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Prove that $x=0.1234567891011\cdots$ is irrational

Proof: We argue by contradiction. Suppose $x$ is rational, then its decimal expansion ultimately periodic. Let $p$ denote the period of this expansion. Now consider a block $B=000 \cdots0$ $ p$ times since any integer of the form $10^k$ with $ k \geq p$ contains $p$ consecutive $0$'s . this block must accure infinitely often in the decimal expansion of $x$ by our assumption that this expansion is ultimately periodic with period $p$ , this implies that $B$ must be the repeating period block. Which means that the sequence consists of all $0$'s from some point onwards. but this clearly contradicts the construction of the sequence.

This proof is correct? If it is not, what's the issue.

hbghlyj
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erfan soheil
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2 Answers2

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The statement "this implies $B$ must be the repeating block" bothers me a bit.

Surely $C = 111 \cdots 1$ $p$ times also occurs infinitely often, as does every other string of length $p$.

Daniel Pietrobon
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Assume the period is $P=1234567891011$, then somewhere in the decimals you are sure to find $P$ followed by $P+1$, i.e. $1234567891011123456789101\color{red}2$, by the definition of the Champernowne constant. This contradicts periodicity.