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As the title indicates, I'm curious why direct proofs are often more preferable than indirect proofs.

I can see the appeal of a direct proof, for it often provides more insight into why and how the relationship between the premises and conclusions works, but I would like to know what your thoughts are concerning this.

Thanks!

Edit: I understand that this question is quite subjective, but that is my intention. There are people who prefer direct proofs more than proof by contradiction, for example. My curiosity is concerning what makes a direct proof preferable to such individuals. In the past, I've had professors grimace whenever I did an indirect proof and showed me that a direct proof was possible, but I never thought to ask them why a direct proof should be done instead. What's the point?

J. Dunivin
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    Unless you are asking for a historic treatment (which might be better suited for HSM then) this is highly subjective. – mrtaurho Sep 01 '19 at 19:41
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    Possible duplicate of Are the "proofs by contradiction" weaker than other proofs? – Carmeister Sep 02 '19 at 04:30
  • I'd view as this, the statement is a aspect of the logical system and the proofs are simply paths which lead to it. Some may be longer then others and thus contain other aspects and statements. – marshal craft Sep 02 '19 at 10:35
  • @ myself, in a formal system. When we get away from formalism we often lose concreteness and what it means to be mathematics, the idea that things can be decided, and that we arrive at an answer the same. But in pioneering new areas of math we often start off in this situation. We have to make our intuitive problems concrete and answerable. How we do that is also a part of math. In this area there may or may not be issues with this, I don't know though. – marshal craft Sep 02 '19 at 10:41
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    In particular beginners often tend to work along this line to proof $A\implies B$: "Assume $\neg B$, but $A$. Then (something following from $A$) ... and so $B$, contradicting the assumption that $\neg B$", i.e., if they look at it carefully, they really do prvide a direct proof and only make it indirect by unnecessary convolution. – Hagen von Eitzen Sep 02 '19 at 13:24

7 Answers7

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I just did a quick lookup and it suggested that the two flavors of indirect proof was contraposition and contradiction. What I'm about to say is criticizing contradiction, because contraposition seems fine to me.

Imagine you have a 1000 statement direct proof. Then every step along that way is provable. Maybe somebody reads your proof and realizes that an observation you made halfway through is exactly the idea they need to solve a problem they have. Mathematical history has many examples of lemmas that are more famous than the theorems they originally supported.

By contrast, a 1000 statement proof by contradiction starts out with two hypotheses that are inconsistent. Everything you're building is a logical house of cards that is intended to collapse at the end. Nothing you wrote can be counted on outside that framework without a separate analysis.

If it truly takes both hypotheses to get you to the result, then so be it. But I was rightfully dinged by my professors when I wrote a proof by contradiction that could easily be modified into a direct proof by contraposition.

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    +1 I like the fact that you described a situation where one employs a direct proof and makes an observation along the way that in itself could help someone else with their work. Very nice! – J. Dunivin Sep 01 '19 at 19:57
  • @Jeremiah Thank you. I respect that mrtaurho may be concerned about this devolving into a flamewar about the Law of the Excluded Middle, and I respect that this might be closed as a result. But I wanted to get this in to offer the idea that there is a values-free reason why a mathematician might consider a direct proof to be more elegant. –  Sep 01 '19 at 20:10
  • I think you are being much too negative about the value of a proof by contradiction. The framework of the proof of F.L.T. was by contradiction but the methods developed were still of great importance outside the proof itself. –  Sep 01 '19 at 20:24
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    The equivalence $(A \rightarrow B) \Leftrightarrow (\lnot B \rightarrow \lnot A)$ (i.e. contraposition) only holds if we assume the law of excluded middle, which is arguably on the same level as proof by contradiction from an intuitionistical logic POV. – ComFreek Sep 02 '19 at 08:23
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    Proofs by contradiction are also somewhat overdone. People hear that that's how one proves that $\sqrt 2$ is irrational, and thinks it's much more useful than it actually is. I have come across many a post on this site that roughly follows these three steps: 1) Assume for contradiction that $A$ cannot be done. 2) Proceed with a direct proof that $A$ can be done, in no way using assumption 1. 3) Finish with "But this contradicts our assumption, so $A$ can be done". – Arthur Sep 02 '19 at 08:43
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    @Arthur it's relevant here that the common “$\sqrt 2$ is irrational” proof is not a proof by contradiction but just a proof by negation. You want to prove $\neg(\sqrt 2\in\mathbb{Q})$, you proceed to assume for some $r\in\mathbb{Q}$ that $r=\sqrt{2}$ and demonstrate that it's inconsistent – that's just what negation means! Whereas a proof by contradiction would e.g. prove $i\in\mathbb{Q}$ by inconsistency of $i\not\in\mathbb{Q}$. Classically the situations are analogous, but not in e.g. intuitionistic logic. http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/ – leftaroundabout Sep 02 '19 at 09:59
  • @leftaroundabout "Classically the situations are analogous, but not in e.g. intuitionistic logic." And which types of proofs are preferable in mathematical papers is about human thinking and human readability, and the proofs are not formal. So this is much more classical than intuitionistic logic. I don't see how the difference is relevant here. – JiK Sep 02 '19 at 12:06
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    @JiK while most mathematicians do not have any qualms about using classical logic, they still often end up preferring those proofs that are also valid in intuitionistic logic, precisely because they tend to be more direct and human-readable. My point was that this doesn't entail not using contradictions in proofs – when proving a negative then this is usually the right, direct thing to do. But when proving a positive, and specifically when proving existance, then proofs relying on double negation are often quite unsatisfactory. – leftaroundabout Sep 02 '19 at 12:18
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A direct proof often provides more information: if, for example, you want to show that there exists a number with some property, then a direct proof is much more likely than an indirect proof to tell you what that number actually is. This is also true for more complicated statements: if you want to prove "For all $x$ there is some $y$ such that [stuff]," you probably also want to know a particular function spitting out such a $y$ for each given $x$. Again, a direct proof is much more likely to provide that information to you.

This idea really shines when we take it further, and look at logical systems which don't even allow indirect proof; I'm thinking in particular of intuitionism, but there are others.

Most obviously, the above point yields an "applied" motivation for such logics, since it turns out that we can often prove that a proof in such a system (of an appropriate statement) immediately gives us explicit bounds/functions/etc. for that statement. Relevant terms here include "proof mining" (or "unwinding") and "term extraction."

At a vastly higher level of abstraction and technicality, arguments in weaker logics are applicable in broader contexts. Now if you're not interested in nonclassical logic a priori, that doesn't seem like much of a motivation at first, but it turns out that nonclassical logics show up in classical mathematics - via the internal logic of categories. This logic is generally intuitionistic. For an example of how this can translate to a result in the "real world" with classical logic, see this old MSE post. This is quite technical, but here's the gist:

  • We want to prove a fact about modules over sheaves.

  • If we work within an appropriate topos, these look like bog-standard modules over a ring, and the fact we want is classically true for actual modules over actual rings.

  • So all we need is to show that that fact translates to the internal logic: the topos in question will then also think the same thing about its "modules," which are actually the more general objects we care about.

  • This ultimately means that we can produce a classical proof about a generalization of a certain object by finding an intuitionistically-valid proof about the more specific kind of object.

Noah Schweber
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Imagine you are trying to solve a murder in a classic Sherlock Holmes manner.

An indirect proof is analogous to the following: You can deduce that 7 out of the 8 possible suspects have bullet proof alibis and logically the perpetrator can only be the remaining one. There is just no other way. You are right and he is the murderer and the conviction that would follow from that is just. But you know nothing about his motives and you do not know how it happened. You learn nothing about how murderers think or what makes them commit crimes. You gain no knowledge about psychology or sociology.

A direct proof is analogous to the following: Not only can you find the perpetrator but you understand what transpired before the murder, how the murderer executed the crime and what tools he used. You now understand what makes humans commit crimes in certain situations and your expanded understanding can be used in other areas of psychology.

problemofficer - n.f. Monica
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    I like the use of Sherlock Holmes here, given that one of his more famous lines is; “When you have eliminated the impossible, whatever remains, however improbable, must be the truth.” That's quite literally the description of an indirect proof. Of course Sherlock Holmes would not stop at that point, but also deduce the actual details of the crime. – celtschk Sep 02 '19 at 14:32
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    @celtschk This is what inspired my answer. – problemofficer - n.f. Monica Sep 02 '19 at 14:34
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This is rooted in the fact that "true" and "provable from a certain set of axioms with a legit sequence of deductions" can be different things, and in most cases a direct proof can be easier to generalize than an indirect proof. A couple of examples might be useful in explaining this.

Theorem. If the elements of $\mathbb{N}$ are colored with $c$ colors, there are infinite monochromatic 3-APs (arithmetic progressions with $3$ terms).

Sketch of indirect proof: we may invoke the https://en.wikipedia.org/wiki/Poincaré_recurrence_theorem and the principle of combinatorial compactness, by putting a tolopogy over the space of ultrafilters.

Sketch of direct proof: we may study the discrete Fourier series of the set of integers with some color. It turns out that if a coefficient is large enough, there is no way to avoid monochromatic $3$-APs. If that is not the case, we are able to achieve an increase in density by intersecting our set with an infinite AP. By dichotomy, a finite number of steps (only depending on $c$) is able to locate a monochromatic $3$-AP, and prove it appears before $\exp(\exp(\exp c)))$ or something like that.

The indirect case might be considered more elegant, but the direct proof provides an extra quantitative information.

Theorem. Any continuous map $f$ from the unit disk to itself has a fixed point.

Sketch of indirect proof: the existence of a continuous map from $D^2$ to $D^2$ without fixed points would give a retraction of $D^2$ over $S^1$, which does not exist due to the fact that $\pi_1(D^2)$ is trivial while $\pi_1(S^1)=\mathbb{Z}$.

Sketch of direct proof: $D^2$ can be continuously deformed into a triangle $T$, and we may consider a triangulation of $T$ with mesh $\varepsilon$. We may assign a color to each vertex of the triangulation according to the value of our function at such point, and check that the hypothesis of Sperner's lemma are fulfilled. The full-colored Sperner's triangles associated to the meshes $\frac{\varepsilon}{2},\frac{\varepsilon}{4},\frac{\varepsilon}{8},\frac{\varepsilon}{16},\ldots$ accumulate towards a fixed point for $f$.

The (combinatorial) direct proof not only proves a fixed point exists, it provides an algorithm for locating it.

Jack D'Aurizio
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Proof by contradiction (especially in the form of proof by minimal counterexample) is an essential method of modern mathematics.

However, a drawback, (especially for students) is that an error made in the proof is likely to give a rapid, but erroneously obtained, contradiction. With a direct proof an error is much more likely to be picked up because it will lead to a result that was not wanted.

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    "Proof by contradiction (especially in the form of proof by minimal counterexample)" -- I don't think those are the same; a minimal counterexample is a direct/constructive proof that a for-all statement is false, whereas proof by contradiction is indirect/non-constructive. – rlms Sep 02 '19 at 12:41
  • Proof by contradiction is based upon assuming there is a counterexample. Proof by minimal counterexample is simply a stronger form of this in the (many) cases where we can put a 'size' upon the possible counterexamples. –  Sep 02 '19 at 12:59
  • No. Proof by contradiction involves assuming a hypothesis is true, then using that fact to 'prove' some absurdity - something that is demonstrably false. If the steps in the chain are good, then the 'fault' in the proof must be the original assumption.

    Finding a counterexample to a statement doesn't involve assuming the statement is true at any point. That's an entirely different method.

     – Paul Smith Sep 02 '19 at 16:34
  • They are strongly connected ideas. The following is a common structure for many proofs. 1:-To prove all maps are 4-colourable. 2. Assume this is false. 3. Let map M be a counterexample 4. Obtain a contradiction. Please note that I do not disagree with you that finding a c.e. need not involve assuming the statement is true. It's just that in many of the most involved proofs it does and having the device of the minimal c.e. is a vital part of many indirect proofs. –  Sep 02 '19 at 17:21
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Direct proof is kind of proof which don't depend on number of values your logic can take - in 2-value logic contradiction is just shortcut to take all the option at once. Such proof will need certain modifications before using it to 2< value logic which means you need new proof for new environment.

user700367
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Even if direct proof is preferable, it's not always possible. How could you prove, for example, that primes are infinitely many, except by showing that any finite number we assign to them entails the existence of yet another prime (Euclid Elements, IX, 20)--[edit] or contradicts something else known to be true, e.g. the divergence of the harmonic series (Euler)?

Proof of any kind of unlimited multitude, it seems, has got to be indirect, i.e. by contradiction.

Edward Porcella
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  • Hmm, maybe by showing, through Euler's product, that the series of the reciprocal of primes is divergent? – Jack D'Aurizio Sep 02 '19 at 00:43
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    There is a direct proof "for every finite set of primes $S$, there is a prime $p \not\in S$", which can even be turned into a computable function, based on Euclid's proof. This issue is complicated by the fact that the way of proving a negative statement $\lnot \phi$ is to deduce a contradiction from $\phi$, so if we want to prove that the set of primes is infinite, i.e. not finite, then we must use contradiction. But it is also possible to prove "there is a bijection from $\mathbb{N}$ to the set of primes", and this doesn't use contradiction. – Robert Furber Sep 02 '19 at 04:41
  • @Jack D'Aurizio--Doesn't it go the other way around--Euler's proof that the series of reciprocals of primes is divergent presupposes that primes are infinitely many, which he has proven (indirectly) from the divergence of the harmonic series? But you rightly point out, there is more than one way to prove the infinitude of primes. – Edward Porcella Sep 02 '19 at 13:33
  • @EdwardPorcella: the fact that $\mathbb{Z}$ is a UFD can be stated as $$\sum_{n\geq 1}\frac{1}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}$$ for any $s>1$, indipendently from the fact that the primes are finite or infinite. On the other hand, if they were finite, the harmonic series would be convergent. – Jack D'Aurizio Sep 02 '19 at 16:08