Irrational Square Roots
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From: Pubkeybreaker on 6 Aug 2010 14:17 On Aug 5, 7:32 pm, bill <b92...(a)yahoo.com> wrote: > On Aug 4, 4:56 pm, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> > wrote: > To say that 2.0 is a finite decimal is to take > advantage of a deficiency in the definition of > a finite decimal. Oh? Please explain this "deficiency" and why you believe that the standard definition is deficient. Also explain how you would correct the standard definition so that (in your eyes) it is NOT deficient.
From: Dave L. Renfro on 6 Aug 2010 15:02 bill wrote (in part): > 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. This seems unclear to me. Your argument appears to rely on the following: B^2 is a factor of A^2 <==> B is a factor of A. The <== direction is clear: B is a factor of A ==> A = KB for some integer K ==> A^2 = (KB)^2 = (K^2)(B^2) for some integer K ==> A^2 = (L)B^2 for some integer L (choose L = K^2) ==> B^2 is a factor of A^2 However, the ==> direction is not so clear: B^2 is a factor of A^2 ==> A^2 = (L)B^2 for some integer L ==> ??? How do we obtain the desired integer K from L? If you intend to take the square root of L, you need to show that L is a perfect square . . . Here's about the simplest way I know (done in high school precalculus classes I taught in the late 1990s): Suppose I is not a perfect square. Then at least one prime factor of I has odd multiplicity. But now we get a contradiction from A^2 = I(B^2) (square both sides of sqrt(I) = A/B, then multiply both sides by B^2), since that prime factor has even multiplicity on the left side (since every prime factor of the square of an integer has even multiplicity) and odd multiplicity on the right side (because odd + even is odd, where 'odd' is the multiplicity of that prime factor in I and 'even' is the multiplicity of that prime factor in B^2). The same method works for n'th roots, except instead of 'even' and 'odd' you have 'divisible by n' and 'not divisible by n'. Dave L. Renfro
From: Arturo Magidin on 6 Aug 2010 15:25 On Aug 6, 2:02 pm, "Dave L. Renfro" <renfr...(a)cmich.edu> wrote: > > Here's about the simplest way I know (done in high school > precalculus classes I taught in the late 1990s): The rational root test also works: the only possibly rational roots of x^n - a are integers that divide a. So if a is not a perfect n-th power, then no rational can be a root of x^n-a, hence no nth square root of a is rational. Of course, the rational root test relies on (the generalization of) "Euclid's Lemma": if a|bc and (a,b)=1, then a|c. -- Arturo Magidin
From: bill on 6 Aug 2010 16:18 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)
From: Dave L. Renfro on 6 Aug 2010 16:38
Arturo Magidin wrote: > The rational root test also works: the only possibly rational > roots of x^n - a are integers that divide a. So if a is not > a perfect n-th power, then no rational can be a root of x^n-a, > hence no nth square root of a is rational. Yep, and I also did that in those precalculus classes. For anyone interested, I've posted a .pdf file of the handout I used back then: http://mathforum.org/kb/message.jspa?messageID=6886991 By the way, one way to take care of the challenge I mention in a footnote at the bottom of p. 3 is to subtract 3 from both sides of the equation and then multiply both sides by a^2 + b^2 + c^2 + ab + bc - ac where a = (x^2 - 3) b = 2x*(cube root of 2) c = (cube root of 4) Dave L. Renfro |