[Math] Non-rigorous reasoning in rigorous mathematics

big-listheuristicssoft-question

I was wondering what role non-rigorous, heuristic type arguments play in rigorous math. Are there examples of rigorous, formal proofs in which a non-rigorous reasoning still plays a central part?

Here is an example of what I am thinking of. You want to prove that some formula $f(n)$ holds, and you want to prove this by induction. Based on heuristic arguments, you conjecture what the correct formula is. Then you prove it by induction. But, if you had just given the induction proof on its own, then you would have to pluck this mysterious formula out of thin air.

I am interested in situations in which there is a heuristic argument which is valid and can be formalized. I am more interested in cases in which there is a heuristic argument and a separate (or complementary) rigorous argument, but the heuristic argument is more enlightening and more explanatory.

Best Answer

It feels a bit surprising that this already has a dozen answers, and none of them is the following fact from noncommutative algebra, which I had thought was a canonical example of "non-rigorous reasoning in rigorous mathematics":

Proposition. In a ring with identity if $1-ab$ is invertible then so is $1-ba$.

Proof: let $c = (1-ab)^{-1}$. Then $1+bca = (1-ba)^{-1}$, because $$ (1-ba)(1+bca) = 1 - ba + bca - babca = 1 - b(1-(1-ab)c)a = 1 - b(1-1)a = 1, $$ and likewise $$ (1+bca)(1-ba) = 1 - ba + bca - bcaba = 1 - b(1-c(1-ab))a = 1 - b(1-1)a = 1. $$

Non-rigorous explanation: $$ c = (1-ab)^{-1} = 1 + ab + (ab)^2 + (ab)^3 + \cdots = 1 + ab + abab + ababab + \cdots, $$ so $$ (1-ba)^{-1} = 1 + ba + (ba)^2 + (ba)^3 + \cdots = 1 + ba + baba + bababa + \cdots, $$ which can be written as $$ 1 + b(1 + ab + abab + \cdots)a = 1+bca. $$

I don't know the original source of the Proposition and the motivation (nor even whether the "motivation" was discovered after the fact!). I do know that Google already guesses the right context from just "1-ab", even before I have the chance to enter "invertible", let alone "1-ba". One of the top hits is this mathoverflow question from almost four years ago.