I want to know if this holds:
An irrational number cannot be written on the form $a/b$ where $a,b \in \mathbb{Z}$, and $b \neq 0$. Assume there is a finite set of irrational numbers $\{r_1,r_2,…r_n\}$ in ascending order. Then $r_n$ would be the largest possible irrational number.
$r_n + 1$ is however larger and outside the set of irrational numbers, thus we can write it on the form $a/b$. Then $r_n = \frac{a}{b}-1 = \frac{a-b}{b}$. But as stated earlier both $a$ and $b$ are integers therefore $a-b$ is as well. We could let $a-b = c \in \mathbb{Z} \Rightarrow \frac{c}{b} \in \mathbb{Q}$. This is clearly a contradiction as $r_n\notin \mathbb{Q}$.
Consequently the set of irrational numbers has to be infinite.
Best Answer
Yes, this is a clear and concise proof that there are infinitely many irrational numbers.