Irrational Square Roots
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From: Pubkeybreaker on 5 Aug 2010 18:01 On Aug 5, 5:55 pm, bill <b92...(a)yahoo.com> wrote: > On Aug 4, 7:47 pm, Bart Goddard <goddar...(a)netscape.net> wrote: > > > bill <b92...(a)yahoo.com> wrote in news:d698bbe6-93d4-4b13-bf88- > > d81debce6...(a)u38g2000prh.googlegroups.com: > > > >> 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". > > > So pick one, and then use it to tell us whether > > 3.99999..... is a finite decimal. > > > -- > > Cheerfully resisting change since 1959. > > From Mathworld: > > "A regular number, also called a finite decimal (Havil 2003, p. 25), > is a positive number that has a finite decimal expansion." > > I don't know what "....." means. If it means > "expand infinitely", then 3.99999..... is not > a finite decimal. So 4 is not a finite decimal??? Let me give a hint to the clueless out there. Every rational number that has a finite decimal representation ALSO has an infinite decimal representation.
From: Bart Goddard on 5 Aug 2010 18:14 bill <b92057(a)yahoo.com> wrote in news:6d6d4901-27da-4ecc-aa97- 10c9bd2a94ba(a)q35g2000yqn.googlegroups.com: >> > I'll rely on any of the several published definitions >> > of "finite decimal" �or "terminating decimal" or >> > "regular number". >> >> So pick one, and then use it to tell us whether >> 3.99999..... is a finite decimal. >> >> -- >> Cheerfully resisting change since 1959. > > From Mathworld: > > "A regular number, also called a finite decimal (Havil 2003, p. 25), > is a positive number that has a finite decimal expansion." > > I don't know what "....." means. If it means > "expand infinitely", then 3.99999..... is not > a finite decimal. And yet 2 squared is 3.9999..... Worse, 1.999.... squared is 4. Your "proof" doesn't work. -- Cheerfully resisting change since 1959.
From: Gerry Myerson on 5 Aug 2010 18:52 In article <41adcf83-f3b0-4869-bcd5-bc4b4fb5afed(a)m1g2000yqo.googlegroups.com>, Butch Malahide <fred.galvin(a)gmail.com> wrote: > On Aug 4, 7:17�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. > > 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. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: bill on 5 Aug 2010 19:28 On Aug 4, 8:39 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote: > In article > <d698bbe6-93d4-4b13-bf88-d81debce6...(a)u38g2000prh.googlegroups.com>, > > > > 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. > > Then you have a lot of catching up to do. It has been proved > that if n is an integer and the cube root of n is not an integer > then the cube root of n is irrational. Similarly for the 4th root, > 5th root, etc. > > Indeed, it has been proved that if you have a polynomial > with integer coefficients and leading coefficient one, and > you have a root of the polynomial and the root is not an > integer, then the root 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. > > > > 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". > > "Regular number" I never heard of. > "Terminating decimal;" by the definitions with which I am familiar, > 2.00000... is a terminating decimal. So, if R is 4, then > the square root of R is a terminating decimal, but the square > of the square root of R is an integer. > So, if "finite decimal" means the same as "terminating decimal," > then your statement is flat out wrong. > > > > 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. > > No kidding. I'm just pointing out that when you say "square root > of N is an infinite decimal," you are not saying "square root of N > is irrational." > > -- > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) 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. 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? regards, Bill J.
From: bill on 5 Aug 2010 19:32
On Aug 4, 4:56 pm, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote: > On 4 Aug, 22:17, 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. > > 4.0 is a real number with sqrt = 2.0 > 2.0 is a finite decimal > 2.0x2.0 = 4.0, an integer To say that 2.0 is a finite decimal is to take advantage of a deficiency in the definition of a finite decimal. |