# Proof verification: There are infinitely many irrational numbers.

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.

• It seems correct to me. Your solution is clear and straightforward
– Gabe
Nov 15, 2019 at 3:22
• Absolutely perfect. I see no flaws or ways to improve upon the proof. Great work! Nov 15, 2019 at 3:32
• A little fun fact: not only is there infinitely many irrational numbers, but because the irrationals are dense in $\Bbb{R}$, there are infinitely many irrational numbers between any two real numbers Nov 15, 2019 at 3:36
• Your solution is correct only if you already know that there is at least one irrational. Otherwise, you need to prove that first. Nov 15, 2019 at 4:01
• @Timber I am honestly not sure about that part, but I believe that Cantor most likely used his research on the irrationals to show that there are "more" irrationals than rationals, which at the time received lots of criticism from people like Leopold Kronecker and may have contributed to Cantor's depression and institutionializations. The whole story about Cantor's set theory is an interesting read if you ever get a chance to that shows a compromise between mathematics and philosophy Nov 15, 2019 at 4:02