Simple question about rational and irrational numbers
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From: Craig Feinstein on 3 Jun 2010 09:54 On Jun 2, 2:32 pm, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote: > Craig Feinstein <cafei...(a)msn.com> writes: > > Here is another question. Is the following true? > > > Let f(x) be a minimal polynomial (of degree n-1) of an algebraic > > irrational element alpha of a field extension Q[alpha] over the > > rationals Q. Then the equation y^n =3D f(x) has no solutions (x,y) in > > the rationals, Q x Q. > > No. In fact, take any rational r. Since f is irreducible, f(r) is not 0. > Now g(x) = f(x)/f(r) is also a minimal polynomial of alpha of degree n-1, > so we may assume it is f(x), i.e. that f(r) = 1. Then y^n = f(x) has > the solution x = r, y = 1. > -- > Robert Israel isr...(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada What if we change the theorem to: Let f(x) be a monic minimal polynomial (of degree n-1) of an algebraic irrational element alpha of a field extension Q[alpha] over the rationals Q. Then the equation y^n = f(x) has no solutions (x,y) in the rationals, Q x Q. Now is this theorem true? Craig
From: Pubkeybreaker on 3 Jun 2010 10:25 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]. Yes. I should have said that Q[a,b] over Q[a] (or Q(b)) has degree > 1. Now, Q[a] and Q[b] must be different (i.e. non-isormorphic) fields.
From: Arturo Magidin on 3 Jun 2010 10:26 On Jun 3, 8:54 am, Craig Feinstein <cafei...(a)msn.com> wrote: > On Jun 2, 2:32 pm, Robert Israel > > > > <isr...(a)math.MyUniversitysInitials.ca> wrote: > > Craig Feinstein <cafei...(a)msn.com> writes: > > > Here is another question. Is the following true? > > > > Let f(x) be a minimal polynomial (of degree n-1) of an algebraic > > > irrational element alpha of a field extension Q[alpha] over the > > > rationals Q. Then the equation y^n =3D f(x) has no solutions (x,y) in > > > the rationals, Q x Q. > > > No. In fact, take any rational r. Since f is irreducible, f(r) is not 0. > > Now g(x) = f(x)/f(r) is also a minimal polynomial of alpha of degree n-1, > > so we may assume it is f(x), i.e. that f(r) = 1. Then y^n = f(x) has > > the solution x = r, y = 1. > > What if we change the theorem to: It wasn't a theorem the first time. > > Let f(x) be a monic minimal polynomial (of degree n-1) of an algebraic > irrational element alpha of a field extension Q[alpha] over the > rationals Q. Then the equation y^n = f(x) has no solutions (x,y) in > the rationals, Q x Q. > > Now is this theorem true? No. Take f(x) = x^2-8, which is the minimal polynomial of alpha=sqrt(8). It is monic of degree 3-1. But (3,1) is a solution to y^3=f(x). -- Arturo Magidin
From: Pubkeybreaker on 3 Jun 2010 10:28 On Jun 1, 10:09 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 1, 3:50 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > > > > > > > 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? > > *No*, for exactly the same reason. > > Pick *your favorite irrational number k*. Then set a=kc^2 and b=1/k. > Then ab=c^2, and if c=/=0 then a is irrational; b is always > irrational, and you have two possibliities: > > (i) If k is algebraic, then Q[a,b] = Q[k], which has the same degree > as the minimal polynomial of k over Q. I clearly have some confusion. Are not Q(k) and Q(1/k) isomorphic? Are not Q(a) and Q(k) isormorphic?
From: Pubkeybreaker on 3 Jun 2010 10:30
On Jun 2, 4:34 am, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-01, Craig Feinstein <cafei...(a)msn.com> wrote: > > > Thank you. You are correct. My real goal here is to come up with > > general conditions where a, b are irrational, c is rational, and > > ab=c^2 are impossible. Anyone aware of any such conditions? > > The condition c = 0 is an example. No condition on a alone will > suffice, as you can always find b = c^2 / a for any irrational a and > nonzero rational c. So you need a conjunction of conditions on both a > and b. > > There are infinitely many possible conditions that suffice, some less > elegant than others. For an inelegant example, suppose a must be > negative while b must be positive. Perhaps it might be better to ask: Let I = R - Q (i.e. I is the irrationals) Can one characterize the largest possible subset M of I, such that M is closed under multiplication? |