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When is proof by contradiction necessary?

It’s been a while since I have written a post in the “somewhat philosophical” category, which iswhere I put questions like “How can one statement be stronger than an another, equivalent,statement?” This post is about a question that I’ve intended for a long time to sort out in my mind

but have found much harder than I expected. It seems to be possible to classify theorems into threetypes: ones where it would be ridiculous to use contradiction, ones where there are equally sensibleproofs using contradiction or not using contradiction, and ones where contradiction seems forced.But what is it that puts a theorem into one of these three categories?

This is a question that arises when I am teaching somebody who comes up with a proof like this.“Suppose that the sequence is not convergent. Then … a few lines of calculation … which

implies that . Contradiction.” They are sometimes quite surprised when you point out thatthe first and last lines of this proof can be crossed out. Slightly less laughable is a proof that is morelike this. “We know that . Suppose that . Since the derivative of

has absolute value at most 1 everywhere, it follows that , which is a contradiction.

Therefore, .” There, it is clearly better to work directly from the premise that

via the lemma that to the conclusion that

. However, the usual proof of the lemma does use contradiction: one assumes

that the conclusion is false and applies the mean value theorem.

The result of all this is that I don’t have a good tip of the form, “If your theorem is like this then try aproof by contradiction, and otherwise don’t.” For the remainder of this post I’ll discuss anothercouple of examples that show some of the complications that arise.

Example 1. The irrationality of the square root of 2.

This is of course the classic proof by contradiction, and one can even give a kind of quasi-proof thatit must use contradiction. The reason is that the word “irrational” means “not equal to for any

pair of integers “. If that is the definition, then let us suppose that the last two lines of the proof

went: therefore has property ; therefore is irrational. We could then ask, “Why does

having property imply that a number is irrational?” It might be obvious that property impliesirrationality, but to prove it it would still be necessary to say, “Well, take any rational number

… therefore does not have property .” (Why would that be necessary? For precisely the

same reason! Perhaps this is an induction on proof length or something like that.)

With those thoughts in mind, consider the following argument. We begin by calculating the

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With those thoughts in mind, consider the following argument. We begin by calculating thecontinued-fraction expansion of . We find that . The

denominator of the fraction is , so we see that the continued-fraction

expansion repeats itself, and, in one reasonably standard notation, is . In particular, it is

infinite. Therefore, is irrational.

At first sight, this looks like a direct argument rather than a proof by contradiction: we used thehypothesis that to deduce that has a property that obviously implies irrationality.However, as I mentioned in the general remarks, one could question this by asking, “Why is it thata number with an infinite continued-fraction expansion has to be irrational?” The answer? It’sobvious that a rational number has a terminating continued fraction, because as you work it out thedenominators keep decreasing … oops, sorry, that was a proof by contradiction.

So perhaps the answer is indeed that if you are trying to prove a negative statement, then you haveto use a proof by contradiction. But what is a “negative statement”? How about the followingtheorem?

Theorem. If and are integers, then .

Aha, you say, the “not equals” makes that negative. But we can deal with that by a quickreformulation.

Theorem. If and are integers, then .

What’s negative about that? If you think that it’s somehow bound up in the notion of “strictlygreater than”, then how about this?

Theorem. If and are integers, then there exists a real number such that .

That looks pretty positive to me, since it is asserting the existence of something.

It becomes less positive if you imagine how you would establish the existence of such an . Theobvious thought would be, “Well, the only thing that could possibly go wrong is if , so all

we have to prove is that .” And that’s negative again. So does that mean that a statement is

negative if the only sensible way of proving it is to reduce it to a statement that includes the word“not”? Even if something like that is correct, it seems quite hard to formalize.

Here’s another example of that last type of question. Is being infinite a negative property? Onemight say yes, because it means not being finite. But when we were talking about continuedfractions, all we cared about was sequences, and we can define a sequence to be infinite if its termscan be put in bijection with the natural numbers. (And we could define a set to be infinite if there isan injection to a proper subset. But is a proper subset a negative concept because it means notincluding all elements?)

Example 2. Continuous functions on closed bounded intervals.

Until recently I “knew” that the following was the case. If you want to prove something using thecompactness of , then you can either prove it directly using the Heine-Borel theorem or you can

prove it by contradiction by reformulating everything in terms of sequences and applying theBolzano-Weierstrass theorem. For example, to prove that a continuous function on is

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bounded, you either find a neighbourhood of each point on which is bounded (by the definition ofcontinuity) and cover by finitely many such neighbourhoods (by the Heine-Borel theorem), or

you assume that is unbounded, construct a sequence such that for every , apply

the Bolzano-Weierstrass theorem, and reach a contradiction.

I had also tacitly assumed that there is an algorithm for converting proofs of one kind into proofs ofthe other, though I had never actually tried to work out the details.

But recently, a colleague and I had a conversation that led to the following proof of the theorem,which disturbs my cosy view of how things work. The idea is to try to imitate the above proof bycontradiction as much as possible, but without actually bothering to prove the result bycontradiction. Specifically, one builds a sequence that is the most likely sequence to cause trouble,and then proves that it does not cause trouble. Here is how the argument goes.

Let be the supremum of . By the definition of a supremum, we can

find a sequence such that . Pick a convergent subsequence , by Bolzano-

Weierstrass. Then is a subsequence of , so . But if is the limit of the

sequence , then . So . So is an upper bound for . (Note that this

proof also shows that the bound is attained.)

There seems to be no contradiction involved. But again, if we dig a little deeper it starts to look asthough what is really going on is that the contradiction is hidden in the “obvious” steps of the proof.For instance, how do we know that we can find a sequence such that ? We’ll need to

split into two cases (unless we want to define the topology we are implicitly putting on the extendedreal line). The case where is not really worth investigating, since it instantly gives us that isbounded (though it was by unnecessarily doing this step that we obtained the added informationthat attains its upper bound). And if we then look at the case , how is what we are doingany different from assuming that is unbounded? I find myself rather confused.

Final remarks.

One thing that seems to be coming out of these examples is that the notion of a proof bycontradiction is relative to the definitions you use and the small results that you take for granted. Forinstance, I could define a number to be irrational if its continued-fraction expansion is non-terminating. I wouldn’t actually advocate doing that, but if one did, then the “direct” proof I gave ofthe irrationality of would be just that — direct. And if I do not allow myself to assume that

then the proof that whenever

appears to stop being direct and require a contradiction.

In which case, perhaps the advice that I give to students — proof by contradiction is a very usefultool, but try not to use it unless you really need it — is, though not completely precise, about thebest one can do.

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53 Responses to “When is proof by contradiction necessary?”

Terence Tao Says: March 28, 2010 at 4:21 pm | Reply

Nice post, as always

One can use the continued fraction expansion of to give a constructive algorithm to solve theequation with an which is guaranteed to be well-defined (basically, x will be another continuedfraction, in which all denominators can be shown to be non-zero). Once one can solve and , onecan of course solve , though it is amusing that this proof of a purely rational fact proceeds usingthe reals (via continued fractions). Though it may be possible to terminate the continued fractionafter a finite time and make this a purely “rational” proof as well (though this may require aninfinite descent, leading to another way for proof by contradiction to sneak in).

At some point one has to use the fact that the sum of two positive numbers is positive (and thushas a reciprocal), but if such a statement falls under the category of “proof by contradiction”then really there’s no escaping that category. There may also be subtleties in the construction ofthe real numbers, and the demonstration that every continued fraction converges to a realnumber.

For me, the purest examples of proofs by contradiction are those of the “non self-defeatingobject” type, where the existence of an overpowered object is crucially needed to demonstrate thenon-existence of said object. See my post at

http://terrytao.wordpress.com/2009/11/05/the-no-self-defeating-object-argument/

Even in those cases, though, one can often extract a large component of the argument whichdoes not involve proof by contradiction (Euclid’s theorem on the infinitude of primes being agood example of this).

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Terence Tao Says: March 28, 2010 at 5:10 pmActually, I just realised that the proof I had in mind that the algorithm to solve terminatesrelies on infinite descent, which of course uses contradiction. But… one can takecontrapositives and prove the solubility of, say, by strong induction and a division into casesinstead. Indeed, if p is odd, or if p is even and q is odd, then is either odd, or twice an oddnumber, in which case solubility is clear; and if are even then one can appeal to the stronginduction hypothesis.

The division into cases would be problematic if one were to adopt an intuitionistic stance onlogic, though. (I believe that one can still prove irrationality of sqrt(2) intuitionistically, butnow proof by contradiction is definitely required.)

Ben Lund Says: April 29, 2010 at 3:41 amFollowing Gowers’ logic above, you still need a contradiction to prove the irrationality ofsqrt(2).For any rational number a, is soluble. This is not the case for , so is irrational.

porton Says: March 28, 2010 at 4:36 pm | ReplyField medalist Gowers, it is really simple indeed.

A direct proof “A1 => A2 => … => An” is equivalent to proof by contradiction “not An => notA(n-1) => … => not A1″.

What is more natural depends on what statements are more natural “Ai” or “not Ai”.

No philosophy.

gowers Says: March 28, 2010 at 4:39 pmEr … thanks.

kazek Says: March 29, 2010 at 9:52 amAxiom1: “X={A,B,C}”Axiom2: “[ (s =A) or (s= B) or (s = C) ] => p(s)”

A1= “s=A”A2= “p(s)”

A1=>A2 (axiom 2)

not A1 = “not (s = A)” or equivalent not A1 = “s=/= A”not A2 = “not p(s)” ( if s=B then s=/=A and p(s) is still true).

So as things are different at least from that You wrote.

kazek Says: March 29, 2010 at 10:49 am

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Last sentence should be: “So things are different at least from that You wrote.” I supposeYou are saying that proof by contradiction and direct are equivalent, and that is true, but away You describe equivalence is probably not correct.

daniel. Says: April 16, 2010 at 10:18 amYou are thinking of ‘contrapositive’ and not contradiction.

sincerely, non-Fields medalist nobody.

Qiaochu Yuan Says: March 28, 2010 at 5:07 pm | ReplyAs far as exercises in classes go, I’ve found it most natural to use proof by contradiction if thestatement I want to prove is universally quantified, since its negation is an existential statement. Ican then use the object the existential statement gives me to “power up” my proof in a way thata “direct” proof (checking that some condition really does hold everywhere) cannot do, at leastas easily. I think I’ve heard similar sentiments expressed somewhere; maybe MO.

As far as the larger question at hand, if “proof by contradiction” means “using the law of theexcluded middle” then the problem is, as you say, that many definitions commonly in use inmathematics are equivalent only if the law of the excluded middle is assumed. Without it onemust make some extra choices, and then one can ask what is provable in intuitionistic logic andwhat is not. (Or something like that.)

gowers Says: March 28, 2010 at 6:28 pmWhat I’m looking for is a way of recognising in advance which statements are best tackledby contradiction. I don’t think your suggestion captures it — there are just too manycounterexamples. For example, suppose I need to prove that there is no smallest positive realnumber. Then I am proving that for every positive real number there is a smaller one. If Ifollowed your advice, the proof I would write out would go like this.

1. Suppose the result is false, and let be the smallest positive real number.

2. The number is positive and smaller than .

3. But this contradicts the assumption that was the smallest positive real number.

It seems to me more natural to say this.

1. Let be a positive real number.

2. Then is positive and smaller than .

The point in this example is that the assumption that is the smallest positive real number is ofno help in finding a smaller positive real number. I think a sensitivity to something like thatshould be part of the general principle I am looking for.

Another example is this: prove that the product of any four consecutive integers is divisibleby 24. Contradiction doesn’t come into it. One can either argue that is the binomialcoefficient , or one can argue that the set must contain at least one multiple of 4, one other

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even number, and one multiple of 3. (Of course, as above it may be that justifying thesefurther facts eventually requires contradiction to prove some very low-level assertion.)

Qiaochu Yuan Says: March 29, 2010 at 6:57 pmHmm. Maybe I should give an example and leave the extraction of the general principle toothers. Here’s a typical example of when I found it useful to negate statements in topology.

Proposition: Suppose a topological space has the property that every open cover has a finitesubcover. Then closed subsets of with the finite intersection property have non-emptyintersection.

Proof. Suppose that has the first property but not the second. Then there exists a collection ofclosed subsets of with the finite intersection property with empty intersection. The collectionis then an open cover of , so it admits a finite subcover, but this contradicts the finiteintersection property.

Maybe it’s equally easy to phrase this proof without the use of contradiction and I ammissing something. But the hypotheses don’t seem to match up as nicely if you don’t usecontradiction.

Qiaochu Yuan Says: March 29, 2010 at 7:02 pmAh, I am missing something: the statement I wrote above can be proven directly by using thecomplement of the intersection to build an open cover. Let me see if I can find an examplewith more quantifiers…

Tom Ellis Says: March 28, 2010 at 5:31 pm | Reply“It’s utterly obvious because if it has a terminating continued-fraction expansion then you canjust work out what rational number”

Am I missing something, or did you mean the converse here?

gowers Says: March 28, 2010 at 5:55 pmWhoops — no, that was my mistake. I’ll change it right away.

Joshua Green Says: March 28, 2010 at 5:32 pm | ReplyIn your first example “Suppose that the sequence (a_n) is not convergent. Then … a few lines ofcalculation … which implies that a_n\rightarrow a. Contradiction.” it isn’t necessarily the casethat the first and last lines can be removed, as that first line might be necessary for one of theimplications along the way. On the other hand, it is often the case that the last line can bedropped, replaced instead by a contradiction with the line before. Terence Tao points out myfavorite example of this, Euclid’s proof of the infinitude of primes, where arguably the finalcontradiction is with the Fundamental Theorem of Arithmetic but many people use that theoremto instead derive a contradiction with the opening statement.

gowers Says: March 28, 2010 at 5:59 pm

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I’d be inclined to say that the assumption that fails to converge is rather unlikely to be helpfulin proving that . The main point is that I really have seen proofs written by students in whichit really is the case that the initial “Suppose not” and final “contradiction” can be crossed outwith no loss to the argument.

Emmanuel Kowalski Says: March 28, 2010 at 7:37 pm | ReplyIn the proof of irrationality of sqrt(2), I find it interesting to see how it enters in a proof of the apriori stronger, and “direct”-looking, statement that gives a positive lower-bound for $|\sqrtt{2}-p/q|$ for any integers $p$ and $q$, as a function of $q$. This is a special case of Liouville’sinequality, and the idea is to compare an upper bound for $|f(p/q)|=f(\sqrt{2}-f(p/q)$, where$f(x)=x^2-2$, which comes from the mean-value theorem and involves $|\sqrt{2}-p/q|$, with alower bound coming from the fact that $q^2f(p/q)$ is an integer, which is non-zero… because$\sqrt{2}$ is not rational.

Klas Markström Says: March 28, 2010 at 8:37 pm | ReplyThis is certainly biased due to my own way of thinking about things but I have found thatproofs by contradiction are often another way of stating a construction algorithm for an objectwith given properties . E.g. the proof by contradiction for the infinitude of primes can be turnedinto an algorithm which constructs an infinite list of primes. The proof of the four colour theoremis another proof by contradiction, which can be turned around and made into a polynomial timealgorithm for constructing a four-colouring of a planar graph.

Mark Bennet Says: March 28, 2010 at 10:40 pmThe Four Colour Theorem is an interesting example – it is necessary to prove that thealgorithm for four-colouring works for all planar graphs, and this seems to me to be done bycontradiction? [Any planar graph must contain one of a finite list of special subgraphs, findsuch a subgraph and you can reduce your problem to a smaller case. And to prove itcontains one of the subgraphs - assume it doesn't and establish a contradiction]

Fermat’s method of descent can be structured to use contradiction in a similar way to manyapproaches to 4CT – by assuming a minimal counterexample and then showing that theremust be a smaller one – contradiction. There are alternate proofs of results proved by thismethod – but now I’ve been tuned to see contradiction where I didn’t expect it, I can’tconfidently cite one where the alternative is definitely positive.

Sobre las demostraciones por contradicción Says: March 28, 2010 at 10:49 pm | Reply[...] publicado en su blog una reflexión sobre las demostraciones por contradicción. Se losrecomiendo: When is proof by contradiction necessary? Clase: Matemáticas | Etiqueta: [...]

Arnaud Spiwack Says: March 29, 2010 at 12:21 am | ReplyI’m not quite sure, but when I read the sentence: “How can one statement be stronger than ananother, equivalent, statement?”. It seem to me that you want to know more about constructivemathematics.

Also it seems that by “proof by contradiction” you mean two things that are logically rather

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Also it seems that by “proof by contradiction” you mean two things that are logically ratherdifferent. First that too prove that ¬a, it suffices to prove a contradiction while assuming a(dubbed “negation introduction”). The second is that ¬¬a → a (or ¬(¬q→¬p)→(p→q),equivalently) (dubbed “double-negation elimination”). Now, the latter is not a valid principle inconstructive mathematics (it is equivalent to the principle of excluded middle). This results inhaving a way to prove negated formula, but no real way to use them to prove stuff (apart fromcontradiction, basically). Consequently definition formulated as negative are quite weak,compared to a classically equivalent definition that would be “all positive”.

Now that means that constructive mathematicians avoid negative definitions as much aspossible. I would recommend reading Errett Bishop’s book : Foundations of constructive analysis— which is, unfortunately out of print, and rather hard to find second-hand. He develops arather significant amount of analysis without ever resorting to a proof by contradiction of thesecond kind, and little, if any, of the first kind.

gowers Says: March 29, 2010 at 8:27 amUndoubtedly whether or not a proof is constructive is closely related to whether or not itrelies on contradiction in an essential way. But my question is a less formal one: I want toknow how an ordinary mathematician who doesn’t mind non-constructive proofs can look ata statement and recognise that a proof by contradiction is more or less forced. It sometimeshappens that one proves a result by contradiction and then realizes that it is quite easy toprove all the contrapositives of the steps. Is there some way of recognising in advance thatthis is going to be the case?

Even at the informal level, it may well be that thinking about whether one can provesomething constructively will be relevant, as Terry showed in his comments above.

I can see though that it’s quite likely that my informal question cannot be made formal andanswered except in more or less the way you indicate.

Arnaud Spiwack Says: March 29, 2010 at 11:12 amWell, I doubt it can be made formal, because different people might put the limit of what isreasonably “forced” to be proved by contradiction. My guess was something like: maybereading some constructive-intensive mathematics can help one make up his mind. Especiallyregarding whether this or that definition should contain a negation.

Andrej Bauer Says: March 29, 2010 at 4:05 pmI think what Arnaud is saying is that you are calling two things “proof by contradiction”when only one of them deserves this name. If you are trying to prove a negation of astatement then what you call “proof by contradiction” is unavoidable in a certain sense(there are meta-theorems in logic about this). If you are proving a statement which itself isnot a negation by assuming its negation and getting to a contradiction, then very likely youcan find a direct proof. For example, your Bolzano-Weierstrass example has a direct proof.It’s to long to be posted here, but I wrote a blog post which explains all this more carefullyand contains the proof at http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/.

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gowers Says: March 29, 2010 at 5:21 pmI’ve just read your blog post and found it very interesting. And I perfectly fit your paradigmof somebody who was unaware of the difference between proof by negation and proof bycontradiction. I’ll try to heighten my awareness from now on. (In fact, your analysis of why Iwrote what I wrote was spot on all the way through: I’m so used to cancelling out doublenegations that I didn’t really appreciate the significance of what Arnaud was saying.)

Two small further remarks. The constructive proof you gave of the boundedness ofcontinuous functions on [0,1] was essentially the same as one that my colleague came upwith, and the argument in my blog post was a different argument that I produced inresponse.

Finally, with reference to your comment about countable choice, one could obtain thesequence by imitating the proof of the intermediate value theorem: for each one wants to findsuch that , so one lets be the set of all such numbers and takes to be the infimum of ; one thenchecks, using the continuity of , that .

Andrej Bauer Says: March 29, 2010 at 5:39 pmYour suggestion for eliminating countable choice works very well classically.Intutitionistically there is trouble (how do you know these infima exist?) so people whousually worry about choice (namely constructive mathematicians) don’t think of such tricks.Anyhow, I think my blog was not explicit enough, so I’ll reiterate my answer to yourquestion: proofs of negation cannot be eliminated, except via previously proved lemmaswhich eventually reduce to a proof of negation. Proofs by contradiction can almost always beeliminated (in analysis), except for some well-known theorems that are equivalent to (weakforms of) excluded middle.

Kenny Easwaran Says: April 10, 2010 at 11:15 pmWhen I started reading the post I was going to make a point similar to Andrej Bauer, thatwhen a statement begins with a negation, then negation-introduction is going to benecessary.

But in the original post, I think Tim Gowers raises some issues that make thischaracterization a bit problematic – whether or not the formal version of an informalstatement begins with a negation depends on how you formalize it. Statements with strictinequalities look like positive statements, but are often equivalent to negated equations. Isuppose constructivists/intuitionists (I’m not always clear on the distinction) avoid this worryby allowing that and are not equivalent statements. So a classical mathematician can't justuse this particular distinction.

I think instead the phenomenon of interest has to be about the informal proofs (whichmathematicians actually use), and not the formalized counterparts (which logicians study).There's something displeasing about using proofs by contradiction in certain contexts, butoften it can only be eliminated by using some lemma that hides it, or by mutilating the proofin some worse way. I'm not certain if there's any way to characterize the set of statements

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whose proofs will be like this. (I suppose in some sense, this problem seems like it might bestrictly harder than characterizing the set of statements that are provable at all, which is ofcourse impossible given the formalized notions of "characterizing" and "provable".)

Zed Norwood Says: October 12, 2011 at 4:14 pmAndrej: Would you mind indicating what meta-theorems you’re referring to in your firstreply to this comment?

Sorry to resurrect an old conversation.

andrejbauer Says: October 12, 2011 at 4:52 pm@Zed: what I have in mind is the theorem which states that in the intuitionistic propositionalcalculus proofs can be converted to normal forms (eliminations followed by introductions).You can understand a normal form proof as “optimal and direct” in the sense that it has nounecessary steps and does not use any lemmas. Because “not A” is the same thing as “Aimplies false”, the normal form proof of a negation is of the form “assume A then provefalse”.

porton Says: March 29, 2010 at 12:24 am | ReplyYou ask: “How can one statement be stronger than an another, equivalent, statement?”

Consider an algorithm (with or without the requirement that it terminates on every input) whichproves a class of mathematical theorems.

The we can define the statement A to be stronger than B if (A=>B) can be proved by thisalgorithm.

Sam Says: March 29, 2010 at 12:05 pm | ReplyI think the reason why it is so difficult for a mathematician to recognise a priori that a proof bycontradiction is forced, is that it is even difficult a posteriori, after writing down the proof, torecognise whether or not a proof “really”, “essentially” is a proof by contradiction.

So I think a question that needs to be answered prior to the one you are posing here would be:what are the conditions a proof has to satisfy in order for it to be called “a proof essentially bycontradiction”?

S Says: March 29, 2010 at 2:00 pm | ReplyThe approach I take is usually “Which direction has more information?”, or rather, “Whichapproach gives me more to work with?”.

Proof of negation and proof by contradiction « Mathematics and Computation Says: March 29, 2010 at 4:00 pm | Reply[...] have been meaning to write for a while. It was finally prompted by Timothy Gowers’s blogpost “When is proof by contradiction necessary?” in which everything seems to be called “proofby [...]

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DC Says: March 30, 2010 at 3:47 am | ReplyIgnoring formal mathematics, this is a thorny pedagogical issue. We develop intuition for howproofs of statements of various “kinds” will “go,” and want to transmit it to students— butmuch of what we experience as “intuition” is really hindsight.

The heart of any tip of the form “If your theorem is like [blank] then try a proof bycontradiction, and otherwise don’t” is not the blank, but the word “theorem.” We ask studentsto find proofs of statements that are known to be true— and known to have proofs using whatwe have taught them. This is often more significant than anything specific to the phrasing of astatement being proved. Surely, sometimes we deliberately phrase a statement so that it conveysinformation about a proof (e.g. “prove that the polynomial x^2 – 2 has no rational roots” maysuggest a classroom theorem), but in general, this information is not explicit— making therelevant connections is part of the student’s job. Make enough of these connections, and youdevelop intuition… or is that hindsight?

What does your intuition say about a proof of “the Euler-Mascheroni constant is irrational”? Justto speak of a proof, we need to know more than we know.

This seems bound up with the meta-mathematical issue of “knowing something is true” versus“having a proof”. These are certainly different concepts in the classroom: I can know Liouville’stheorem is true, but not remember any proof during an exam. Outside the classroom it is harderto say what the difference is. It is the basis of a very unfunny math joke.

Q: What’s the best way to start a proof of X?A: Know that X is true.

Jonathan Vos Post Says: March 31, 2010 at 6:51 pm | ReplyI remembered Feynman telling me the last quote here, and googling it I saw this nice page:

http://infoproc.blogspot.com/2006/12/keynes-and-planck.htmlFound in the comments on Economist’s View:

“Professor Planck, of Berlin, the famous originator of the Quantum Theory, once remarked tome that in early life he had thought of studying economics, but had found it too difficult!Professor Planck could easily master the whole corpus of mathematical economics in a few days.He did not mean that! But the amalgam of logic and intuition and the wide knowledge of facts,most of which are not precise, which is required for economic interpretation in its highest form is,quite truly, overwhelmingly difficult for those whose gift mainly consists in the power to imagineand pursue to their furthest points the implications and prior conditions of comparatively simplefacts which are known with a high degree of precision.” (Keynes, Essays in Biography 1951158n)

It’s true: powerful mathematical minds are not necessarily comfortable with the messiness of thereal world. This observation might be applied as well to string (or more formal or mathematical)theorists vs theorists who are more data- or intuition-driven (often called phenomenologists, in aterrible use of terminology). Of course, some people (like Feynman, although he wasn’t verymathematical by today’s standards) are good at everything…

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“We know a lot more than we can prove” — Feynman.

Posted by Steve Hsu

John Says: April 1, 2010 at 4:37 am | ReplyYour final comments resonate with some of my own informal observations – it depends on whatkind of objects and statements you are working from.

For example, like I think you were saying, it seems that you can `bundle’ some of the non-constructive or inherently `negative’ aspects into definitions and give direct proofs based on thesedefinitions, while different definitions might require a `negative’ proof, given the definitions youare using.

In practice, I sometimes like to think about contradiction as about consistency and elimination ofalternatives, usually arising when you think, `this MUST be true, because it can’t not be’ vs `thisIS true because I know this’. Whenever you feel like saying `MUST’ you might think aboutcontradiction.

John Says: April 1, 2010 at 4:43 amOr `MUST not be true because if it was…’

gowers Says: April 1, 2010 at 11:31 amI like that “must” test. I’ll try it out on a few examples and see whether it seems to work.

fan Says: April 1, 2010 at 10:26 pm | ReplyIn graph theory the easiest way to get your induction right is a proof by contradiction. Ratherthan trying to figure out in what ways small graphs can extend to bigger ones (messy and error-prone), you grab a “smallest counterexample”.

Anonymous Says: April 2, 2010 at 1:29 am | ReplyAs a non-mathematician can I ask why proof by contradiction is so disliked? I get that it wouldseem to be inherently non-constructive and that might annoy some people. I suppose if youarrive at a contradiction you know that something is wrong – but is it necessarily the thing youwere trying to disprove, or is the the problem with the method of proof/disproof? Perhapsequally you might be suspicious of a direct constructive proof as it could be an accident offlawed logic or other false assumptions?

Getting to know your math tools. « Math Society the club Says: April 2, 2010 at 3:28 pm | Reply[...] Getting to know your math tools. Timothy Gowers is a Fields Medal winner who writes ablog about math. He has written a post asking “When is proof by contradiction necessary?” [...]

Paulo Oliva Says: April 14, 2010 at 10:32 am | Reply

Two points. First, in [R. L. Goodstein, Proof by reductio ad absurdum. The Mathematical

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Two points. First, in [R. L. Goodstein, Proof by reductio ad absurdum. The MathematicalGazette, Vol 32, No. 300, pp 198-204, 1948] the same issue of this blog is also discussed, and aproof that square root of 2 is irrational is given without using proof by contraction. The article isshort and easy to read. Regarding the question “How can an ordinary mathematician look at astatement and recognise that a proof by contradiction is more or less forced?”, I would say, givenmy background in proof theory, that if the theorem states that an object x has property P(x),where P(x) is computer-checkable then proof-by-contradiction can normally be avoided. Forinstance, existence of infinitely many prime (object x in this case is a number, and P(x) says thatx is prime and bigger than n), or square root of 2 being irrational (object x is an epsilon, and P(x)says that square root of 2 is away from p/q by that epsilon). In these two cases n and p/q arearbitrary parameters, which can be thought of as inputs. If, on the other hand, the theoremstates that an object x with property P(x) exists but P(x) cannot be checked “computationally”,then you most certainly will only be able to prove the theorem via a proof by contradiction. Forinstance, in the case of the Bolzano-Weierstrass theorem, the object x is a real number, and theproperty P(x) saying that x is an accumulation point would require someone to check infinitelymany possibilities. Another example is Brouwer’s fixed point theorem on [0,1] for instance, inthis case x is a real in [0,1] and P(x) says that fx = x, which is an equality between real numbersthat cannot be checked effectively by computer (think of how you would check two realnumbers are the same, you might check digit by digit and never know). This is of course aninformal answer to an informal question.

Anonymous Says: April 19, 2010 at 4:29 am | ReplyI agree with Andrej and Paulo generally, the question you are asking is a natural question infoundations of mathematics and logic, and there is a difference between “Reductio adabsurdum” and introduction rule for negation. I just want to add three minor points.

About Andrej’s comment, in Intuitionistic logic, one is mainly concerned with true statement,therefore if you want to show something is false the only thing we can do is to show that byassuming it we can derive contradiction, i.e. it does not study how to refute a statement directly.There are statements that we can refute directly without going through this, example $0 \neq1$. Similarly, to show that $\forall x, \varphi(x)$ is false we can give a specific $x$ such that$\varphi(x)$ is false. We don’t need to derive a contradiction form it. Intuitionistic logic is notsymmetric with respect to true and false statements.

About Paulo’s comment, his trick work in many cases because computable mathematics is amodel of constructive mathematics. But it is not faithful with respect to constructive reasoning,i.e. there are statements that hold in computable mathematics which can not be provenconstructively. As a result, a statement can be both true in classical mathematics andcomputable mathematics, and still you may need to use proof by contradiction to derive it.

Finally, I want to emphasize the seemingly obvious but not trivial point that to prove astatement using proof by contradiction can be much easier than proving it without it.

If you are interested, I would suggest taking a look at Beeson’s book:Foundations of Constructive Mathematics: Metamathematical Studies, Springer,Berlin/Heidelberg/New York, 1985.It is a nice book, and it is not an ideological one so it should make sense to classicalmathematicians.

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Andrej Bauer Says: April 19, 2010 at 5:51 amDear Anonymous, your examples of direct refutations ($0 \neq 1$ and giving acounterexample to a universal statement) work intuitionistically just as well as clasically, andare frequently used in intuitionistic mathematics, just as in classical mathematics. If you boilthem down to their formal proofs, you will discover however, that they (a) either rely on anaxiom which has the form of a negation, such as $0 \neq 1$ in the theory of fields, or $n+1\neq 0$ in Peano arithmetic, or (b) your “direct” method of proving a negation relies on alemma of the form $A \implies \lnot B$ (for example $(\exists x, \lnot \phi) \implies \lnot\forall x. \phi)$) whose proof then contains a proof of negation.

Your view of truth in intuitionistic logic is somewhat strange. What is “in intuitonistic logicone is concerned mainly with true statements” supposed to mean? I thought allmathematicians, no matter what party they belong to, are mostly concerned with truestatements. The trouble is, we don’t know which one are true

Anyhow I think this question is not about intuitionistic logic. It’s just that having practicewith intuitinstic logic helps one distinguish various logical forms which classicalmathematicians autoamatically view as “the same”.

Anonymous Says: April 20, 2010 at 4:00 amHi Andrej,

In classical logic we have symmetry between truth and falsity, so the point I made does noteffect it. Stating something is true is the same as stating its (classical) negation is false. Thissymmetry is broken in intuitionistic logic, if we need constructions for proving a statement istrue, we can also have constructions to show that a statement is false. Of course if we don’ttreat falsity similar to truth and formalize them as intuitionistic negation we will end up withwhat you said. But similar to the situation with true statements, we can have directobservations showing that a statement is false. To show that 2^2=5 is false, we can justcompute 2^2 and get 4 and compare the normal forms of them to conclude that thisstatement is false, where as in intuitionism since falsity is replaced with intuitionisticnegation, what we end up is that assuming 2^2=5 we derive 0=1 (and either define $\bot$ tobe just $0=1$ or have an axiom that states $0=1 implies \bot$) and conclude with $\lnot2^2=5$. Even a computer does not need to find a proof of contradiction from $2^2=5$ toclaim it is a false statement. My point is there is a more natural way to establish that $2^2=5$is false. This is a toy example but this holds in general, in place of coming up with aconstruction for $\lnot \varphi$ we can show directly $\varphi$ is false by giving aconstruction showing its falsity. “A white raven” is enough to show the universal statement“every raven is black” is false, there is no need to assume it and derive a contradiction. Butfor this to make sense one has to distinguish between falsity and intuitionistic negation,similar to the situation in linear logic.

As far as I remember, the negation and falsity in intuitionism was considered problematiceven by some pioneers (Gilevenko?).

Andrej Bauer Says: April 20, 2010 at 6:51 am | Reply

I will reiterate, but then I hope we can close discussion because intuitionistic mathematics is not

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I will reiterate, but then I hope we can close discussion because intuitionistic mathematics is notrelevant to this post. You give examples of how a classical mathematician proves negations andthat equations such as $2^2 = 5$ do not hold. But all of your examples and methods of proof arebadly chosen because precisely the same methods work for the same reasons intuitionistically. Iwrite papers in constructive mathematics, and I assure you we do not derive false every time wewant to prove a negation. In fact, good math is written in such a way that it is largely irrelevantwhether it is classical or intuitionistic.

If on the other hand we speak about formal proofs (without cut and in “normal form” as far asthat is possible for classical math, so we cannot hide things inside lemmas), then proofs ofnegated statements will generally end with an introduction rule for negation, both in classicaland intuitionistic mathematics. You speak of a symmetry in classical mathematics between truthand falsity, but even that symmetry has to be proved somehow, does it not? The rules ofinference for negation (in natural deduction style) are the same classically and intuitionistically.The symmetry you speak of is proved for classical logic from the law of excluded middle. But therules of inference come first, and they do not include the symmetry. (You can build thesymmetry into the logic if you use classical sequent calculus, but that’s not how mathematicianswrite proofs.)

Bivek Says: April 30, 2010 at 6:22 am | ReplyHi there,

Why we use this proving technique?

General Application?

Thank you.

Some Mathematical Gifts « Gödel’s Lost Letter and P=NP Says: December 20, 2010 at 2:40 pm | Reply[...] that does not use “proof by contradiction.” See also Tim Gower’s interesting discussion onthis and [...]

Mike Says: January 26, 2012 at 4:55 pm | ReplyI was just wondering, if you don’t get a contradiction when using a proof by contradiction, whatdoes it mean?I’m asking this question because in the books I’ve read so far, usually a counterexample wasused to show that a proposition is false. But can we also use the proof by contradiction and notget a contradiction to show that the proposition is in fact false?Thanks.

Anonymous Says: February 29, 2012 at 4:27 am | ReplyAccording to his biography packet (available on Wikipedia) the late, great statistician DavidBlackwell was able to provide a positive approach to the proof of the irrationality of the squareroot of 2. Unfortunately, no details were provided.

Quora Says:

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September 5, 2012 at 3:26 am | ReplyWhat is the most beautiful theorem proof, and why?…

Stephen: To answer “What specific advantages are there to viewing this as a constructive proof”,one answer is that it will no longer be a false statement, historically speaking (i.e., it will really becloser to “Euclid’s proof” rather than a lat…

How proof by contradiction differs of direct proof | Victor Porton's Math Blog Says: February 26, 2014 at 4:03 pm | Reply[…] famous mathematician Timoty Gowers asked this question: What is the difference betweendirect proofs and proofs by […]

Friday, February 28 | Discrete Mathematics Says: February 28, 2014 at 5:09 am | Reply[…] Check out the post on Gowers’ blog Is contradiction necessary? […]

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