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I know of two situations resulting from asserting that a false mathematical statement is true (by this we assume that the statement has been made to be a mathematical axiom and that it must be true and only true):

First, if one forces two unequal real numbers a and b to be equal, then all real numbers can be made to equal each other.

Secondly, if one changes the definition of the derivative itself by taking away the removable discontinuity in derivatives of piecewise constant functions (thereby making it always 0), then the antiderivative will change such that antiderivitives no longer vary by constants, but rather, piecewise constants.

Of course if one keeps working through changes in mathematical laws as a result of changing a statement to true, one will eventually either reach a contradiction or end up with something far weirder than one might expect at a first glance.

My question now is:

What other situations are there where we find it useful or interesting to force false statements to be true, and what would be their consequences?

(Of course, most of these will break mathematics, so let's only look at the very close by changes to properties. Obviously these statements were false for a reason.)

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    $\begingroup$ One can consider the theory of fields of characteristic 2 to be exactly the result of taking $0=2$. $\endgroup$ – MJD Jun 19 '16 at 15:23
  • $\begingroup$ @MJD Interesting, and I was always thought that if you make 0 = 2, then multiplication will cause all numbers 0, and by transitivity all numbers equal each other. That is very interesting that it works out differently than I thought. $\endgroup$ – The Great Duck Jun 19 '16 at 21:24
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    $\begingroup$ No, because to prove say $0=1$, you would need to divide both sides of $0=2$ by 2, but you can't divide by 2, for the same reason that you can't divide by 0. Of course you have to abandon the idea that there is such a number as $\frac12$. $\endgroup$ – MJD Jun 19 '16 at 22:32
  • $\begingroup$ @MJD ah, I see. Interesting. $\endgroup$ – The Great Duck Jun 19 '16 at 22:40
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    $\begingroup$ Take the complex numbers, and pretend there are two numbers $i$ and $j$ with $i^2=j^2=-1$, but $j\ne-i$. Insist that everything except for zero has a multiplicative inverse. This should be impossible; $(i+j)(i-j)$ multiplies out to $0$, so $i+j$ shouldn't be able to have a multiplicative inverse. So… let's throw out commutativity, too. Have $ij$ equal $-ji$. (Define $k=ij$.) Let's keep norm ($\lvert a\rvert\lvert b\rvert=|ab|$) and associativity $a(bc)=(ab)c$ in, though; we're not that crazy. What this gets us is the quaternions, numbers of the form $a+bi+cj+dk$ with $i^2=j^2=k^2=ijk=-1$. $\endgroup$ – Akiva Weinberger Jun 19 '16 at 23:30
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If you force two integers $a$ and $b$ to be equal, you get $\mathbb{Z}_{b-a}$, or modulo arithmetic.
If you pretend that $x^2$ can be negative, you get complex numbers.
If you say $3$ can be divided evenly by $2$, you get fractions.
If you pretend that $p^n$ is large when $n$ is negative instead of positive, you get $p$-adic numbers.
If you pretend there is no such thing as parallel lines, you get non-euclidean geometry.

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    $\begingroup$ @user21820 but making 2=3 does change the ordering, so modular arithmetic would make sense. Also, parallel lines do exist. We have many examples of them, but even if that isn't strong enough, we accept it as true, so it's still something that counts IMO. I never thought of fractions that way though. $\endgroup$ – The Great Duck Jun 19 '16 at 17:42
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    $\begingroup$ Anyway, not study but use, this is quite important in reasoning by the absurd. $\endgroup$ – Piquito Jun 19 '16 at 18:01
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    $\begingroup$ @WetSavannaAnimalakaRodVance I might look into that, thanks! $\endgroup$ – The Great Duck Jun 20 '16 at 3:05
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    $\begingroup$ @TheGreatDuck we actually have no examples of lines in reality, and even if we did we wouldn't be able to recognize them. Lines are infinitely long and infinitesimally thin; even if we could detect them, we would have no way of knowing if they ever ended. Lines are purely hypothetical. It makes no more sense to say parallel lines exist than to say that they don't. $\endgroup$ – Matt Samuel Jun 20 '16 at 3:15
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    $\begingroup$ @MathematicsStudent1122, five hundred years ago, $\exists x:x^2=-1$ was a false statement. Someone had to invent $\mathbb{C}$ to make it true. $\endgroup$ – Empy2 Jun 20 '16 at 8:59
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There is a lot of subtlety within the term "false statement". In the simplest sense, if we are studying a particular model or structure, each sentence in the appropriate formal language is true or false in that structure. There is little reason I can think to to try to pretend that a statement that is known to be false in a structure is true in that same structure.

But there are other senses of "true" and "false". For example, most mathematicians (if you give them no other context) would agree that "$2 + 2 = 6$" is false. If pressed, they might say they mean "false in the real numbers" - false in a particular structure. But, if we don't start talking about multiple structures, and we just talk in an informal way, most people who know basic arithmetic would say that "$2 + 2 = 6$" is false.

However, there are structures where $2 + 2 = 6$ is true. The simplest example is the finite field with two elements, $F_2$, in which $1 + 1 = 0$ and so $2 = 0$, and also $6 = 3 \times 2 = 3 \times 0 = 0$. So in this field, $2 + 2$ does equal $6$. On the other hand, we still have $0 \not = 1$ in this field - it is not true that all numbers must be equal just because we assume that $2$ and $6$ are equal. And there is a great use in studying finite fields like $F_2$ in many areas of mathematics. The key point is that $2 + 2 = 6$ is true in some other structure, not in the real numbers.

Similarly, there are other axioms which mathematicians, given no other context, would typically regard as "false". For example, most mathematicians accept there are Lebesgue nonmeasurable sets, which contradicts an axiom known as the Axiom of Determinacy. Only a vanishingly small number of mathematicians who know about the Axiom of Determinacy regard it as a "true" axiom, as far as I can tell. In fact, the Axiom of Determinacy is disprovable in ZFC set theory. But it is somewhat common for these same mathematicians to assume the Axiom of Determinacy in the study of descriptive set theory, because it has very beautiful consequences. People regularly enough publish peer reviewed papers which include theorems that assume the Axiom of Determinacy. One could say that these are examples of theorems proved from a "false" axiom.

In both cases (assuming $2 + 2 = 6$ in the context of fields, or assuming the Axiom of Determinacy in the context of set theory), we don't break mathematics. We just end up studying structures other than the usual ones.

In some cases, we can show there are no structures of a given kind that satisfy a particular axiom (e.g. there is no field with only one element). In that case, there would be little benefit in trying to assume the axiom. This situation can occur, for example, if we have already assumed other axioms that allow you to prove that a given axiom is false.

But, when some structures of a certain kind satisfy the axiom and others don't, just because we think the axiom is "false" in our favorite structure doesn't automatically make it uninteresting to study other structures where the axiom is true, provided that our other assumptions don't already prove the axiom is false.

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  • $\begingroup$ I think that the majority of set theorist who study AD, and are willing to take some sort of Platonic statement about true and false, are likely to opt for $\mathsf{AD}^{L(\Bbb R)}$ inside a universe of sets satisfying choice. $\endgroup$ – Asaf Karagila Jun 19 '16 at 22:19
  • $\begingroup$ @Asaf : Thanks, I hope I have expunged all the $5$s now. It's true that people can try to save face philosophically, or something like that, by working with submodels where AD is true (is really just a more specific way of noting that AD is consistent with ZF?). But, when I read theorems that assume AD or Turing determinacy, in my limited experience, they often enough just assume AD and then work formally in ZF + DC after that, rather than by first fixing a model of ZFC and then working within a submodel that satisfies AD. Here I am reading the theorems "as stated" without retrofitting them. $\endgroup$ – Carl Mummert Jun 19 '16 at 22:24
  • $\begingroup$ This is why I like the utilitarian approach: work in whatever approach that fits the best. If it's easier to argue in ZF+DC+AD, just do that. The translation from this to "AD in inner models" is very robust, so it doesn't change anything, but the complexity of the proof (for the reader, not in the a proof-theoretic way). $\endgroup$ – Asaf Karagila Jun 19 '16 at 22:29
  • $\begingroup$ Yes, of course. It's another example of how some math is not what it seems. Similarly, constructive math can use the same basic words as regular math but mean something completely different. So if a constructive mathematician writes "assume that all functions $f\colon \omega \to \omega$ are computable," I can mentally retrofit their theorem to be about a model where that happens to occur. But if I read them "as stated" without retrofitting, they assume something that is just... false. It doesn't cause them trouble because their other assumptions are compatible. AD is similar in that respect. $\endgroup$ – Carl Mummert Jun 19 '16 at 22:34
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    $\begingroup$ Yes, I get what you're saying. And I guess that if you're reading a paper about constructive mathematics, or sitting in such a lecture, you should be prepared to hear these sort of statements, because the implicit context is fitting. $\endgroup$ – Asaf Karagila Jun 19 '16 at 22:36
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There is no situation where it is meaningful to assume that some false statement is true. The reason is that we would be asserting a falsehood, which means that we can derive any statement at all, not just nonsense equalities like $1 = 2$. It arises as follows:

Let $P$ be a false mathematical statement, and assume that $P$ is true.

Take any mathematical statement $Q$.

If $Q$ is false:

  $P$ is both true and false, which is a contradiction.

Therefore $Q$ is true. [See this for why proof by contradiction is valid.]

So it is completely useless to treat/force any false statement to be true. A common objection by some people is that perhaps we don't really know whether a statement is true or false anyway, but this is an illogical objection; if you don't know whether it is false, then you can't assert that it is a false statement and so it is irrelevant to the above argument. Furthermore, if one wants to capture human knowledge then one ought to use a predicate for that, say $K$, where "$K(P)$" means "We know that $P$ is true.". It is of course possible that both $K(P)$ and $K(\neg P)$ are both false as of now, but that does not change the fact that always either $P$ or $\neg P$ is true.

"Secondly, if one changes the definition of the derivative itself by taking away the removable discontinuity in derivatives of piecewise constant functions (thereby making it always 0), then the antiderivative will change such that antiderivitives no longer vary by constants, but rather, piecewise constants."

We don't change things. We use a different variable to refer to a modified object. Indeed, you can define a kind of operation that is the usual differentiation followed by removing all removable discontinuities. Then yes its inverse will be different from the usual anti-differentiation, and as you say this kind of anti-derivatives will differ by piecewise constant functions. This has nothing to do with forcing false statements to be true. Whatever had been true for derivatives and anti-derivatives remain true. You just have a new structure that has different properties than the old structure.

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  • $\begingroup$ And I don't recall what I told you before, but I suggest you learn logic, since your question shows that your misconceptions come from not knowing how all mathematical statements can be stated and proven in a formal logic system. See math.stackexchange.com/a/1684208 for references and math.stackexchange.com/a/1782071 for a game-semantics viewpoint. $\endgroup$ – user21820 Jun 19 '16 at 15:13
  • $\begingroup$ It isn't change the derivative of anything. It's changing the derivative itself. The derivative operator. And no, you completely twisted my question to try to prove some point. I asked where the consequences (that is to say the resulting new identities and/or properties) of making a false statement are interesting. You've twisted this to say I don't know how to prove things or work in math. That is not true. I stated two cases where claiming a false statement to be true results in something interesting rather than just a purely egregious statement. $\endgroup$ – The Great Duck Jun 19 '16 at 17:38
  • $\begingroup$ And by an egregious statement I mean that claiming something is true might not actually cause anything else to change as no algebra and/or logic could make another statement that was also false but is now true. Actually, the answer the other guy gave is what I was looking, except maybe complex numbers. $\endgroup$ – The Great Duck Jun 19 '16 at 17:40
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    $\begingroup$ Nothing interesting comes out of treating a false statement as a true one. It is as simple as that. $\endgroup$ – Mariano Suárez-Álvarez Jun 19 '16 at 22:31
  • $\begingroup$ @MarianoSuárez-Alvarez while the answer isnt always giving example of treating false statements as true, I'd have to say that the answer I accepted proves you wrong by giving many examples of how treating false statements as true resulted in new definitions and ideas. Granted, those are creating new objects... So it isnt quite the same, but a very close approximation. After all, if they originated by studying "false statements made true" and then made into actual objects in math... Then their properties had to be pretty useful and interesting! $\endgroup$ – The Great Duck Jun 19 '16 at 22:57

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