Simple question about rational and irrational numbers
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From: Virgil on 1 Jun 2010 15:10 In article <d0a5b861-32e4-4a3a-9a3c-a20eab19b06e(a)v37g2000vbv.googlegroups.com>, Craig Feinstein <cafeinst(a)msn.com> wrote: > Is the following possible? > > a,b are irrational. c is rational. ab=c^2. > > Why or why not? > > Craig Consider a = sqrt(2), b = 1/a
From: Pubkeybreaker on 1 Jun 2010 15:22 On Jun 1, 2:27 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > On Jun 1, 2:17 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > > > On Jun 1, 1:56 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > > > > On Jun 1, 1:44 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > > By general conditions, I mean the largest set of nontrivial conditions > > > > > possible. > > > > > This is still undefined. What is a 'trivial condition' in this > > > > context? > > > > > The most general condition is that ab = c^2 where c is rational is > > > > impossible for all a,b, such that a*b is irrational. > > > > That is trivial. > > > Then DEFINE what *you* mean by 'trivial' and 'non-trivial'. > > > >I'm looking for nontrivial conditions. For instance, > > > what you said above about a transcendental number and an algebraic > > > number is what I'm looking for, except I don't want transcendental > > > numbers. > > > Then for &*(#&*#! sake, give us a RIGOROUS DEFINITION of the > > conditions > > that you do want. > > > You can't do it, can you? This makes the question meaningless. > > > Would you settle for the following: ab = c^2 is impossible with a,b, > > irrational > > and c rational if the extension degree of Q[a,b] is greater than 1??? > > Yes! That's the answer I was looking for. It appears that I don't need > to give you a rigorous definition, since you seem to be able to read > my mind. > > Can you prove that ab=c^2 is impossible with a,b, irrational and c > rational if the extension degree of Q[a,b] is greater than 1?- Hide quoted text - Yes, I can. The proof is immediate and follows from the definition.
From: Craig Feinstein on 1 Jun 2010 15:34 On Jun 1, 3:22 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > On Jun 1, 2:27 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > > > > > > > On Jun 1, 2:17 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > On Jun 1, 1:56 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > > > > > On Jun 1, 1:44 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > > > By general conditions, I mean the largest set of nontrivial conditions > > > > > > possible. > > > > > > This is still undefined. What is a 'trivial condition' in this > > > > > context? > > > > > > The most general condition is that ab = c^2 where c is rational is > > > > > impossible for all a,b, such that a*b is irrational. > > > > > That is trivial. > > > > Then DEFINE what *you* mean by 'trivial' and 'non-trivial'. > > > > >I'm looking for nontrivial conditions. For instance, > > > > what you said above about a transcendental number and an algebraic > > > > number is what I'm looking for, except I don't want transcendental > > > > numbers. > > > > Then for &*(#&*#! sake, give us a RIGOROUS DEFINITION of the > > > conditions > > > that you do want. > > > > You can't do it, can you? This makes the question meaningless. > > > > Would you settle for the following: ab = c^2 is impossible with a,b, > > > irrational > > > and c rational if the extension degree of Q[a,b] is greater than 1??? > > > Yes! That's the answer I was looking for. It appears that I don't need > > to give you a rigorous definition, since you seem to be able to read > > my mind. > > > Can you prove that ab=c^2 is impossible with a,b, irrational and c > > rational if the extension degree of Q[a,b] is greater than 1?- Hide quoted text - > > Yes, I can. The proof is immediate and follows from the definition.- Hide quoted text - > > - Show quoted text - Can you generalize your statement to this? a_1*a_2*...*a_n = c^n is impossible with a_1,a_2,...,a_n irrational and c rational if the extension degree of Q[a_1,a_2,...,a_n] is greater than n-1.
From: Arturo Magidin on 1 Jun 2010 15:43 On Jun 1, 2:22 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > Would you settle for the following: ab = c^2 is impossible with a,b, > > > irrational > > > and c rational if the extension degree of Q[a,b] is greater than 1??? > > > Yes! That's the answer I was looking for. It appears that I don't need > > to give you a rigorous definition, since you seem to be able to read > > my mind. > > > Can you prove that ab=c^2 is impossible with a,b, irrational and c > > rational if the extension degree of Q[a,b] is greater than 1?- Hide quoted text - > > Yes, I can. The proof is immediate and follows from the definition. I don't understand where you are going here... If a or b is irrational, then Q[a,b] =/= Q. If they are algebraic, then Q[a,b] is a field and has a degree over Q. Otherwise, Q[a,b] may be merely a ring, isomorphic to some subring of Q[x,y]. But in any case: there certainly *are* solutions to ab=c^2 with a,b irrational and c rational (necessarily c different from 0); in *all* such solutions, Q[a,b]=/=Q, and so the "extension degree" (the dimension of Q[a,b] as a Q-vector space) will be greater than 1. So why do you say that it is impossible to have it? Now, I focused on a given c, you are focusing on a and b; but in any case, say b = c^2/a. If a is irrational, then so is b. If a is algebraic, then so is b, and Q(a,b) = Q(a) = Q[a] has degree greater than 1 over Q (because a is not rational). Do it with your favorite nonrational algebraic. What is the problem? -- Arturo Magidin
From: Craig Feinstein on 1 Jun 2010 16:50
On Jun 1, 3:43 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 1, 2:22 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > Would you settle for the following: ab = c^2 is impossible with a,b, > > > > irrational > > > > and c rational if the extension degree of Q[a,b] is greater than 1??? > > > > Yes! That's the answer I was looking for. It appears that I don't need > > > to give you a rigorous definition, since you seem to be able to read > > > my mind. > > > > Can you prove that ab=c^2 is impossible with a,b, irrational and c > > > rational if the extension degree of Q[a,b] is greater than 1?- Hide quoted text - > > > Yes, I can. The proof is immediate and follows from the definition. > > I don't understand where you are going here... > > If a or b is irrational, then Q[a,b] =/= Q. If they are algebraic, > then Q[a,b] is a field and has a degree over Q. Otherwise, Q[a,b] may > be merely a ring, isomorphic to some subring of Q[x,y]. > > But in any case: there certainly *are* solutions to ab=c^2 with a,b > irrational and c rational (necessarily c different from 0); in *all* > such solutions, Q[a,b]=/=Q, and so the "extension degree" (the > dimension of Q[a,b] as a Q-vector space) will be greater than 1. So > why do you say that it is impossible to have it? > > Now, I focused on a given c, you are focusing on a and b; but in any > case, say b = c^2/a. If a is irrational, then so is b. If a is > algebraic, then so is b, and Q(a,b) = Q(a) = Q[a] has degree greater > than 1 over Q (because a is not rational). Do it with your favorite > nonrational algebraic. What is the problem? > > -- > Arturo Magidin Perhaps he meant ab=c^2 is impossible with a,b, irrational and c rational if the extension degree of Q[a,b] is greater than 2. This is true, correct? Similarly, my generalization (above) would be: a_1*a_2*...*a_n = c^n is impossible with a_1,a_2,...,a_n irrational and c rational if the extension degree of Q[a_1,a_2,...,a_n] is greater than n. This is true too, correct? Craig |