Another Australian question: Proving that e is irrational

Attached.

Attached.

Nice question. Is this a question you want answering btw?

I have a formal proof by contradiction

Well you can have a go. The examiner's report says that there were "virtually no totally successful solutions".

Well I don't do the way as probably intended, but hey-ho.

Idea: All transcendental numbers are irrational. Therefore if $\ e$ is transcendental then $\ e$ is irrational.

Claim: $\ e$ is transcendental.

Proof: Assume $\ e$ is not transcendental. Then

$\ a_me^m +...+a_1e + a_0 =0$

Without the loss of generality we can assume $\ a_j \in \mathbb{Z}, \forall j$ and $\ a_0 \neq 0$

Define $\ f(x)=\frac{x^{p-1}(x-1)^p(x-2)^p...(x-m)^p}{(p-1)!}$

Where p is an arbitrary prime.Then $\ f$ is a polynomial in $\ x, \partial =mp + (p-1)$. Put

$\ F(x)=f(x) + f^{\prime}(x) + ... + f^{(mp + p -1)}(x)$

and note that $\ f^{(mp + p -1)}(x) =0$

$\ \frac{d}{dx}(e^{-x}F(x)) = e^{-x}(F^{\prime}(x) - F(x)) = -e^{-x}F(x)$

Hence for any $\ j$



$\ = a_j F(0) - a_j e^{-j}F(j)$

Multiply by $\ e^j$ and sum over $\ j$ to get

$\displaystyle \sum_{j=0}^{m} \left( a_je^j \int_{0}^{j} e^{-x}f(x)dx \right) = F(0)\sum_{j=0}^{m} a_je^j - \sum_{j=0}^{m} a_jF(j)$

$\displaystyle = -\sum_{j=0}^{m} \sum_{i=0}^{mp+p-1} a_j f^{(i)}(j)$ $\ (\Phi)$

from the equation supposedly satisfied by by $\ e$

We claim that for each $\ f^{(j)}(j)$ is an integer, and that this integer is divisible by $\ p$ unless $\ j=0$ and $\ i=p-1$. To establish the claim we just apply Liebniz's rule; the only non-zero term arising when $\ j \neq 0$ come from the factor $\ (x-j)^p$ being differentiated exactly p times. Since $\ \frac{p!}{(p-1)}=p$, all such terms are integers divisible by $\ p$. In the exceptional case $\ j=0$, the frist non-zero term occurs when $\ i=p-1$, and then

$\ f^(p-1)(0)=(-1)^p...(-m)^p$

The next non-zero terms are all multiples of $\ p$. The value of equation $\ (\Phi)$ is therefore

$\ K_p +a_0(-1)^p...(-m)^p$

for some $\ K \in \mathbb{Z}$. If $\ p>max(m,|a_0|)$, then the integer $\ a_0(-1)^p...(-m)^p$ is not divisible by $\ p$. So fot sufficiently large primes $\ p$ the value of the equation $\ (\Phi)$ is an integer not divisible by $\ p$, hence not zero.

Now estimating the integral. If $\ 0 \leq x \leq m$ then

$\ |f(x)| \leq \frac{m^{mp+p-1}}{(p-1)!}$

so

$\displaystyle \left|\sum_{j=0}^{m} a_je^j \int_{0}^{j} e^{-x}f(x)dx \right| \leq \sum_{j=0}^{m} |a_je^j| \int_{0}^{j} \frac{m^{mp+p-1}}{(p-1)!}dx$

$\displaystyle \leq \sum_{j=0}^{m}|a_je^j|j\frac{m^{mp+p-1}}{(p-1)!}$

which tends to zero as $\ p \rightarrow +\infty$, thus we have a contradiction . Therefore $\ e$ is transcendental which implies it is also irrational.

(In no way the method they would want for the answer but was fun LaTeXing it )

The intended solution: page 11 of exampaper.web.fc2.com/test/hs/y12/maths/HSC4USolutions/2001.pdf, starting at (b)(i).

Yup, knew they wanted a solution linking to the question, but why be so boring?

Especially when they didn't say "hence"

The best is the 'hence or otherwise' question, and you actually take the otherwise option