Simple question about rational and irrational numbers
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From: Arturo Magidin on 1 Jun 2010 22:09 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. -- Arturo Magidin
From: Arturo Magidin on 1 Jun 2010 22:26 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
From: Tim Little on 2 Jun 2010 03:59 On 2010-06-01, Craig Feinstein <cafeinst(a)msn.com> wrote: > Is the following possible? > > a,b are irrational. c is rational. ab=c^2. Sure. Rearrange slightly to a = c^2 / b. If c is a nonzero rational and b is irrational, then a must be irrational. - Tim
From: Tim Little on 2 Jun 2010 04:04 On 2010-06-01, Craig Feinstein <cafeinst(a)msn.com> wrote: > 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. Still yes with an example almost as simple: a = 2^(1/3), b = 1/a, c = 1. - Tim
From: Tim Little on 2 Jun 2010 04:34
On 2010-06-01, Craig Feinstein <cafeinst(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 the most elegant conditions would have a and b equally restricted to some easily described set of irrationals. For example, roots of squarefree integers. This example still leaves the feeling of being too restrictive, in that this set is countable. Is there some easily described uncountable set of irrationals such that no pair of members a,b yields ab = c^2 for any rational c? - Tim |