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单词 ExampleOfTranscendentalNumber
释义

example of transcendental number


The following is a classical application of Liouville’s approximation theorem. For completeness, we state Liouville’s result here:

Theorem 1.

For any algebraic numberMathworldPlanetmath α with degree m>1, there exists a constant c=c(α)>0 such that:

|α-pq|>cqm

for all rationals p/q (with q>0).

Next we use the theorem to construct a transcendental numberMathworldPlanetmath.

Corollary 1.

The real number

ψ=n=1110n!=0.1100010

is transcendental.

Proof.

Clearly, the number ψ is well defined, i.e. the series convergesPlanetmathPlanetmath. Indeed,

110n!<110n

and n=110-n=1/9. Thus, by the comparison test, the series converges and 0<ψ<1/9.

Suppose, for a contradictionMathworldPlanetmathPlanetmath, that ψ is algebraic of degree m. We will construct infinitely many rationals p/q such that

|ψ-pq|<cqm

where c=c(ψ) is the constant given by the theorem above. Let k be such that 1/2k<c. Then, in fact, we will show that there are infinitely many rationals p/q with q2 such that

|ψ-pq|<1qm+k<12k1qm<cqm

For all j>k+m we define a rational number pj/qj by:

pj=10j!n=1j10-n!,qj=10j!

then pj and qj are relatively prime integers and we have:

|ψ-pjqj|=n=j+1110n!
<110(j+1)!(1+110+1102+)
=10/91qj(j+1)
<1qjj
<1qj(k+m)

where in the last inequalityMathworldPlanetmath we have used the fact that j>k+m. Therefore, all rationals {pj/qj}j=k+m+1 satisfy the desired inequality, which leads to the contradiction with the theorem above. Thus ψ cannot be algebraic and it must be transcendental.∎

Many other similar transcendental numbers can be constructed in this fashion.

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更新时间:2025/5/5 2:19:26