Dirichlet's approximation theorem
In
Here represents the
is satisfied by infinitely many integers p and q. This shows that any irrational number has irrationality measure at least 2.
The
Simultaneous version
The simultaneous version of the Dirichlet's approximation theorem states that given real numbers and a natural number then there are integers such that [1]
Method of proof
Proof by the pigeonhole principle
This theorem is a consequence of the
Proof outline: Let be an irrational number and be an integer. For every we can write such that is an integer and . One can divide the interval into smaller intervals of measure . Now, we have numbers and intervals. Therefore, by the pigeonhole principle, at least two of them are in the same interval. We can call those such that . Now:
Dividing both sides by will result in:
And we proved the theorem.
Proof by Minkowski's theorem
Another simple proof of the Dirichlet's approximation theorem is based on Minkowski's theorem applied to the set
Since the volume of is greater than , Minkowski's theorem establishes the existence of a non-trivial point with integral coordinates. This proof extends naturally to simultaneous approximations by considering the set
Related theorems
Legendre's theorem on continued fractions
In his Essai sur la théorie des nombres (1798), Adrien-Marie Legendre derives a necessary and sufficient condition for a rational number to be a convergent of the simple continued fraction of a given real number.[4] A consequence of this criterion, often called Legendre's theorem within the study of continued fractions, is as follows:[5]
Theorem. If α is a real number and p, q are positive integers such that , then p/q is a convergent of the continued fraction of α.
Proof
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Proof. We follow the proof given in An Introduction to the Theory of Numbers by G. H. Hardy and E. M. Wright.[6] Suppose α, p, q are such that , and assume that α > p/q. Then we may write , where 0 < θ < 1/2. We write p/q as a finite continued fraction [a0; a1, ..., an], where due to the fact that each rational number has two distinct representations as finite continued fractions differing in length by one (namely, one where an = 1 and one where an ≠ 1), we may choose n to be even. (In the case where α < p/q, we would choose n to be odd.) Let p0/q0, ..., pn/qn = p/q be the convergents of this continued fraction expansion. Set , so that and thus, where we have used the fact that pn−1 qn - pn qn−1 = (-1)n and that n is even. Now, this equation implies that α = [a0; a1, ..., an, ω]. Since the fact that 0 < θ < 1/2 implies that ω > 1, we conclude that the continued fraction expansion of α must be [a0; a1, ..., an, b0, b1, ...], where [b0; b1, ...] is the continued fraction expansion of ω, and therefore that pn/qn = p/q is a convergent of the continued fraction of α. |
This theorem forms the basis for
See also
- Dirichlet's theorem on arithmetic progressions
- Hurwitz's theorem (number theory)
- Heilbronn set
- Kronecker's theorem (generalization of Dirichlet's theorem)
Notes
- ^ Schmidt, p. 27 Theorem 1B
- ^ http://jeff560.tripod.com/p.html for a number of historical references.
- ^ "Dirichlet theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
- ^ Legendre, Adrien-Marie (1798). Essai sur la théorie des nombres (in French). Paris: Duprat. pp. 27–29.
- JSTOR 26273940.
- ^ Hardy, G. H.; Wright, E. M. (1938). An Introduction to the Theory of Numbers. London: Oxford University Press. pp. 140–141, 153.
- doi:10.1109/18.54902– via IEEE.
References
- ISBN 978-3-540-38645-2.
- Schmidt, Wolfgang M. (1991). Diophantine Approximations and Diophantine Equations. Lecture Notes in Mathematics book series. Vol. 1467. Springer. S2CID 118143570.