Irrational Square Roots
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From: bill on 6 Aug 2010 18:02 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.
From: bill on 6 Aug 2010 18:11 On Aug 6, 11:15 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > On Aug 5, 5:55 pm, bill <b92...(a)yahoo.com> wrote: > > > > > "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. > > Sigh. From your quote above, I extract the words: > > "has a finite decimal expansion" > > Note the quantifier, "a". The integer 4 has a finite decimal > expansion. It fact, it has infinitely many different finite > decimal expansions. 4.0, 4.00, 4.000, ..... > > It also has an infinite decimal expansion: 3.999999999........ > > Note that the text you quoted did NOT say "a UNIQUE finite decimal > expansion". You should take this up with the editors of MathWorld.
From: Gerry on 6 Aug 2010 18:42 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
From: Bart Goddard on 6 Aug 2010 19:10 bill <b92057(a)yahoo.com> wrote in news:565a678d-2104-41db-bc33- 5f470e07c56b(a)m35g2000prn.googlegroups.com: > On Aug 6, 11:15�am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: >> On Aug 5, 5:55�pm, bill <b92...(a)yahoo.com> wrote: >> >> >> >> > "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. >> >> Sigh. �From your quote above, I extract the words: >> >> "has a finite decimal �expansion" >> >> Note the quantifier, �"a". �The integer 4 has a finite decimal >> expansion. �It fact, it has infinitely many different finite >> decimal expansions. �4.0, �4.00, �4.000, ..... >> >> It also has an infinite decimal expansion: �3.999999999........ >> >> Note that the text you quoted did NOT say "a UNIQUE finite decimal >> expansion". > > You should take this up with the editors of > MathWorld. > No, MathWorld has it right. You have it wrong. By the MathWorld definition, 3.9999..... is a finite decimal. You said you were using this defintion, yet declared that 3.9999.... is NOT a finite decimal. This shows that you're using it incorrectly. If your definitions are munged, then there's no way your proof can be correct. -- Cheerfully resisting change since 1959.
From: Virgil on 6 Aug 2010 20:02
In article <38d15a0a-0045-419e-bb28-3c5bafb41373(a)o7g2000prg.googlegroups.com>, bill <b92057(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. |