 pm, Gerry <ge...(a)math.mq.edu.au> wrote: > On Aug 7, 8:02 am, bill <b92...(a)yahoo.com> wrote: > > > > > On Aug 5, 6:22 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > > wrote: > > > Ah. How do you get from b^2 divides a^2 to b divides a? > > > That's the crux of the matter - it's not hard, but it does > > > require proof. > > Given If B^2/A^2 is an integer, then B/A is an > > integer. > > > If B is prime, B^2 is divisible only 1, B or B^2. > > Then A is 1, or B > > > If B is not prime, then B is the product of several > > prime factors. > > > The prime factors of B^2 consist solely of the > > prime factors of B. > > > If B^2 is divisible by A^2, then all of the prime factors of A^2 are > > prime factors of B^2. > > > The prime factors of B^2 are prime factors of B. > > > The prime factors of A^2 are prime factors of A. > > > Then, the prime factors of A are prime factors of B. > > > Therefore, A is a factor of B. > > You had me until that last line. > > The prime factors of 18 are prime factors of 12. > Therefore, 18 is a factor of 12. > > Back to the drawing board, Bill. > -- > GM Recall that all B/A > 1 and (B^2)/(A^2) is an integer (18^2)/(12^2) is not an integer. If A is not a factor of B, we can reduce the fraction B/A so that none of the primes in B are factors in A. Then A^2 cannot be a factor of B^2. Wait an minute, that's it! If B and have any common prime factors, these can be divided out until B & A have NO common prime factors! If B & A have NO common prime factors, then B^2 & A^2 have NO common prime factors. Therefore, (B^2)/(A^2) cannot be an integer. BJ
From: bill on 7 Aug 2010 18:34 On Aug 6, 5:02 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <38d15a0a-0045-419e-bb28-3c5bafb41...(a)o7g2000prg.googlegroups.com>, > > > > bill <b92...(a)yahoo.com> wrote: > > On Aug 5, 6:22 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > > wrote: > > > In article > > > <bf978706-879b-4062-80de-dcb8c13c2...(a)h40g2000pro.googlegroups.com>, > > > > bill <b92...(a)yahoo.com> wrote: > > > > So I need to prove that Sqrt(I) cannot be a > > > > repeating decimal. > > > > > Let Sqrt(I) = A/B > > > > > If A & B are integers, then (A^2)/(B^2) is an > > > > integer iff (B^2) is a factor of (A^2). > > > > Then B is a factor of A. > > > > Ah. How do you get from b^2 divides a^2 to b divides a? > > > That's the crux of the matter - it's not hard, but it does > > > require proof. > > > > > Then A/B is an integer. > > > > > Therefore,if Sqrt(I)is not an integer, it is not rational. > > > > > I suppose that if I looked hard enough, I would > > > > learn that this analysis has already been > > > > accomplished? > > > > Really, you don't have to look very hard at all. > > > Many many many intro number theory texts > > > will tell you all about this. > > > > -- > > > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) > > > Given If B^2/A^2 is an integer, then B/A is an > > integer. > > > If B is prime, B^2 is divisible only 1, B or B^2. > > Then A is 1, or B > > > If B is not prime, then B is the product of several > > prime factors. > > > The prime factors of B^2 consist solely of the > > prime factors of B. > > > If B^2 is divisible by A^2, then all of the prime factors of A^2 are > > prime factors of B^2. > > > The prime factors of B^2 are prime factors of B. > > > The prime factors of A^2 are prime factors of A. > > > Then, the prime factors of A are prime factors of B. > > > Therefore, A is a factor of B. > > Suppose A = 12 and B = 6. > > The prime factors of 12 are all prime factors of 6 but 12 is not a > factor of 6. > > Oops. If B/A < 1, Then (B^2)/(A^2) << 1. Therefore your example is invalid. B should be 12 and A should be 6. Then B/A = 2
From: bill on 7 Aug 2010 18:45
On Aug 7, 10:35 am, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: > Gerry <ge...(a)math.mq.edu.au> wrote: > > On Aug 7, 8:02 am, bill <b92...(a)yahoo.com> wrote: > >> erry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>> wrote: > > >>> Ah. How do you get from b^2 divides a^2 to b divides a? > >>> That's the crux of the matter - it's not hard, but it does > >>> require proof. > > >> Given If B^2/A^2 is an integer, then B/A is an > >> integer. > > >> If B is prime, B^2 is divisible only 1, B or B^2. > >> Then A is 1, or B > > >> If B is not prime, then B is the product of several > >> prime factors. > > >> The prime factors of B^2 consist solely of the > >> prime factors of B. > > >> If B^2 is divisible by A^2, then all of the prime factors of A^2 are > >> prime factors of B^2. > > >> The prime factors of B^2 are prime factors of B. > >> The prime factors of A^2 are prime factors of A. > >> Then, the prime factors of A are prime factors of B. > >> Therefore, A is a factor of B. > > > You had me until that last line. > > > The prime factors of 18 are prime factors of 12. > > Therefore, 18 is a factor of 12. > > Back to the drawing board, Bill. > > He's overlooking multiplicity, e.g. p^n & p have same prime factors > but not p^n|p if n>1. It fails precisely when multiple factors occur: > > LEMMA TFAE for naturals M, N > > 1) M|N if prime p|M => p|N > > 2) M is squarefree, i.e. p|M => not pp|M > > PROOF (1 => 2) if pp|M then N = M/p is contra 1) > > (2 => 1) e.g. p,q,r|N => pqr|N for nonassoc primes p,q,r > > --Bill Dubuque What does the "|" in "M|N" mean? |