I stumbled upon this proof of $\pi$ being rational (coincidentally, it's Pi Day). Of course I know that $\pi$ is irrational and there have been multiple proofs of this, but I can't seem to see a flaw in the following proof that I found here. I'm assuming it will be blatantly obvious to people here, so I was hoping someone could point it out. Thanks.
Proof:
We will prove that pi is, in fact, a rational number, by induction on
the number of decimal places, N, to which it is approximated. For
small values of N, say 0, 1, 2, 3, and 4, this is the case as 3, 3.1,
3.14, 3.142, and 3.1416 are, in fact, rational numbers. To prove the rationality of pi by induction, assume that an N-digit approximation
of pi is rational. This number can be expressed as the fraction
M/(10^N). Multiplying our approximation to pi, with N digits to the
right of the decimal place, by (10^N) yields the integer M. Adding the
next significant digit to pi can be said to involve multiplying both
numerator and denominator by 10 and adding a number between between -5
and +5 (approximation) to the numerator. Since both (10^(N+1)) and
(M*10+A) for A between -5 and 5 are integers, the (N+1)-digit
approximation of pi is also rational. One can also see that adding one
digit to the decimal representation of a rational number, without loss
of generality, does not make an irrational number. Therefore, by
induction on the number of decimal places, pi is rational. Q.E.D.
Best Answer
Let's apply this technique to a more transparent question.
CLAIM: $0.333\ldots < 1/3$
Proof: We induct on the number of decimal digits. Clearly, $0.3 < 1/3$. Now, by induction, if $n$ digits of $0.333\ldots 3 < 1/3$, than in particular $3 \cdot 0.333\ldots3 = 0.999\ldots900 < 1$, and so $0.999\ldots 990$ (i.e. with one more $9$ digit) $<1$, and thus it holds for $n+1$ as well. So by induction, the claim is proven.
What's wrong with this? Induction is a proof for all natural numbers, not for $\infty$. It's clear that $0.333\ldots = 1/3$. But any finite decimal representation is less than $1/3$. And the induction only shows that any finite decimal representation is, in fact, less than $1/3$.
This is the same flaw at the heart of the $\pi$ rational argument.