Rational points on a line
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From: Arturo Magidin on 6 Jul 2010 17:59 On Jul 6, 3:03 pm, Kaba <n...(a)here.com> wrote: > James Waldby wrote: > > On Tue, 06 Jul 2010 21:38:28 +0300, Kaba wrote: > > > > Let R be the set of reals and Q be the set of rationals. Given x in > > > R^n\{0}, is it always possible to find t in R, t > 0, such that x in > > > Q^n? > > > You don't specify any relation between t and x, and apparently > > are using x in different ways in its two instances. > > Oops, I meant: > > Given x in R^n\{0}, is it always possible to find t in R, t > 0, such > that tx in Q^n? Consider the case n=2, and (x,y) in R^n, xy=/=0 (this is stronger than your condition, since (x,0) with x=/=0 does lies in R^n\{0}); if you have (x,y) and (tx,ty) lies in Q^2, with t>0, then x/y must be a rational. In particular, if you choose x to be rational and y to be nonrational, then no such t can exist. Even if both x and y are nonrational, no such t need exist: pick y such that y is not in Q(x), the smallest subfield of R that contains Q and x; since Q(x) is countable, there are uncountably many choices of y which satisfy this. Since y is not in Q(x), you cannot have x/y in Q. Clearly, the condition will fail for uncountably many tuples for any n>2 as well. -- Arturo Magidin
From: Tim Little on 6 Jul 2010 21:01 On 2010-07-06, Kaba <none(a)here.com> wrote: > Given x in R^n\{0}, is it always possible to find t in R, t > 0, such > that tx in Q^n? No. E.g. (1, sqrt(2)) in R^2. - Tim
From: OwlHoot on 6 Jul 2010 21:44 On Jul 6, 7:38 pm, Kaba <n...(a)here.com> wrote: > > Let R be the set of reals and Q be the set of rationals. Given > x in R^n\{0}, is it always possible to find t in R, t > 0, such that > x in Q^n? > > --http://kaba.hilvi.org No - For example, let x = (1, pi, pi^2, ..., pi^(n-1)). If there was some t > 0 in R with t.pi, t.pi^2, .. all rational, then some linear combination of these involving only integer multiples, not all zero, would be zero. But then dividing out t (> 0) would leave a polynomial equation of finite degree and integer coefficients with a root pi, contrary to the fact that pi is transcendental. Cheers John Ramsden
From: Kaba on 7 Jul 2010 05:47
Thanks everyone for the replies, I got it!:) -- http://kaba.hilvi.org |