A question on proving that $e$ is irrational

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.

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 $$|aq+bp|\geq 1$$.