4
$\begingroup$

This is a four part question on proving that $e$ is irrational that I've attempted

i) Show that $\int_0^1{x^{\alpha}e^x}dx<\frac{3}{\alpha+1}$, you may assume $e<3$

ii) Prove using induction that there exists, for all $n=0,1,2,3,\dots$, integers $a_n$ and $b_n$ such that $\int_0^1{x^{n}e^x}dx=a_n+b_ne$

iii) Suppose that $r$ is a positive rational in the form of $\frac{p}q$ where $p$ and $q$ are both positive integers. Show that, for all integers $a$ and $b$, that either $|a+br|=0$ or $|a+br|≥\frac1q$.

iv) Prove that $e$ is irrational.

I have shown parts i) and ii). Can someone please explain how to do iii) and and double check the following proof for iv) which I think is faulty.

Assume that $e$ is rational, i.e $e=\frac{p}{q}$ where $p,q\in\mathbb{Z}$ and where the HCF of $p$ and $q$ is 1. Using the result from part ii) and iii) $\int_0^1{x^{n}e^x}dx=0$ or $\int_0^1{x^{n}e^x}dx≥\frac1q$.

For $\int_0^1{x^{n}e^x}dx=0$, this statement is false as $x^ne^x>0$ for all $0≤x≤1$, therefore there is always area under the curve $f(x)=x^ne^x$ between the limits of integration.

For $\int_0^1{x^{n}e^x}dx≥\frac1q$, using the result in part i), $\frac{3}{n+1}>\frac1q$

Hence $q>\frac{n+1}{3}$. Since in part ii), the statement was shown to be true for all $n$ then $q>\frac{n+1}{3}$ should be true for all $n=0,1,2,\dots$. However, when $n\to\infty$ that means $q>\infty$, which is a contradiction since $q\in\mathbb{Z}$.

Therefore there exists no $e$ such that $e=\frac{p}{q}$ where $p,q\in\mathbb{Z}$, hence $e$ is irrational by proof by contradiction.

$\endgroup$

1 Answer 1

3
$\begingroup$

Your proof of iv) is correct. Assuming $e=\frac{p}{q}$ you may take $p,q\in\mathbb{N}^+$.

As regards iii), note that if $a+br\not=0$ then $$0<|a+br|=\frac{|aq+bp|}{q}\geq \frac{1}{q}$$ because $aq+bp$ is a nonzero integer and therefore $|aq+bp|\geq 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .