rational points on one dimensional sphere
Let
𝕊1={(x,y)∈ℝ2|x2+y2=1} |
be a one dimensional sphere.
We will denote by
𝕊1ℚ={(x,y)∈ℚ2|(x,y)∈𝕊1} |
the rational sphere. We shall try to describe 𝕊1ℚ in terms of Pythagorean triplets.
Theorem. Let (x,y)∈ℝ2. Then (x,y)∈𝕊1ℚ if and only if there exists a Pythagorean triplet a,b,c∈ℤ (i.e. |a|,|b|,|c|∈ℕ is a Pythagorean triplet) such that x,y are of the form ac and bc.
Proof. ,,⇐” If (for example) x=ac and y=bc for a Pythagorean triplet a,b,c∈ℤ, then we have
x2+y2=a2c2+b2c2=a2+b2c2=c2c2=1 |
and thus (x,y)∈𝕊1ℚ.
,,⇒” Assume that (x,y)∈𝕊1ℚ. Then x=pq for some p,q∈ℤ. It follows, that
1=x2+y2=p2q2+y2 |
and this is if and only if
y=√q2-p2q |
(up to a sign of course). Therefore y∈ℚ if and only if √q2-p2=n is an integer. In this case we have
x=pq,y=nq. |
Note, that
q2-p2=n2 |
and thus
q2=n2+p2, |
so n,p,q∈ℤ is a Pythagorean triplet, which completes the proof. □
Corollary. The rational sphere 𝕊1ℚ is an infinite set.
Proof. Let m,n∈ℕ be natural numbers such that n is fixed and even. Let m run through primes. Then (due to the theorem in parent entry)
2mn,m2-n2,m2+n2∈ℕ |
is a Pythagorean triplet. Let
xm=2mnm2+n2=2nm+n2m. |
Ir follows from the theorem, that there exists ym∈ℚ such that (xm,ym)∈𝕊1ℚ. It is easy to see, that xm=xm′ if and only if m=m′ and thus we generated infinitely many rational points on sphere. This completes the proof. □
Title | rational points on one dimensional sphere |
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Canonical name | RationalPointsOnOneDimensionalSphere |
Date of creation | 2013-03-22 19:07:49 |
Last modified on | 2013-03-22 19:07:49 |
Owner | joking (16130) |
Last modified by | joking (16130) |
Numerical id | 7 |
Author | joking (16130) |
Entry type | Definition |
Classification | msc 11-00 |
Related topic | RationalSineAndCosine |