I remember hearing several times the advice that, we should avoid using a proof by contradiction, if it is simple to convert to a direct proof or a proof by contrapositive. Could you explain the reason? Do logicians think that proofs by contradiction are somewhat weaker than direct proofs?

Is there any reason that one would still continue looking for a direct proof of some theorem, although a proof by contradiction has already been found? I don’t mean improvements in terms of elegance or exposition, I am asking about logical reasons. For example, in the case of the “axiom of choice”, there is obviously reason to look for a proof that does not use the axiom of choice. Is there a similar case for proofs by contradiction?

**Answer**

To this MathOverflow question, I posted the following answer (and there are several other interesting answers there):

*With good reason*, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)

What is the reason? The reason is the *fecundity* of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r_{1}, r_{2}, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r_{1} and r_{2} and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds.

Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r_{1}, r_{2}, and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r_{1} and r_{2} and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r_{1} implies q), we learn about many different hypotheses that all imply q.

These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q).

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r_{1}, r_{2}, and so on, before arriving at a contradiction. The statements r_{1} and r_{2} are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistant, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.

For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid’s theorem that there are infinitely many primes. In a common proof by contradiction, one assumes that p_{1}, …, p_{n} are *all* the primes. It follows that since none of them divide the product-plus-one p_{1}…p_{n}+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner’s falsely expect after this argument that whenever p_{1}, …, p_{n} are prime, then the product-plus-one is also prime. But of course, this isn’t true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p_{1}, …, p_{n} were *all* the primes. If one organizes the proof, however, as a direct argument showing that whenever p_{1}, …, p_{n} are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p_{1}…p_{n}+1 has merely a prime divisor not on the original list. (And Michael Hardy mentions that indeed Euclid had made the direct argument.)

**Attribution***Source : Link , Question Author : AgCl , Answer Author : JDH*