# 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.

• This just shows each number in the sequence $3$, $3.1$, $3.14, \ldots$ is rational, but it certainly doesn't show that $\pi$ is rational (and it's not even a particularly clever fake proof imo) – ShawnD Mar 15 '12 at 2:55
• One way to state the error here is that $\pi$ is not obtained by adding one more digit to any decimal number. – Jonas Kibelbek Mar 15 '12 at 3:02
• Is it fair to say that there are more decimal places than there are natural numbers? – Greg L Mar 15 '12 at 6:17
• @Greg: No, there's a 1st decimal place, a 2nd, a 3rd, a 4th and so on. So for each natural number there's a decimal place (and vice versa). Thus you can say that there are as many natural numbers as decimal places. – Hendrik Vogt Mar 15 '12 at 7:18
• To state briefly what the error is, the limit of a sequence of rational numbers is not necessarily rational. In fact this is the key analytic difference between $\mathbb{Q}$ and $\mathbb{R}$: $\mathbb{R}$ is complete in the sense that every Cauchy sequence converges, but $\mathbb{Q}$ is not, with this "proof" providing a counterexample. – William Mar 15 '12 at 15:20

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.

• Ahh, I completely forgot that induction only applied for the natural numbers! Thanks! – gsingh2011 Mar 15 '12 at 3:02
• User jasonk wishes to add: To clarify - the statement "assume that an N-digit approximation of pi is rational". This 'proof' has proved that the approximation of N-digit pi is rational, not that pi itself is rational. 0.33333 is in fact rational, but it is only an approximation and is not 1/3. – anon Mar 15 '12 at 9:31

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..."

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.