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In the proof that $\sqrt{p}$ is irrational where $p$ is a prime number:

We first assume $\sqrt{p}$ is rational.

From this we deduce $\sqrt{p}=\dfrac{a}{b}$, where $a$ and $b$ are co-prime.

Then using other reasonings we deduce $a$ and $b$ are not co-prime.

That is from the wrong statement (i.e. $\sqrt{p}$ is rational), we deduced two contradictory statements.

From a statement, if we apply correct reasonings, we may deduce two results which contradict each other. How can this be reasonable? Can anyone explain with simple examples?

Anyway; from a statement, if we apply correct reasonings, we may deduce a result which contradicts another established result. This seems reasonable to me. And this shows that the statement is wrong.

Hanul Jeon
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Joe
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    I don't understand the problem. In principle, from a false statement you can deduce anything at all, so infinitely many contradictions. Is that what you are asking? – lulu Jul 21 '19 at 10:14
  • Yes, I mean how is it reasonable that we can deduce infinitely many contradictions from a false statement? – Joe Jul 21 '19 at 10:16
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    The statement "if $P$ then $Q$" is always true if $P$ is false. – lulu Jul 21 '19 at 10:17
  • But your question isn't nearly as deep as that. If my false assumption is $2>3$ then I can deduce that $0>1$ or $3>4$ or $4>5$ or... – lulu Jul 21 '19 at 10:19
  • Let us make a wrong assumption $2>3$. We can then deduce that $3>4$. Then can we deduce something from the same assumption $2>3$ which contradicts our previous result, i.e. $3>4$? If yes, how would it be reasonable? – Joe Jul 21 '19 at 10:27
  • @Joe If you ask that, how can assuming $2>3$ be reasonable to begin with? – Hagen von Eitzen Jul 21 '19 at 10:30
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    Sorry, I'm not seeing your problem. The false assumption leads to all sorts of false conclusions. Why would you think that there was somehow only one other false statement that could follow from the false belief? – lulu Jul 21 '19 at 10:31
  • all we have here is the inconsistency of it being rational, but then failing to play well with rational arithmetic rule. That is, it doesn't act like a rational would. –  Jul 21 '19 at 12:25
  • Along the lines of previous comments, it is generally deemed to be the case that a false premise can be used to prove any conclusion (true or false) without making logical errors in reasoning. A famous example can be found here: http://ceadserv1.nku.edu/longa//classes/mat385_resources/docs/russellpope.html – Keith Backman Jul 21 '19 at 16:05

5 Answers5

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The issue is that the false statement we're considering doesn't exist in a vacuum - what we're really talking about is taking an existing theory $T$ and adding to it a new statement, $p$, which $T$ already knows is false. The resulting theory $T+\{p\}$ - call it "$T_{stupid}$" - already has two results that contradict each other built right into it.

So it's not quite true to say that we can deduce lots of contradictory stuff from something false - rather, we should say that we can deduce lots of contradictory stuff from something disprovable. For example, the sentence $$(*)\quad \exists x\forall y(x>y)$$ is false in the natural numbers, but $\{(*)\}$ is perfectly consistent (for example, it has a $17$-element model where "$>$" holds everywhere). It's only when we consider $(*)$ in the context of a theory which disproves it that it winds up being "explosive."

Noah Schweber
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Yes we can deduce two conflicting results from a false assumption.

For example from $$3=5$$ we conclude $$5=3$$ which is false and upon adding both sides we conclude $$8=8$$ which is true.

In logic the implication $$P\implies Q$$ is considered true if $P$ is false regardless of the truth value of $Q$

Thus the statement

If a triangle has four sides then $3=4$ is logically true.

Mohammad Riazi-Kermani
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That's kind of the whole point of proof by contradiction.

If the starting premise is true, we will not be able to find any inconsistencies.

But if the starting premise is false, we should soon run into a contradiction, thereby proving that the starting premise is false, and therefore the opposite of the starting premise must be true.

I think you might find the premise that there are only finitely many prime numbers to be a much simpler example. The classic proof is not a proof by contradiction, some will yell, but that doesn't matter here.

Let's say 2 and 7 are the only prime numbers. Thus $2 \times 7 + 1 = 15$ should be not be prime, and it isn't prime, but it's not divisible by either of the only two primes that exist, 2 and 7.

Therefore, there must be other primes, and indeed we find that 3 and 5 are also prime. But let's insist that there are only four primes, 2, 3, 5, 7. Thus 211 should not be prime, but it actually is.

So the only primes are 2, 3, 5, 7, 211. And 44311 should not be prime, and it isn't: $44311 = 73 \times 607$.

It should then become obvious that whenever we assert that a finite selection of primes consists of all primes, we can always find at least one other prime not already on our list by multiplying the known primes and adding 1.

We can continue to hold to our false premise indefinitely, slowly increasing our list of primes, or we can admit our starting premise is false and move on from there.

Robert Soupe
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I read your comment: "Let us make a wrong assumption $2>3$. We can then deduce that $3>4$. Then can we deduce something from the same assumption $2>3$ which contradicts our previous result, i.e. $3>4?$ If yes, how would it be reasonable?"

I guess what you mean is the following:

We have a wrong statement $A$

You deduce statement $B$ from it

Then from statement $A$ and $B$, you deduce a statement $C$. AND statement $C$ contradicts statement $B$

You seem to say this is unreasonable.

First ask yourself "Why on earth its negation is true?"

There is no reason why its negation is true because we have counterexamples.

Next let me show you a counterexample (somewhat similar to the answer by Mohammad Riazi-Kermani):

$3=5 \{ \text{statement}A\} \tag1$

$\Rightarrow 5=3 \{ \text{statement}B\} \tag2$

Adding $(1)$ and $(2)$:

$\Rightarrow 8=8$

$\Rightarrow 8-3=8-3$

$\Rightarrow 5=5$

$\Rightarrow 5 > 5-2$

$\Rightarrow 5 > 3 \{ \text{statement}C\} \tag3$

Observe that $(2)$ and $(3)$ are contradictory.

Thus I have shown you a counterexample.

lorilori
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The statement that $\sqrt{p}$ (with $p$ a prime number) is rational is not just a false statement, but a statement that contradicts the basic axioms for the real numbers. Indeed, it is from the statement that $\sqrt{p}$ is rational together with those axioms that you derive an explicit contradiction.

Bram28
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