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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?

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It is irrational, because there is no repeating block.

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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.

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