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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.)

Andrés E. Caicedo
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user64742
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    One can consider the theory of fields of characteristic 2 to be exactly the result of taking $0=2$. – MJD Jun 19 '16 at 15:23
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    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 $\frac12$. – MJD Jun 19 '16 at 22:32
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    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 $i^2=j^2=k^2=ijk=-1$. – Akiva Weinberger Jun 19 '16 at 23:30
  • Well, I have no idea how Hamilton came up with them. It was probably some stroke of genius (he did carve the definition into a bridge in his excitement, after all). (If you haven't heard that story, that's essentially the long and short of it.) But what I wrote above seems to make some sense, at least. – Akiva Weinberger Jun 20 '16 at 03:02
  • If you keep commutativity, and are fine with $i+j$ having no inverse, by the way, then I'm pretty sure that what you get is isomorphic to $\Bbb C\times\Bbb C$; map $i\mapsto(i,i)$ and $j\mapsto(i,-i)$. Which explains why $i+j$ has no multiplicative inverse, because $i+j\mapsto(2i,0)$. – Akiva Weinberger Jun 20 '16 at 03:16

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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.

Carl Mummert
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  • 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. – Asaf Karagila Jun 19 '16 at 22:19
  • @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. – Carl Mummert Jun 19 '16 at 22:24
  • 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). – Asaf Karagila Jun 19 '16 at 22:29
  • 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. – Carl Mummert Jun 19 '16 at 22:34
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    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. – 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.

Community
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user21820
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  • 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 http://math.stackexchange.com/a/1684208 for references and http://math.stackexchange.com/a/1782071 for a game-semantics viewpoint. – user21820 Jun 19 '16 at 15:13
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    Nothing interesting comes out of treating a false statement as a true one. It is as simple as that. – Mariano Suárez-Álvarez Jun 19 '16 at 22:31
  • @TheGreatDuck: You're still making a false statement. The answer you've accepted isn't about treating false statements as true, but is about modifying the axiom system so that we can add previously disprovable axioms. Of course, people make such modifications that are meaningful some way or another. If that is what you're looking for, then you asked the wrong question. It is as simple as that. If you edit your question to ask about interesting modified axiom systems, then of course the answer you accepted is very relevant, but my answer was correct for your original question as stated. – user21820 Jun 20 '16 at 04:04
  • @TheGreatDuck: By the way I honestly think you're not helping yourself by avoiding the study of logic, since you won't be able to understand clearly the distinction between provability and truth, which is a crucial point here. I'll give you a gist, but you really should study it in detail. [continued] – user21820 Jun 20 '16 at 04:07
  • @TheGreatDuck: No. I'm talking about logic proper, not elementary logic. See those references I linked to earlier. – user21820 Jun 20 '16 at 04:10
  • @TheGreatDuck: [continued] We need to work in a meta-system when discussing different axiom systems. Truth is defined in the meta-system with respect to models of the axiom system (a statement may be true in one model but false in another). Provability on the other hand depends only on the axiom system. Some axioms are independent of (neither provable nor disprovable from) the others, such as parallel postulate in Euclidean geometry. That means that we can replace it by its negation and there will still be a model. Of course, every axiom is true in any model, so there was no false axiom. – user21820 Jun 20 '16 at 04:11
  • @TheGreatDuck: What happened was that when we modified the axiom system (such as removing the parallel postulate), we changed what was provable, so what was previously disprovable (equivalent to being false in every model) may now be independent. The bottom-line is that "false" is meaningless without also specifying a model. – user21820 Jun 20 '16 at 04:13
  • @TheGreatDuck: Okay let me explain about modulo arithmetic then. The integers Z are defined axiomatically as a model of some axiom system (similar to PA). If one changes the axiom system, Z no longer is a model of the new axiom system. So for example if you wish to add an axiom like "0 = 1+1+1+1+1+1", then the ordering properties contradict it so we usually completely remove the "<" symbol. The references for logic that I linked to are the best free resources I could find on the internet. – user21820 Jun 20 '16 at 04:19
  • Let us continue this discussion in chat. – user21820 Jun 20 '16 at 05:14
  • @TheGreatDuck: I've made a chatroom where we can continue the discussion. See you there! – user21820 Jun 20 '16 at 05:23
  • @TheGreatDuck: By the way I've edited my answer since now your question about derivatives is clear. I didn't get it earlier but didn't twist your question, which said "piecewise constant derivative" at first; now it says "derivative of piecewise constant functions", which has a totally different meaning and makes sense in the way you described. – user21820 Jun 20 '16 at 07:35
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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.

Empy2
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  • Technically this doesn't answer the question, since each of this requires changing the definitions. Just for example, going from $\mathbb{Z}$ to $\mathbb{Z}_n$ loses ordering, same for reals to complex numbers, integers to rationals loses discreteness (and well-ordering), p-adic numbers involves a different metric than reals. Finally, existence of parallel lines was never proven to be true, so assuming its negation is not assuming a false statement. If one accepts that example then we could do the same for every axiom in any formal system (which can sometimes be interesting). – user21820 Jun 19 '16 at 16:52
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    Anyway, not study but use, this is quite important in reasoning by the absurd. – Ataulfo Jun 19 '16 at 18:01
  • @TheGreatDuck If you've never thought of fractions that way, then you may enjoy reading Chapter 1 of Roger Penrose's "Road To Reality" wherein he describes grappling with the notion of fractions as a little child and being very fazed by the notion of common factor cancellation. Showing early signs of his strong Platonistic leanings, he described being highly alarmed that one child essentially proposed defining fractions as pairs of integers above and below a horizontal line and that cancellation would just be one of the rules of the "game" that we make up. – Selene Routley Jun 20 '16 at 01:38
  • @WetSavannaAnimalakaRodVance But then you need to show that what was just defined has nice things like a total order, which I doubt this other child thought of ^_^ – Akiva Weinberger Jun 20 '16 at 03:11
  • @AkivaWeinberger Of course, I doubt it too: but it's fascinating to see the germ of the idea of constructing axiom systems taking root in the child's mind. – Selene Routley Jun 20 '16 at 03:13
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    @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. – Matt Samuel Jun 20 '16 at 03:15
  • @TheGreatDuck: In fact, that the number 4 in some sense does not physically exist is very true! It is merely part of the larger concept of natural numbers, which concept seems to reflect the structure of some physical entities (like internal states of a computer). For more, see my explanation of interfaces (abstractions) and implementations (models) at http://math.stackexchange.com/a/1809216. – user21820 Jun 20 '16 at 05:13
  • None of these examples are false, though. The statement $\exists x \in \mathbb{C}$ such that $x^2 = -1$ is a true statement. Replace $\mathbb{C}$ with $\mathbb{R}$, only now do you have something false. – MathematicsStudent1122 Jun 20 '16 at 07:36
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    @MathematicsStudent1122, five hundred years ago, $\exists x:x^2=-1$ was a false statement. Someone had to invent $\mathbb{C}$ to make it true. – Empy2 Jun 20 '16 at 08:59
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Andreas Blass's Seven Trees in One (1991) is not just about graph theory, but also explains how to reconcile intuitive false ideas with more rigorous reasoning based on the same underlying concepts. The example here is the "eighteenth-century spirit" of analysis, where an argument that makes no literal sense is presented, the reader bears with the manipulations, and the ideas underlying it (manipulating equations of powers of sets) get tied down to more rigorous ideas (isomorphisms in ring of polynomials given the $T=1+T^2$ equality).

To see why seven is special, we first give an argument to establish Theorem 1 in the style of eighteenth-century analysis, where meaningless computations (e.g., manipulating divergent series as though they converged absolutely and uniformly) somehow gave correct results. This argument begins with the observation that a tree either is $0$ or splits naturally into two subtrees (by removing the root). Thus the set $T$ of trees satisfies $T=1+T^2$. (Of course equality here actually means an obvious isomorphism. Note that the same equation holds also for the variant notions of tree mentioned at the end of Section 1.) Solving this quadratic equation for $T$, we find $T=\frac12\pm i\frac{\sqrt{3}}{2}$. (The reader who objects that this is nonsense has not truly entered into the eighteenth-century spirit.) These complex numbers are primitive sixth roots of unity, so we have $T^6=1$ and $T^7=T$. And this is why seven-tuples of trees can be coded as single trees.

Then, why the above argument is valuable:

Although this computation is nonsense, it has at least the psychological effect of suggesting that something like Theorem 1 has a better chance of being true for seven than for five or two. To improve the effect from psychological to mathematical, we attempt to remove the nonsense while keeping the essence of the computation.

C7X
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