Can A be proved using B when B was proved using A?

Question

Many questions have been posted on this site about the irrationality of $\pi$; I'll be referring to one such question here.

The accepted answer mentions that $\pi$ is irrational because it is the quotient of two numbers, namely the circumference and the diameter, with at least one of them being irrational. This makes complete sense to me, since the quotient of an irrational number with a rational one cannot be rational.

A rational number is a number that can be expressed as $p/q$, where $p$ and $q$ are integers. The number $\pi$ cannot be expressed in this form; hence it is irrational.

In other words, the definition of "fraction" does not include ratios like "circumference/diameter" in which the numerator and denominator are arbitrary numbers, not necessarily integers.

Now, later on in the answer, the author proves that the circumference or the diameter in a circle has to be irrational.

In the case of "circumference/diameter" (which you denoted $\pi = C/D$), it will always be the case that if the diameter is an integer, the circumference ($C = \pi D$) is not an integer, and if the circumference is an integer, the diameter ($D = C/\pi$) is not an integer: precisely because $\pi$ is irrational.

In this case, $\pi$ is being used to prove that $c$ or $d$ is irrational, and the irrationality of $c$ or $d$ is being used to prove that $\pi$ is irrational.

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This above method does not makes sense to me; IMHO, you cannot prove A using B when you've proved B using A. In this case, I've read alternate proofs proving that $\pi$ is irrational, so I know that the circumference or diameter of a circle is also irrational.

However, are proofs like the above valid? Could I use B to prove A when I've proved B using A? Am I going wrong somewhere in my line of thinking?

Answer 1

Two points:

1. The question asked why $\pi$ is irrational if it can be represented as $\frac pq$(when $p=\pi$ and $q=1$), and the answer was right: p and q must both be integers.

2. If you proved A=>B, then you can use that as a fact to try to prove B=>A. However this is not always true: consider the statement

A:someone's name is Andy and

B:he is a human.

If someone's name is Andy then he is a human (unless you are naming a dog:) ), but it's not true that everyone's name is Andy.

Answer 2

The author of that answer is only trying to prove that, since $\pi$ is irrational, at least one of $C$ or $D$ must be irrational. They're not trying to prove the other direction, they're taking it as a given that $\pi$ is irrational and just explaining why that isn't inherently contradictory with the identity $\pi=\frac CD$.

The fact that a number is a ratio of two numbers, at least one of which is irrational, doesn't prove the number is irrational anyway. A ratio of two irrationals can be rational, like $\frac {\sqrt 8} {\sqrt 2}$.

However, of course in general you can't prove $A$ using $B$ if your only proof of $B$ relies on $A$. For example, there are many results that famous open conjectures like the Riemann hypothesis are equivalent to some other result. Those papers read something like "If the Riemann hypothesis is true, then $X$ is true, [proof] and if $X$ is true, then the Riemann hypothesis is true [proof]". But that doesn't prove the Riemann hypothesis.