# 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 number^{} $\alpha $ with degree $m\mathrm{>}\mathrm{1}$, there exists a constant $c\mathrm{=}c\mathit{}\mathrm{(}\alpha \mathrm{)}\mathrm{>}\mathrm{0}$ such that:

$$|\alpha -\frac{p}{q}|>\frac{c}{{q}^{m}}$$ |

for all rationals $p\mathrm{/}q$ (with $q\mathrm{>}\mathrm{0}$).

Next we use the theorem to construct a transcendental number^{}.

###### Corollary 1.

The real number

$$\psi =\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{10}^{n!}}=0.1100010\mathrm{\dots}$$ |

is transcendental.

###### Proof.

Clearly, the number $\psi $ is well defined, i.e. the series converges^{}. Indeed,

$$ |

and ${\sum}_{n=1}^{\mathrm{\infty}}{10}^{-n}=1/9$. Thus, by the comparison test, the series converges and $$.

Suppose, for a contradiction^{}, that $\psi $ is algebraic of degree $m$. We will construct infinitely many rationals $p/q$ such that

$$ |

where $c=c(\psi )$ is the constant given by the theorem above. Let $k\in \mathbb{N}$ be such that $$. Then, in fact, we will show that there are infinitely many rationals $p/q$ with $q\ge 2$ such that

$$ |

For all $j>k+m$ we define a rational number ${p}_{j}/{q}_{j}$ by:

$${p}_{j}={10}^{j!}\sum _{n=1}^{j}{10}^{-n!},{q}_{j}={10}^{j!}$$ |

then ${p}_{j}$ and ${q}_{j}$ are relatively prime integers and we have:

$|\psi -{\displaystyle \frac{{p}_{j}}{{q}_{j}}}|$ | $=$ | $\sum _{n=j+1}^{\mathrm{\infty}}}{\displaystyle \frac{1}{{10}^{n!}}$ | ||

$$ | $\frac{1}{{10}^{(j+1)!}}}(1+{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{{10}^{2}}}+\mathrm{\dots})$ | |||

$=$ | $10/9\cdot {\displaystyle \frac{1}{{q}_{j}^{(j+1)}}}$ | |||

$$ | $\frac{1}{{q}_{j}^{j}}$ | |||

$$ | $\frac{1}{{q}_{j}^{(k+m)}}$ |

where in the last inequality^{} we have used the fact that $j>k+m$. Therefore, all rationals ${\{{p}_{j}/{q}_{j}\}}_{j=k+m+1}^{\mathrm{\infty}}$ satisfy the desired inequality, which leads to the contradiction with the theorem above. Thus $\psi $ cannot be algebraic and it must be transcendental.
∎

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

Title | example of transcendental number |
---|---|

Canonical name | ExampleOfTranscendentalNumber |

Date of creation | 2013-03-22 15:02:45 |

Last modified on | 2013-03-22 15:02:45 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 7 |

Author | alozano (2414) |

Entry type | Example |

Classification | msc 11J82 |

Classification | msc 11J81 |

Related topic | LiouvillesTheorem |

Related topic | RothsTheorem |