Simple question about rational and irrational numbers
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From: Arturo Magidin on 3 Jun 2010 13:15 On Jun 3, 9:48 am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > 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. And again: there is no "the" in such a construction. You can, at best, talk about "the elements of *such a* set". Using the definite article implies uniqueness of some kind. > I see no way to do that. The existence of maximal subsets closed under multiplication follows from Zorn's Lemma (as expected). Consider the collection of all subsets of I that are closed under multiplication, partially ordered by inclusion. The collection is non-empty since it contains {pi^n : n>0 a natural number}. Suppose that C is a chain of subsets of I, all of which are closed under multiplication, and let S =\/C; this is a subset of I, and if a,b are in S, then there exist X and Y in C such that a in X and b in Y; since C is a chain, either X is contained in Y or Y is contained in X; thus, a and b are in the same element of C, and therefore their product ab is also in that element of C, hence in S. Thus, S is closed under multiplication, and is clearly an upper bound for C in the collection. By ZL, the collection of subset of I that are closed under multiplication has maximal elements. By the observation made above, it has *multiple* maximal elements. -- Arturo Magidin
From: Robert Israel on 3 Jun 2010 13:25 > On Jun 3, 10:38=A0am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > On Jun 3, 9:30=A0am, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > Perhaps it might be better to ask: > > > > > Let =A0I =3D =A0R - Q =A0 =A0(i.e. I is the irrationals) > > > > > Can one characterize the largest possible subset =A0M =A0of I, =A0such > > > 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' =3D {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. Such an M must be a nonmeasurable set. As Solovay showed, it is consistent with Zermelo-Frankel set theory (without the Axiom of Choice) that so such subset of the reals exists. In particular, you can't have an explicit construction of M. Here is the proof that M is nonmeasurable. Suppose M is measurable. Case 1: M has measure 0. Then so does (r/M)^(1/j) = {t: r/t^j is in M} for any nonzero rational r and positive integer j. So we can take some t in I that is not in M and not in any (r/M)^(1/j), and M t^N = {x t^j: x in M, j nonnegative integer} is a larger subset of I that is closed under multiplication, contradiction. Case 2: M has positive measure. Let u be a point of density of M. If r is rational and sufficiently close to u, and d>0 is sufficiently small, the measures of M intersect (u-d,u+d) and (r/M) intersect (u-d,u+d) will both be greater than d, so their intersection is nonempty. But that says there are x and y in M such that x = r/y, i.e. x y = r, contradiction. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Pubkeybreaker on 3 Jun 2010 13:36 On Jun 3, 1:25 pm, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote: > > On Jun 3, 10:38=A0am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > On Jun 3, 9:30=A0am, 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 > > > > 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' =3D {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. > > Such an M must be a nonmeasurable set. As Solovay showed, it is consistent > with Zermelo-Frankel set theory (without the Axiom of Choice) that so such > subset of the reals exists. In particular, you can't have an explicit > construction of M. > > Here is the proof that M is nonmeasurable. Suppose M is measurable. > > Case 1: M has measure 0. Then so does > (r/M)^(1/j) = {t: r/t^j is in M} for any nonzero rational r and positive > integer j. So we can take some t in I that is not in M and not in > any (r/M)^(1/j), and M t^N = {x t^j: x in M, j nonnegative integer} > is a larger subset of I that is closed under multiplication, contradiction. > > Case 2: M has positive measure. Let u be a point of density of M. If r > is rational and sufficiently close to u, and d>0 is sufficiently small, > the measures of M intersect (u-d,u+d) and (r/M) intersect (u-d,u+d) will > both be greater than d, so their intersection is nonempty. But that says > there are x and y in M such that x = r/y, i.e. x y = r, contradiction.. > -- > Robert Israel isr...(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada- Hide quoted text - Very nice!
From: Tim Little on 3 Jun 2010 23:35 On 2010-06-03, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: > Here is the proof that M is nonmeasurable. [...] That is very neat. So in particular it must be uncountable, which nicely answers my question. - Tim
From: adamk on 3 Jun 2010 20:57
> On Jun 1, 1:15 pm, Pubkeybreaker > <pubkeybrea...(a)aol.com> wrote: > > On Jun 1, 1:09 pm, Craig Feinstein > <cafei...(a)msn.com> wrote: > > > > > > > > > > > > > On Jun 1, 12:45 pm, "Mike Terry" > > > > > <news.dead.person.sto...(a)darjeeling.plus.com> > wrote: > > > > "Craig Feinstein" <cafei...(a)msn.com> wrote in > message > > > > > > >news:bb2d0bb1-139b-4ada-b42a-f9d3077247b6(a)s41g2000vba > .googlegroups.com... > > > > > > > On Jun 1, 12:24 pm, José Carlos Santos > <jcsan...(a)fc.up.pt> wrote: > > > > > > On 01-06-2010 17:19, Craig Feinstein wrote: > > > > > > > > > Is the following possible? > > > > > > > > > a,b are irrational. c is rational. > ab=c^2. > > > > > > > > Sure. a = sqrt(2), b = 1/sqrt(2) and c = 1. > > > > > > > > Best regards, > > > > > > > > Jose Carlos Santos > > > > > > > Thank you, that was too simple. I really > meant to ask is the following > > > > > possible? > > > > > > > a^2 and b^2 are irrational. c is rational. > ab=c^2. > > > > > > Well, there was nothing very special about the > number 2 in José's example - > > > > e.g. replace the number 2 with an irrational > number like Pi... > > > > > > Mike.- Hide quoted text - > > > > > > - Show quoted text - > > > > > 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? If I understood you well: take two square-free integers a',b', define: a=a'^1/2 b=b'^1/2 ; a'=/b' Then ab will never be rational. > > > > Define what you mean by "general conditions". It > is not well defined. > > > > And yes, I know of such conditions. Consider, e.g. > one of a, b is > > algebraic, > > and the other is transcendental. Their product is > transcendental, > > hence > > c can not be rational.- Hide quoted text - > > > > - Show quoted text - > > By general conditions, I mean the largest set of > nontrivial conditions > possible. > > What about conditions in which both a, b are > algebraic and irrational, > c is rational, and ab=c^2 is impossible? > > Craig > |