Irrational Square Roots
in [Math]
|
Prev: CREATION OF NUMBERS AND THE CONSISTANT 0.999 factor PATENT HOPE RESEARCH SHARED WITH THE "FERMATISTS" IN GOOD FAITH
Next: An exact simplification challenge - 102 (Psi, polylog)
From: bill on 5 Aug 2010 19:35 On Aug 5, 4:08 am, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote: > On 5 Aug, 02:38, bill <b92...(a)yahoo.com> wrote: > > > On Aug 4, 5:17 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > > wrote: > > > > In article > > > <6f46e88a-7728-4bed-acd3-59671f5b3...(a)g21g2000prn.googlegroups.com>, > > > > 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. > > > This proof? is for the dilletante. I know that it was proved for N = > > 2. I did not that it had been proved > > for all integers. > > > > > 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. > > > > Again, you had better give a precise definition of "finite decimal." > > > I'll rely on any of the several published definitions > > of "finite decimal" or "terminating decimal" or > > "regular number". > > > > You might also want to note that there are infinite decimals > > > that are not irrational, e.g., .33333333333333.... > > > True, but the squares of such "rational" numbers are > > never exact integers. > > The square of 2.000000... > is 4.00000.. > > Is 4.000.. and exact integer or not ? No, in the same sense that 2.0000 is not an exact integer.
From: Bill Dubuque on 5 Aug 2010 21:04 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. --Bill Dubuque
From: Gerry Myerson on 5 Aug 2010 21:18 In article <l2ceiectuto.fsf(a)shaggy.csail.mit.edu>, 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. Euclid's elements are available online. Book X is available (in translation) at http://aleph0.clarku.edu/~djoyce/java/elements/bookX/bookX.html Proposition 9 says, in part, "squares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either." -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Gerry Myerson on 5 Aug 2010 21:22 In article <bf978706-879b-4062-80de-dcb8c13c27e9(a)h40g2000pro.googlegroups.com>, bill <b92057(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 (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Gerry Myerson on 5 Aug 2010 21:25
In article <1956dc10-5e54-4374-b5c1-3ef246e62edb(a)m35g2000prn.googlegroups.com>, bill <b92057(a)yahoo.com> wrote: > On Aug 5, 4:08�am, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> > wrote: > > On 5 Aug, 02:38, bill <b92...(a)yahoo.com> wrote: > > > > > On Aug 4, 5:17�pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > > > wrote: > > > > > > In article > > > > <6f46e88a-7728-4bed-acd3-59671f5b3...(a)g21g2000prn.googlegroups.com>, > > > > > > �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. > > > > > This proof? is for the dilletante. I know that it was proved for N = > > > 2. I did not that it had been �proved > > > for all integers. > > > > > > > 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. > > > > > > Again, you had better give a precise definition of "finite decimal." > > > > > I'll rely on any of the several published definitions > > > of "finite decimal" �or "terminating decimal" or > > > "regular number". > > > > > > You might also want to note that there are infinite decimals > > > > that are not irrational, e.g., .33333333333333.... > > > > > True, but the squares of such "rational" numbers are > > > never exact integers. > > > > The square of 2.000000... > > is 4.00000.. > > > > Is 4.000.. and exact integer or not ? > > No, in the same sense that 2.0000 is not an > exact integer. And what sense is that? It is incontrovertible that 2.0000... = 2 I mean, I can almost understand someone denying that .999... = 1, but in all my years I've never seen anyone deny that 2.0000... = 2 Isn't 2 an exact integer? -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |