Simple question about rational and irrational numbers
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From: Bill Dubuque on 2 Jun 2010 10:43 Arturo Magidin <magidin(a)member.ams.org> wrote: > On Jun 1, 12:09 pm, Craig Feinstein <cafei...(a)msn.com> wrote: >> On Jun 1, 12:45 pm, "Mike Terry" >>> "Craig Feinstein" <cafei...(a)msn.com> wrote: >>>>>> 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 = ac^2. >> >>>>> Sure. a = sqrt(2), b = 1/sqrt(2) and c = 1. >> >>>> 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... >> >> Thank you. You are correct. My real goal > > So, the third try is your "real" goal? When this fails, will you come > up with some new "what you really meant"? > >> 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? > > Yes: ab=c^2 with a,b irrational and c rational is impossible if and > only if c=0. The "if" clause if hopefully clear. > > Simply pick your favorite irrational number k; then let a = c^2*k, > which is of course irrational whenever c is rational and nonzero, and > let b=1/k, which is of course irrational. This immediately generalizes from Q to any multiplicative subgroup - recall my prior posts on the "complementary view of a subgroup" THEOREM A nonempty subset G of commutative group H is a subgroup iff G * ~G = ~G, where ~G = G complement. See my prior posts for further details http://groups.google.com/group/sci.math/msg/da23f94a09e7611b http://google.com/groups?selm=y8zvdkdkvj8.fsf%40nestle.csail.mit.edu http://groups.google.com/group/sci.math/browse_frm/thread/76082c277c480e05/68efd6ddda213962 http://google.com/groups?selm=y8zllr72kqg.fsf%40nestle.ai.mit.edu --Bill Dubuque
From: Craig Feinstein on 2 Jun 2010 10:53 On Jun 1, 10:26 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 1, 9:09 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. Since there are polynomials > > over Q of arbitrarily high degree which are irreducible, the degree of > > Q[a,b] over Q can be arbitrarily large, and you *still* have a > > solution. > > > (ii) If k is transcendental, then Q[a,b] = Q[k,1/k], which is > > isomorphic to the ring of polynomials in x and 1/x with coefficients > > in Q. This ring has infinite degree over Q, and you still have a > > solution with a dimension which is greater than 2. > > > > 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? > > > No, this is trivially false as well, using the exact same argument. > > Pick your *favorite* irrational number k, then set a_1 = c^n*k, > > a_2=....=a_{n-1}=k, and a_n = 1/k^{n-1}. The equation holds, and the > > extension is either the field Q[k] or Q[k,1/k], according to whether k > > is algebraic or transcendental, resp. In the first case, the degree is > > finite but can be arbitrarily large; in the second the degree is > > infinite. > > Since he is thinking on placing conditions on a and b, it's possible > he meant "for a and b algebraic over Q, ab is not the square of an > algebraic number if the degree of the extension Q[a,b] OVER Q[a] is > greater than 1"; note the missing clause saying we would be looking at > the extension Q[a,b] / Q[a], and not merely the degree of Q[a,b] > (which means, the degree over Q). > > *This* is true, because if ab is rational, then Q[a,b]=Q[a]. But the > condition is stronger than needed, because for example, if b=3/a, then > Q[a,b]=Q[a], so the extension is of degree 1, but ab is not a rational > square. > > In your alleged "generalization", the same thing happens: you if > (a1*...*a(n-1)) * an is rational, then > Q[a_1,...,a_{n}]=Q[a_1,...,a_{n-1}]; so a necessary and sufficient > condition for the product to be *rational* (but not necessarily a > square) is that for each i, you have Q[a_1,...,a_{i-1},a_{i > +1},...,a_n] = Q[a_1,...,a_n]. > > -- > Arturo Magidin- Hide quoted text - > > - Show quoted text - Thank you very much Arturo. Your responses have answered all of my questions. Craig
From: Craig Feinstein on 2 Jun 2010 13:25 On Jun 2, 10:53 am, Craig Feinstein <cafei...(a)msn.com> wrote: > On Jun 1, 10:26 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > > > On Jun 1, 9:09 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. Since there are polynomials > > > over Q of arbitrarily high degree which are irreducible, the degree of > > > Q[a,b] over Q can be arbitrarily large, and you *still* have a > > > solution. > > > > (ii) If k is transcendental, then Q[a,b] = Q[k,1/k], which is > > > isomorphic to the ring of polynomials in x and 1/x with coefficients > > > in Q. This ring has infinite degree over Q, and you still have a > > > solution with a dimension which is greater than 2. > > > > > 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? > > > > No, this is trivially false as well, using the exact same argument. > > > Pick your *favorite* irrational number k, then set a_1 = c^n*k, > > > a_2=....=a_{n-1}=k, and a_n = 1/k^{n-1}. The equation holds, and the > > > extension is either the field Q[k] or Q[k,1/k], according to whether k > > > is algebraic or transcendental, resp. In the first case, the degree is > > > finite but can be arbitrarily large; in the second the degree is > > > infinite. > > > Since he is thinking on placing conditions on a and b, it's possible > > he meant "for a and b algebraic over Q, ab is not the square of an > > algebraic number if the degree of the extension Q[a,b] OVER Q[a] is > > greater than 1"; note the missing clause saying we would be looking at > > the extension Q[a,b] / Q[a], and not merely the degree of Q[a,b] > > (which means, the degree over Q). > > > *This* is true, because if ab is rational, then Q[a,b]=Q[a]. But the > > condition is stronger than needed, because for example, if b=3/a, then > > Q[a,b]=Q[a], so the extension is of degree 1, but ab is not a rational > > square. > > > In your alleged "generalization", the same thing happens: you if > > (a1*...*a(n-1)) * an is rational, then > > Q[a_1,...,a_{n}]=Q[a_1,...,a_{n-1}]; so a necessary and sufficient > > condition for the product to be *rational* (but not necessarily a > > square) is that for each i, you have Q[a_1,...,a_{i-1},a_{i > > +1},...,a_n] = Q[a_1,...,a_n]. > > > -- > > Arturo Magidin- Hide quoted text - > > > - Show quoted text - > > Thank you very much Arturo. Your responses have answered all of my > questions. > > Craig- Hide quoted text - > > - Show quoted text - 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 = f(x) has no solutions (x,y) in the rationals, Q x Q. Craig
From: Robert Israel on 2 Jun 2010 14:32 Craig Feinstein <cafeinst(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 israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Craig Feinstein on 2 Jun 2010 14:42
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 Thank you. Craig |