Proof that $\pi$ is rational 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.

 A: 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. 
A: This "proof" shows that any real number is rational...
The mistake here is that you are doing induction on the sequence $\pi_n$  of approximations. And with induction you can get information on each element of the sequence, but not on their limit.
Or, put in another way, the proof's b.s. is on "therefore, by induction on the number of decimal places..." 
A: This proof also shows that every countably infinite set is finite, including the set of positive integers $\{1, 2, 3, 4, \ldots\}$.  After all $\{1,2,3,\ldots,n\}$ is finite, and so if we add the next number $n+1$, the set we get, $\{1,2,3,\ldots,n,n+1\}$ is finite.  Adding one more member does not make the set infinite, so by induction, we see that the set of all positive integers is finite.
