Simple question about rational and irrational numbers
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From: Craig Feinstein on 3 Jun 2010 10:33 On Jun 3, 10:26 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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- Hide quoted text - > > - Show quoted text - Right. Thank you.
From: Arturo Magidin on 3 Jun 2010 10:36 On Jun 3, 9:28 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > 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? I'm confused here about your confusion... If k is algebraic over Q (and nonzero), then Q(k)=Q(1/k) = Q[k[ = Q[1/ k]. ( If F is a field, K is an overfield, and a lies in K, then F(a) is the smallest subfield of K that contains both F and a, while F[a] is the smallest sub*ring* of K that contains F and a. If a is algebraic over F, it is a theorem that F[a] is actually a field). If k is transcendental over Q, then Q(k) and Q(1/k) are equal, but Q[k] and Q[1/k] are isomorphic and unequal. In my example above, a=c^2k and b=1/k. Thus, if k is algebraic, then b lies in Q(k) = Q(a) = Q[a], so Q[a,b]=Q[a]=Q[k]. -- Arturo Magidin
From: Arturo Magidin on 3 Jun 2010 10:38 On Jun 3, 9:30 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > 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? I don't think there can be a "largest" (though there might be some that are "maximal"). For suppose S is a subset of I that is closed under multiplication. Consider the set S' = {a^{-1} | a in S}. This is clearly contained in I and closed under multiplication as well, but S/ \S' must be empty (otherwise, S contains an element x and x^{-1}, whose product is in Q). So there can be no single "largest" subset. -- ArturO Magidin
From: Pubkeybreaker on 3 Jun 2010 10:44 On Jun 3, 10:36 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 3, 9:28 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > > > 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? > > I'm confused here about your confusion... > > If k is algebraic over Q (and nonzero), then Q(k)=Q(1/k) = Q[k[ = Q[1/ > k]. > > ( If F is a field, K is an overfield, and a lies in K, then F(a) is > the smallest subfield of K that contains both F and a, while F[a] is > the smallest sub*ring* of K that contains F and a. If a is algebraic > over F, it is a theorem that F[a] is actually a field). > > If k is transcendental over Q, then Q(k) and Q(1/k) are equal, but > Q[k] and Q[1/k] are isomorphic and unequal. > > In my example above, a=c^2k and b=1/k. Thus, if k is algebraic, then b > lies in Q(k) = Q(a) = Q[a], so Q[a,b]=Q[a]=Q[k]. Yes. We agree here.
From: Pubkeybreaker on 3 Jun 2010 10:48
On Jun 3, 10:38 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 3, 9:30 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > 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? > > I don't think there can be a "largest" (though there might be some > that are "maximal"). For suppose S is a subset of I that is closed > under multiplication. Consider the set S' = {a^{-1} | a in S}. This is > clearly contained in I and closed under multiplication as well, but S/ > \S' must be empty (otherwise, S contains an element x and x^{-1}, > whose product is in Q). So there can be no single "largest" subset. Well, yes, clearly such a set M can not be unique. Equally clearly, the OP would not accept a description such as "S is (a) maximal subset of I such that S is closed under multiplication". The OP would want a way of characterizing the elements of the set. I see no way to do that. |