# Define a real number by stringing together all positive integers in order $0.123456789101112131415...$

The question is is this number irrational?

I know that an irrational number cannot be written as the quotient of two integers. The only way I can think to begin this problem is to assume to the contrary that that number can be written as $$\frac{m}{n}$$

Still, I don't know how to show this using math or even where to begin. Ideas?

Suppose that number, let's call it $$\alpha$$, were rational with $$\alpha=\frac{n}{m}$$. Let $$k\geq 0$$ be a nonnegative integer with $$10^k>m$$. Then at some point the digits of $$\alpha$$ look like $$\alpha=0.12345\cdots 999\underbrace{1000\cdots00}_{10^k}100\cdots$$ Multiplying by the right power of $$10$$ gives $$10^a\alpha= 12345\cdots 9991\,.\,\underbrace{000\cdots00}_{k\text{ zeros}}100\cdots$$ and so $$0<10^a\alpha-12345\cdots 9991< \frac2{10^{k+1}}<\frac1{10^{k}}$$. Multiplying this by $$m$$ gives $$0<\underbrace{\overbrace{m10^a\alpha}^{=10^an} - m\times 12345\cdots 9991}_{\in\Bbb{N}} = m\times(10^a\alpha-12345\cdots 9991)<\frac{m}{10^{k}}<1$$ This is a contradiction and so $$\alpha$$ is irrational.