Rational points on a line
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From: Kaba on 6 Jul 2010 14:38 Hi, 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
From: James Waldby on 6 Jul 2010 15:48 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. Are you asking if every line in the plane, every plane in 3-space, etc. has non-zero points with all-rational coordinates? If that's the question, y = pi * x is a refutation. -- jiw
From: Kaba on 6 Jul 2010 16:03 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? > Are you asking if every line in the plane, every plane in 3-space, > etc. has non-zero points with all-rational coordinates? If that's > the question, y = pi * x is a refutation. That's the question, although just for a line. That's a good counter- example. It's interesting that there are lines that avoid all-rational points. -- http://kaba.hilvi.org
From: Axel Vogt on 6 Jul 2010 16:07 Kaba 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? > >> Are you asking if every line in the plane, every plane in 3-space, >> etc. has non-zero points with all-rational coordinates? If that's >> the question, y = pi * x is a refutation. > > That's the question, although just for a line. That's a good counter- > example. It's interesting that there are lines that avoid all-rational > points. > n=1 ---> t:= 1/abs(x).
From: Kaba on 6 Jul 2010 17:42
Axel Vogt wrote: > Kaba 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? > > > >> Are you asking if every line in the plane, every plane in 3-space, > >> etc. has non-zero points with all-rational coordinates? If that's > >> the question, y = pi * x is a refutation. > > > > That's the question, although just for a line. That's a good counter- > > example. It's interesting that there are lines that avoid all-rational > > points. > > > > n=1 ---> t:= 1/abs(x). Sure:) -- http://kaba.hilvi.org |