4
$\begingroup$

I once asked here a question about how I can get from a false premise to a true premise, and people noted that it is called "Principle of Explosion" - meaning that from a false premise you can get to whatever you want. Link is can be found here: Using deductive reasoning to prove a true statement with false premises?.

But, now came to my mind the proof that $\sqrt{2}$ isn't rational, which is a proof by contradiction (the famous one) and I thought to myself, assuming $\sqrt{2}$ is rational, is as we now know a false premise as it is not rational, meaning that we have again the principle of explosion! (Or in Latin "Ex falso quodlibet").

What is going on here? Why can we assume $\sqrt{2}$ is rational and say that we come to a contradiction when in the first place we assumed a false statement!

For me, it is like saying "Assuming $1=2$, now I can prove that $1=3$" (Hey that is a contradiction no? we said $1=2$ ) thus $1 \ne 3$... but as we know $1=2$ isn't true also!.

Maybe the example of $\sqrt{2}$ is pretty obvious that there is a dichotomy between those relations, either it is rational or not. But there are proofs which do no deal with a dichotomic "property" of something... or proof by contradiction is always for dichotomic properties? (like, being even or odd, rational and irrational, etc...)

Thank you!

$\endgroup$
8
  • 4
    $\begingroup$ If you, or someone else, ever proves that $\sqrt2$ is rational, then, after that, you could use this result and apply the principle of explosion to prove everything. On the other hand, if you "temporarily" assume that $\sqrt2$ is rational, you have not proved that it is. You are within the proof, which is not finished. At the end of the proof you get a contradiction, which only shows you that the "temporary" assumption that $\sqrt2$ was rational turned out to be wrong, so the proof now is finished and the result is that $\sqrt2$ is irrational. That is ok, you can't use it to explode anything. $\endgroup$
    – Mirko
    Commented Mar 15, 2021 at 20:35
  • 1
    $\begingroup$ In the root 2 case, you're not assuming a false statement to go prove other things are true. You're assuming a statement, which you don't yet actually know is false, and deriving a contradiction, from which you can conclude that the original statement is actually false $\endgroup$
    – wormram
    Commented Mar 15, 2021 at 20:35
  • 2
    $\begingroup$ Well, proof by contradiction generally involves comparing the statements $A\to B$ and $\lnot B$, which then proves $\lnot A$. In this case, the system doesn't show $B$ to be true, so we don't have explosion. $\endgroup$ Commented Mar 15, 2021 at 20:35
  • $\begingroup$ I dont understand what are you arguing. There are examples that assuming a wrong premise leads to many new stuff. It depends on how "important" the premise is. But you still have other axiosms and premises that constrain the new possibilities. For example 5th Euclids postulate, when assumed differently leads to whole new geometries. $\endgroup$
    – croraf
    Commented Mar 15, 2021 at 20:37
  • $\begingroup$ @MichaelMorrow But in "secret" we know that these proposition have a truth value, so this leads to the question of does it really matter? If I for example don't know whether $1=2$ is true or false, why can't I assume that, and thus I can say "it is not an explosion"? $\endgroup$
    – CSch of x
    Commented Mar 15, 2021 at 20:39

1 Answer 1

19
$\begingroup$

A proof by contradiction has an important step that you omitted: you start from a false premise, reason to a contradiction, and then discard the false premise as incorrect. It is in fact exactly the avoidance of "proof by explosion."

In other words, we assume various proven facts, call them $A,B,C...$, as well as facing a claim that $P \implies Q$. If you start from an assumption of $\neg (P \implies Q)$, and use it to derive $\neg C$, you have shown that $\neg (P \implies Q) \implies \neg C$, and the contrapositive of that is $C \implies (P \implies Q)$. Since you know $C$ is true, you can conclude that $P \implies Q$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .