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Corrected exercise: Irrationality of ln(2)

Here is the statement of an exercise that we are going to correct on an exercise showing the irrationality of the number ln(2) through several questions. It's an exercise that we're going to put in the arithmetic chapter and also that of integers. It is a doable exercise right from the very first in the superior. Here is the statement:

States

Irrationality of ln(2)

If you are interested in irrationality topics, we have also demonstrated

If you are unfamiliar with logarithms, check out our dedicated article !

Proof of the irrationality of ln(2)

As a reminder, a bit of culture: ln(2) ≈ 0.69314718056.

Let's start the demonstration, you will see, it is not very long! We will use the Legendre polynomials. We are going to pose a small variant, by defining Ln by

\forall n \in \N, L_n(x) = \dfrac{1}{n!} \dfrac{d^n}{dx^n} \left(x^n(1-x)^n\right)

Lemma

We will prove the following lemma

\forall a \in \mathbb{Q}, e^a \notin \mathbb Q

To prove this lemma, we will set the following integral

\forall n \in \N, I_n = \int_0^1 e^{at}L_n(t) dt

A bit like we were able to do with the Legendre polynomials and the Hermite polynomials, we perform n integrations to obtain (we derive n times theexponential and we integrate the polynomial n times):

I_n = (-1)^n \left( \dfrac{a^n}{n! } \right) \int_0^1 e^{at} t^n (1-t)^n dt

A classic increase to know is

t(1-t) \leq \dfrac{1}{4}

We can therefore increase In

\begin{array}{ll} |I_n| &\leq \dfrac{a^n}{n! } \displaystyle \int_0^1 e^{at} t^n (1-t)^ndt\\ & \leq \dfrac{a^n}{n! } \displaystyle \int_0^1 e^{at} \left(\dfrac{1}{4}\right)^n dt\\ & \leq \dfrac{a^n}{n! } \displaystyle \int_0^1 e^{a} \left(\dfrac{1}{4}\right)^n dt\\ & \leq \dfrac{e^a}{n! } \left(\dfrac{a}{4}\right)^n \end{array}

From this inequality, we get that

\lim_{n \to + \infty} |I_n| = 0 

Now, on the other hand, expanding with the expression with Newton's binomial and differentiating n times, we get that

L_n(X) = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} (-1)^kX^k

From this expression, we deduce that

L_n(X) \in \Z [X]

Moreover, let's look

\int_0^1 e^{at}t^k dt

Doing k integrations by parts, we get

\exists A_k,B_k \in \Z, \int_0^1 e^{at}t^kdt =\dfrac{1}{a^{k+1}} A_k+ B_ke^a 

Moreover, as

L_n(X) = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} (-1)^kX^k

We obtain

I_n = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} (-1)^k \int_0^1 e^{at}t^kdt

which simplifies to

I_n = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} (-1)^k\dfrac{1}{a^{k+1}} ( A_k + B_ke^a )

And therefore by dividing the sums in two:

I_n = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} \dfrac{a^{nk}}{a^{n+1}}(-1 )^k A_k +e^a \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n} (-1)^k \dfrac{a^{nk}} {a^{n+1}} B_k

What can be rewritten as

\exists C_n, D_n \in \Z, I_n = \dfrac{1}{a^{n+1}}(C_n + e^a D_n)

And now, by absurd, suppose that

\exists p \in \Z, \exists q \in\N^*, ​​e^a = \dfrac{p}{q}

From the previous calculations, we have:

\dfrac{1}{qa^{n+1}} \leq \dfrac{1}{a^{n+1}} |C_n + \dfrac{p}{q}D_n| = |I_n|

This contradicts the increase obtained previously.
Conclusion: ea is irrational

Back to demo

Let us prove the property by the absurd. Suppose that ln(2) is rational, so we have:

\exists p \in \Z, \exists q \in\N^*, ​​\ln(2) = \dfrac{p}{q}

We then have:

e^{\frac{p}{q}} = 2 

However, we have shown in the lemma that the exponential of a rational is irrational. We therefore arrive at a contradiction.

Conclusion : ln(2) is irrational

Demonstration through a subject of Mines-Ponts

Maths subject 2 of Mines-Ponts from 2002 demonstrates the irrationality of ln(2) through a different proof. I let you directly discover the subject

Did you like this exercise?

This proof is largely inspired by this spring

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