# What's wrong with this "proof"? (real number to integer injection)

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?

• What integer would $0.11111...$ correspond to? Jul 30 at 21:56

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 $$+ \infty$$.