It's easy to prove that there's more real numbers between 0 and 1 than there are integers, so in what way is this wrong?

For any number between 0 and 1, you can write it as $0.abcde\ldots$ (where $a, b, c\ldots$ are digits). This number "corresponds" to $2^a3^b5^c7^d11^e\cdots$ (the primes as bases and the digits as exponents). Because of unique prime factorization, every real number corresponds to a different integer. So that would mean that the amount of integers >= the amount of real numbers between 0 and 1.

Which is obviously not true. What am I doing wrong here?

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    $\begingroup$ What integer would $0.11111...$ correspond to? $\endgroup$ Jul 30 at 21:56

Irrational numbers have an infinite decimal expansion, which would mean that number would "correspond" to an infinite product in your coding. natural numbers have all have a FINITE representation as a product of primes.


Consider the expression $$0.111\ldots$$ Then this would correspond to $$\prod_{n = 1}^\infty p_n,$$ where $p_n$ is the $n$th positive prime, which is not a real number, but $+ \infty$. In general, this would map any sequence $(a_n)_{n = 1}^\infty$ that doesn’t just settle to $000000\ldots$ to $+ \infty$.


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