I'm reading Manfredo Do Carmo's differential geometry book. In section 1-7, he discusses the "Isoperimetric Inequality" which is related to the question of what 2-dimensional shape maximizes the enclosed area for a closed curve of constant length. He mentions that

A satisfactory proof of the fact that the circle is a solution to the isoperimetric problem took, however, a long time to appear. The main reason seems to be that the earliest proofs assumed that a solution should exist. It was only in 1870 that K. Weierstrass pointed out that many similar questions did not have solutions.

This line of reasoning would suggest that assuming the existence of a solution might lead to a contradiction (such as an apparent solution that is not in fact valid). Is this actually a problem?

Are there any problems that produce invalid solutions under the (flawed) assumption that a solution exists at all? If so, what is an example and how does it differ from the statement of the isoperimetric problem?

YuiTo Cheng
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    This is at least superficially similar to an inductive argument lacking a valid base case. – Dan Brumleve Dec 28 '17 at 00:14
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    Consider the Fibonacci sequence defined by $F_1=F_2=1, F_{n+2}=F_{n+1}+F_n \mid n\ge1\,$. Assuming the sequence has a limit $F\,$, the recurrence can be passed to the limit giving $F=F+F \implies F=0\,$. But the sequence is increasing from $1$ up, thus a contradiction. – dxiv Dec 28 '17 at 00:21
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    I thought **Proof By Contradiction** was standard in Mathematics by now, though some would question the approach. – Dehbop Dec 28 '17 at 06:40
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    See also [this related question on Math Educators SE.](https://matheducators.stackexchange.com/questions/4351/unique-candidate-that-fails) – Ilmari Karonen Dec 28 '17 at 08:14
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    More interesting: could there be any example of a problem *with* an existing solution, where assuming the existence of a solution (too early in the process) leads to wrong/incomplete solution? (My guess is no.) – Neinstein Dec 28 '17 at 09:57
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    @Neinstein Hm... In my experience, proofs that assume the existence of a solution proceed by deriving conditions on the solution that most of the potential solution space does not satisfy, after which the remaining part can be checked "manually". As long as that's the case, what you're suggesting would require that, from the assumed existence of a solution, you derive some condition which is _not_ satisfied by the actual solution. If you're being careful (and the problem is logically consistent), that probably shouldn't happen. This might make for an interesting followup question though. – David Z Dec 28 '17 at 20:54
  • @Neinstein: The Proof By Contradiction approach works even if the solution space exists or not. It just proves whether if the assumed solution is **NOT** in the solution space, if it exists. (edited) – Dehbop Dec 29 '17 at 14:31
  • @Neinstein: You might be able to prove "assuming there is a solution, the solution is either X or Y" when the solution is X. – gnasher729 Dec 29 '17 at 16:15
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    Yet what @Neinstein describes seems to be similar to what Do Carmo actually says in the quoted claim. He seems to assert that the *correct* assumption that the Isoperimetric Inequality has a solution interfered with proving that a circle in fact is that solution. I find this much more remarkable than the OP's actual question. – John Bollinger Dec 29 '17 at 20:25
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    @JohnBollinger All Do Carmo is saying is that the proof was incomplete, but no one really realized until Weierstrass pointed it out. Presumably, the proof *was* considered satisfactory to most up until that point. – Derek Elkins left SE Dec 29 '17 at 22:41
  • See @KCd ‘s answer [here](https://math.stackexchange.com/questions/348198/best-fake-proofs-a-m-se-april-fools-day-collection/348230#348230). Didn’t want to plagiarise. –  Dec 30 '17 at 12:09
  • @gnasher729 But 'it can be X or Y' doesn't contradict with 'it's X', does it? The only contradiction could be between 'Y is a solution' and 'Y is not a solution', but that's somewhat equivalent to the no-solution examples. – Neinstein Dec 31 '17 at 02:17
  • Unless one has reason to believe that a solution exists, how may one assume that it exists? That is akin to assuming a theorem is true, without any evidence, for the purpose of proving that it is true. – ThisIsNotAnId Jan 01 '18 at 20:45

12 Answers12


Just the first thing that came to my mind... assume $A=\sum_{n=0}^{\infty}2^n $ exists, it is very easy to find $A $: note $A=1+2\sum_{n=0}^{\infty}2^n =1+2A $, so $A=-1$.

Of course, this is all wrong precisely because $A $ does not exist.

  • I also immediately thought of this procedure (though not your specific example) when reading the question. It is a very common error for students to make. – Paul Sinclair Dec 28 '17 at 02:28
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    $\mathbb Q_2$ would like a word... – Artimis Fowl Dec 28 '17 at 06:31
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    @ArtimisFowl: I didn't know fields can talk. Next they would want voting rights, and the right for proper education. Before you know it, all the universities are full of fields. And then we all go back to being farmers. – Asaf Karagila Dec 28 '17 at 06:58
  • @AsafKaragila nice!! Is $\mathbb{Q}_2$ the same as $\mathbb{Z}_2$ but for rationals? – Shashi Dec 28 '17 at 10:12
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    @Shashi: I don't know what you mean by $\Bbb Z_2$, because some people would use it to denote the $2$-adic integers, and some would use it to denote $\{0,1\}$, the quotient of the integers by the even numbers. If you mean the former, then yes, $\Bbb Q_2$ is the field of $2$-adic numbers, which is also obtained as a completion of $\Bbb Q$ with respect to the $2$-adic valuation. – Asaf Karagila Dec 28 '17 at 10:14
  • @SimplyBeautifulArt I regularly watch Numberphile, it is a great effort to popularise one of the hardest things to get the interest of the layman in. (Mathematics.) Those sums they mention are usually (but not necessarily) derived using Riemann zeta function, while, as per OP's question, I wanted to show something that comes out quickly within the bounds of elementary calculus, once you assume that the limit exists. –  Dec 28 '17 at 15:13
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    I like this answer a lot. Although it doesn't discuss why existence assumptions can cause problems in a proof, the answer is fairly self evident from the example given. I'm not an expert in regularizing divergent series, but my understanding is that there is a sense in which this answer is not only unique but correct. That suggests that there is another implicit assumption in this proof which moves the problem from standard arithmetic into another regime (perhaps through an equivocation of notation). Can you address this concern? Am I just over-thinking it? – Geoffrey Dec 28 '17 at 16:32
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    Why does $A = 1 + 2 \sum_{n=0}^{\infty}(2^n)$? – Brian J Dec 28 '17 at 17:02
  • @Geoffrey First, we are talking about the existence of the solution as a real number, with the ordinary $\varepsilon-\delta $ semantics. I did not explicitly state which theorems I used to justify calculation, but you can fill in the gaps, and they are all similar to "if $\lim_{n\to\infty} a_n$ exists and is equal to A, then this other limit e.g. $\lim_{n\to\infty} 2a_n $ exists and is equal to $2A$"... I never leave the realm of the real calculus. This question is separate from a question "can we change rules to make the limit exist in the first place?". –  Dec 28 '17 at 17:30
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    @$\mathbb Q_2$, see my response to @Geoffrey above ;) –  Dec 28 '17 at 17:32
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    Got it. That makes sense. In the context of real analysis we proved the conditional statement "If it exists, then it is -1." And by assuming it exists we deduced it must be -1. This is clear now. – Geoffrey Dec 28 '17 at 18:53
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    @AsafKaragila How do we go back to being farmers if all our fields are in universities? – Joonas Ilmavirta Dec 28 '17 at 20:06
  • @AsafKaragila or maybe Ramanujan would like a word? – user541686 Dec 28 '17 at 22:15
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    @Joonas: Us, the academics, we become the farmers. – Asaf Karagila Dec 28 '17 at 23:34
  • @Mehrdad: No, he's dead... And Ramanujan summation is its own context, just declaring "oh, a sum which is clearly divergent" is not enough to automatically assume that the right context is Ramanujan summation (or any other). Just like you assume I am writing in English and not in Achblungish (which I just made up) and the syntax coincides with English. – Asaf Karagila Dec 28 '17 at 23:36
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    AsafKaragila: Ah whoops, I forgot $\mathbb Q_2$ was alive. (Also, whoops, I auto-completed the wrong person... meant to ping @ArtimisFowl!) – user541686 Dec 28 '17 at 23:57
  • @Mehrdad: There is no proof that $\Bbb Q_2$ is dead. Ramanujan, on the other hand, well... :) – Asaf Karagila Dec 29 '17 at 09:53
  • @BrianJ: Are you questioning the validity of the required operations, namely ( sum = head + sum of tail ) and ( sum of doubled terms = doubled sum of terms ), or asking the equality was derived? – PJTraill Dec 29 '17 at 18:17
  • @PJTraill I understand how $A= \sum$. It's the second step, "note: $A = 1 + 2 * \sum$" that I don't understand. And maybe there are several steps to get to that point that I'm not considering? – Brian J Dec 29 '17 at 18:56
  • You could also point out that it cannot be right as similarly $ A = 3 + 4A $ whence also $ A = $ — oops, it _is_ the same! So it would be better to point out that $ -1 $ is not the limit of the partial sums (which is the definition you are presumably presuming), as none of them are within $ 1/2 $ of it, an $ ε $ for which there is thus no $ δ $. – PJTraill Dec 29 '17 at 18:58
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    @BrianJ, there is a theorem that says, for convergent series $\sum_{n=0}^{\infty}a_n $ and $k\in\mathbb R $, that $\sum_{n=0}^{\infty}ka_n$ is also convergent and $\sum_{n=0}^{\infty}ka_n=k\sum_{n=0}^{\infty}a_n$. I used that with $k=2$, and with $a_n=2^n $, for example. –  Dec 29 '17 at 19:09
  • @BrianJ: The second step is derived by assuming the two operations in my previous comment are valid on such sums and applying them to the series: $ A $ is $ 1 $ plus the remaining-terms and the remaining terms are double the original terms. We can prove these operations work **if** the series does converge, i.e. the partial sums tend to a limit, which they do not. **N.B.** Some of the other comments are about whether one can work with different definitions of infinite sums or different sorts of number whose partial sums _do_ converge. – PJTraill Dec 29 '17 at 19:09
  • @user8734617 that theorem makes sense to me. It's the $A = 1 + 2A$ that I'm lost at. At this point, I'm thinking I may need to take a step back and review infinite series some more rather than continue to blow up this comment thread. – Brian J Dec 29 '17 at 19:17
  • @PJTraill thanks to both you and user8734617 for trying to help me understand :) – Brian J Dec 29 '17 at 19:18
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    wow. -1 is a perfectly fine answer. that's what computers give. if you had a binary register with a decimal point, and no bits set below decimal point, and no bits unset above the decimal point; then you do get -1 as the answer. – Rob Dec 30 '17 at 08:08
  • i don't understand how it's a contradiction. continuing on with -1 as the computer answer (it is!). 2s complement arithmetic gives the same answers as normal arithmetic if you have registers in which you can always add more bits on demand. it is QUITE well defined what a string of one bits to the right of the decimal point means. a loop that runs "forever" adding more 2^n terms will never actually add in a last term, so you can't get contradictions such as adding a term and getting a lower number. iterative definitions and recursive ones are not making identical claims. sum is iterative. – Rob Dec 30 '17 at 08:21
  • is it not the case that putting $\inf$ at the top of the sum is actually ambiguous? the whole reason for the existence of limit notation is that you can't just plug in the infinite value, but can only state where finite terms are headed - usually one answer. for sin(x), it's headed for anything from -1 to 1. (which isn't the same thing as "doesn't exist... you have a well defined set of values that the value must be at after an unbounded amount of time. it's explicit that it will never be at 5. to do otherwise is to throw away information. if you throw away information, it can't be equal) – Rob Dec 30 '17 at 08:47
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    @Rob, the meaning of $\sum_{n=0}^{\infty}2^n $, in real calculus, and in this post, is $\lim_{M\to\infty}\sum_{n=0}^{M}2^n $, which itself is, provably unique (if it exists), number $A\in\mathbb R$ such that $(\forall\varepsilon\gt 0)(\exists M_0\in\mathbb N)(\forall M\gt M_0, M\in\mathbb N)(|A-\sum_{n=0}^M 2^n|\lt\varepsilon) $. There I just rewrote it without using the symbol $\infty $. Now recall all the rules and theorems in calculus that let us manipulate those expressions. No ambiguity there, just a bunch of calculus theorems justifying it! –  Dec 30 '17 at 09:24
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    Saying "but this works in $\mathbb Q_2$", or "but it is correct on computers with a finite register size $ is like moving goalposts: you reinvent the meaning of the sum, no wonder you may get different conclusions. –  Dec 30 '17 at 09:42
  • presume that we just start with a recursive equation: $A = 1 + 2A$. $A$ is $-1$. If we substitute equals for equals once: $A = 1 + 2(1 + 2A)$, then we get the expansion. There is no iterative summation here at all, as it's just a recursive definition. $A = 1 + 2(1 + 2(1 + 2A))$. This recursive equation solves for -1 and behaves like a recursively defined summation. This is legitimate algebraic manipulation. Where is the contradiction? – Rob Dec 31 '17 at 00:45
  • another example... say that you perform a bunch of algebra starting from a bunch of axioms that are agreed to be true. but then you derive $0 = 12$. That's not a contradiction. That's a condition that we are working in mod 12 arithmetic. It's like if you encounter a $a/b$ somewhere, then that statement implies "b != 0". – Rob Dec 31 '17 at 01:04
  • @Rob: $$0<1<1+2<1+2+4<\cdots\leq-1.$$ Unless you want to not assume that the real numbers are ordered, or that a monotone limit of positive numbers is not positive. – Martin Argerami Dec 31 '17 at 02:17
  • the number of terms to (iteratively) expand to reach -1 doesn't exist, so -1 is not an item in the sequence. But, $A = 1 + 2A$ still legitimately solves for both -1, and also this sequence. Taking a limit is not adding enough terms to reach -1 and equal the recurrence. No space for it, but there is a way to make the distinction ($n$ is term to start from, $t$ is number of expansions): $A = A[0,t]$, $ A[n,t] = \underbrace { \left \{ \sum_{x}^{0 \leq x < t}{2^{n+x}} \right \} + A[n+t,0] }_{1 + 2 + 4 + 8 + \ldots = -1} $ The diff between the limit and the recurrence is $A[L,0]$ – Rob Dec 31 '17 at 02:49
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    @BrianJ Perhaps you are looking for $$ A = 1 + \sum_{n=1}^\infty 2^n = 1 + \sum_{n=0}^\infty 2^{n+1} = 1 + 2\sum_{n=0}^\infty 2^{n} = 1+2A $$ – zahbaz Dec 31 '17 at 08:45

Here is a "joke" due to Perron showing that assuming the existence of a solution is not always a very good idea:

Theorem. $1$ is the largest positive integer.

Proof. For any integer that is not $1$, there is a method to obtain a larger number (namely, taking the square). Therefore $1$ is the largest integer. $\square$

A good source is V. Blåsjö, The isoperimetric problem, Amer. Math. Monthly 112 (2005), 526-566.

John B
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    https://www.maa.org/programs/maa-awards/writing-awards/the-evolution-of-the-isoperimetric-problem – Will Jagy Dec 28 '17 at 01:13
  • Is this wrong because it assumes the existence of a largest positive integer? – Lug Gian Dec 28 '17 at 01:33
  • @LugGian Kind of. The proof goes "if not A, then not B. Hence A", where A is "the integer is 1" and B is "the integer is the largest integer". This jump is simply unjustified. – Patrick Stevens Dec 28 '17 at 08:16
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    I don't like this example at all, it is too convoluted. One could just as well say $2$ is the biggest integer, because $2 > 1$ and for every number different than $2$, you can find something bigger by squaring it. This only falls in the category of the question if you ignore several, very intuitive things we know about integers and follow exactly the line of reasoning given above. – Ant Dec 28 '17 at 09:49
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    @Ant It's not true that for every number different than $2$, you can find something bigger by squaring it, not even if you mean every integer. The counterexamples are $0$ and $1$. That's the whole point why this fake proof uses $1$: it needs an integer for which squaring does not produce a larger number. But I do think even as a fake proof, it's wrong to not mention $0$. – hvd Dec 28 '17 at 11:02
  • @hvd It is the same thing. We know $2 > 1$ and $2 > 0$, so clearly $1$ and $0$ are out of the game. For every other integer different than $2$, we can find a larger integer by squaring it. Hence $2$ is the largest integer – Ant Dec 28 '17 at 11:35
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    @Ant: Any example of a fallacious proof will be a fallacious proof. But it's the examples where the fallacy is obvious that makes it easy to work out why the proof form is fallacious. Note, incidentally, that "the proof uses an argument that is only strong enough to rule out everything but $1$" is not a fallacy. –  Dec 28 '17 at 15:28
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    I appreciate this answer and the interesting article that goes along with it. However, I think that this isn't a perfect example since it assumes a specific solution which is then proved by a weak proof method rather than assuming the existence of an unknown solution and deducing a specific (but wrong) answer from that assumption. – Geoffrey Dec 28 '17 at 15:54
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    @Geoffrey Precisely, that's what "joke" means. But I completely disagree with you since wrong will always be wrong, there is no way around it. In this example (by the great Perron) only all of us see what is wrong immediately. From this point of view I think that it is very unlikely to get a better example (while it is trivial to give hundreds of examples). – John B Dec 28 '17 at 15:57
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    @Geoffrey: This **is** a perfect example. If a largest integer is assumed to exist, it is easily proven that this integer is 1 (which, in fact, is false) - just like the OP asked. – Meni Rosenfeld Dec 28 '17 at 18:08
  • Oh, I see, I misunderstood the argument the first time I read it. I was comparing it too closely to the verification-style "proof" that the circle has maximal area it is presented next to in the article. Still, it doesn't seem a strong argument since the correct conclusion would seem to be that squaring an integer *doesn't* always produce a larger integer (the negation of the second premise). But at that point the proof of 1 being the largest integer required more than an existence assumption: it also needed a false premise. – Geoffrey Dec 28 '17 at 18:43
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    @Geoffrey: You still misunderstand; the given proof already handles that correctly. It only invokes the squaring argument for positive integers that aren't 1 -- and it is valid on that domain. Thus, it has *correctly* been proven that every positive integer that isn't 1 cannot be a largest positive integer, and thus 1 is the only remaining candidate for the largest positive integer. (and so we conclude that it actually is the largest if we make the existence assumption) –  Dec 28 '17 at 21:50
  • @Hurkyl Okay, yes, I see. Since you exclude all other positive integers and assume there must be a greatest positive integer, that integer must be 1. Obviously, if it were your only proof method, it might seem convincing. – Geoffrey Dec 28 '17 at 23:08
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    To appreciate this "joke", note that in the 19th century some mathematicians claimed they had proved the isoperimetric inequality (that the largest area for a given fixed circumference is obtained by the circle) in the following way: For each Jordan curve which is ___not___ the circle, they devised a technique that modified the curve slightly to "improve" the area it enclosed. In their opinion, it was then proved that the circle was the solution. But they had only proved that _if_ a solution existed, then the circle would be it. Number squaring proves: If a largest number exists, it is $1$. – Jeppe Stig Nielsen Dec 29 '17 at 10:15
  • @JeppeStigNielsen this is the context in which the joke appears in the cited article. I think this is a great example precisely because of the perfect analogy to Steiner's incomplete proofs. – Sasho Nikolov Dec 29 '17 at 21:34
  • But even for 1 there are methods to obtain a larger integer, namely 1+1. I realise I'm missing the point here, would someone explain? – Questions about math Jan 05 '18 at 21:50

The danger Weierstrass points out is similar to the issue that comes up in the following problem:

What is the minimum value of $x^3-3x$ on $\mathbb R$?

You can easily show with calculus that the only local minimum of this function is $x=1$. Therefore, if the function has a minimum, it must be at $x=1$. However, of course, this function has no minimum, so this reasoning has failed.

In the context of the isoperimetric inequality, the fear would be that there could be shapes with the same perimeter as a circle, but greater area - but perhaps as we add more area, the shapes get increasingly weird and we can't somehow take a limit to get a shape of maximum area.

Really, this should be thought of more as a continuity and compactness issue than an existence issue - we are looking for some way to control the behavior of a function (the area) on a set (the shapes of a fixed perimeter) and know that the circle is the only candidate for a minimum. We would like to say that this implies that every other such shape has less area than the circle, but this requires that we know something about the function and its domain to rule out possibilities like the $x^3-3x$ example.

Milo Brandt
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  • Ah, that's interesting. I see now why it would be important to prove existence before proving its actually value. It reminds me of that old line about how the only things worth trying to prove are the ones that seem obvious to begin with. – Geoffrey Dec 30 '17 at 02:36
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    @Geoffrey It's not necessary to prove existence first. It's often [well, sometimes at least] a good strategy to first prove that if a solution (to whatever problem one is investigating) exists, it must have this or that property - in good cases, reduce it to a small finite number of explicitly known options - and only afterwards check existence. The derived properties can greatly help proving or disproving existence of a solution. – Daniel Fischer Dec 30 '17 at 22:07

Let's find the maximum of the the function $f(x) = x^2$.

Assume it does have a maximum. Since $f$ is differentiable, the maximum must happen at some $x$ where $f'$ vanishes.

$x=0$ is the only solution to the equation $f'(x) = 0$.

Therefore, $f(0) = 0$ is the maximum value of $f$.

  • f' vanishing doesn't guarantee a maximum. It just tells you that there could be a maximum or minimum at that point. At the same point, f'' < 0 means a maximum and f'' > 0 means a minimum. In this case f'' (0) = 2 and thus a minimum should occur and that's precisely the case. – Whiskeyjack Dec 30 '17 at 13:37
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    @Whiskeyjack: However, if differentiable function has a maximum, it must happen at a critical point. And if differentiable function has a maximum and exactly one critical point, that critical point must be the maximum. –  Dec 30 '17 at 14:04
  • my bad. I see your point now. :) – Whiskeyjack Dec 30 '17 at 16:58

Literally every problem that has no solution is an example. Indeed, consider any problem which has no solution, and assume $x$ is a solution to the problem. Then the problem has no solution, but it also has a solution (namely, $x$). This is a contradiction, and anything follows from a contradiction. In particular, for instance, it follows that $x=1$, or $x=2$, or any other conclusion you would like to reach.

Of course, this is somewhat artificial, and there are more natural examples. But as far as the raw logic is concerned, this is just as valid as a more natural example, and illustrates why assuming something which turns out to be false is always a problem.

(In fact, this argument is essentially the same as Perron's "joke" in John B's answer. You can think of that example as assuming a solution $x$ exists, and then assuming for a contradiction that $x\neq 1$. Since no solution exists but $x$ is a solution, we have a contradiction, and therefore $x=1$.)

Eric Wofsey
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    While this is a worthy point, I think it misses the spirit of the question - namely, whether assuming that a solution exists *before* knowing whether one exists or not would lead to a reasonable-seeming (but wrong) answer. – Geoffrey Dec 28 '17 at 16:17
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    @Geoffrey I think you're right, but I also think that contradiction and the explosion principle that then comes into play has to be discussed in any full discussion of this question. It was my first thought on reading the title. Questions and their answers are meant to be for general readership and to be helpful to an audience wider than only the OP. – Selene Routley Dec 29 '17 at 23:08

Here is one that actually shook the foundations of mathematics. Assume that there is a set of all sets that are not members of themselves. Symbolically, we define the set $R = \{ x : Set(x) \land x \notin x \}$. Then $R \in R \Leftrightarrow R \notin R$, so either classical logic fails for set membership or we have a contradiction!

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Suppose $x=\log(x)$ has some solution $x=a$.

$0<a$ is obviously true. Then $0<\log(a)$ as well (as $a= \log (a)$ was assumed), so $1<a$. Then $1<\log(a)$, so $10<a$, so ...

Numerous contradictions unfold from this point onward.

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Courtesy of NumberPhile:

ASTOUNDING: $1+2+3+4+5+\dots=-1/12$

Sum of Natural Numbers (second proof and extra footage)

One minus one plus one minus one

(These videos are all misleading, and they assume these series can even exist, then proceeding to treat them under algebraic manipulations and rearrangements)

Simply Beautiful Art
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    Please state clearly in your answer that the three links are rubbish. Otherwise it is misleading. – user21820 Dec 28 '17 at 15:16
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    Please link to at least one video debunking these embarrassments. – Did Dec 28 '17 at 15:46
  • But isn't this channel an educational one? Why do they do misleading things? – Ooker Dec 28 '17 at 15:51
  • @Ooker For popularity and a wider audience (that may not understand what is properly happening with things such as the Riemann zeta function) – Simply Beautiful Art Dec 28 '17 at 15:56
  • @Did Hm... I've never watched any of those videos. Have any particular suggestions? – Simply Beautiful Art Dec 28 '17 at 16:01
  • Every channel needs popularity, but why wouldn't this one say that these are faulty? (If they don't say so of course) – Ooker Dec 28 '17 at 16:03
  • @Ooker Beats me $\ddot\frown$ – Simply Beautiful Art Dec 28 '17 at 16:07
  • @SimplyBeautifulArt https://www.youtube.com/watch?v=jcKRGpMiVTw (admire the classy restraint at 1:40...). – Did Dec 28 '17 at 16:11
  • @Did Haha, thanks. I'll watch the video soon and probably add it in. – Simply Beautiful Art Dec 28 '17 at 16:13
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    Those results are relevant and used in (quantum?) physics. Under strict mathematical rigour, they are divergent series, and should not have a result, though. [ref1](http://physicstoday.scitation.org/do/10.1063/PT.5.8029/full/) - and [physics.se](https://physics.stackexchange.com/questions/3096/what-is-the-relation-between-renormalization-in-physics-and-divergent-series-in) also [math.se](https://math.stackexchange.com/a/39811/142244) – Mindwin Dec 28 '17 at 16:29
  • @Did: There isn't one. There's a screw lose big enough that Wikipedia thinks is true too. https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF – Joshua Dec 29 '17 at 00:19
  • @Ooker: SBA is right that they just want to be popular. It is extremely clear that they do not understand a shred of rigorous mathematics. So there is nothing more else that needs to be said. They are not the only ones out there, but they are currently one of the most popular. – user21820 Dec 29 '17 at 06:36
  • @user21820 oh, so the creators while fascinating on math don't really know it? But they do interview mathematicians, why does no-one say so? – Ooker Dec 29 '17 at 08:21
  • @Ooker "But they do interview mathematicians, why does no-one say so?" Is this supposed to be an objection to what user21820 explained? – Did Dec 29 '17 at 08:57
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    @Joshua Sorry but what do you think WP "thinks" exactly? The WP page takes care to explain that "Many summation methods are used in mathematics to assign numerical values even to a divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of $−1/12$, which is expressed by a famous formula: $1+2+3+4+\cdots =-{\frac {1}{12}},$ where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods ... – Did Dec 29 '17 at 09:03
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    ... and not as the sum of an infinite series in its usual meaning" which is immensely more precise (and less click-baity, I guess) than the ridiculous statements uttered in the Nuberphile's video. If WP take had been emulated by the authors of the Numberphile video, nobody would complain. – Did Dec 29 '17 at 09:03
  • @Did no. I just wonder why interviewed mathematicians in their videos don't say that what they present are not true. Their reputation would be ruined for this, wouldn't it? – Ooker Dec 29 '17 at 09:57
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    @Ooker: We have no way of knowing what exactly the mathematicians said to them. All we know is that the video creators talked nonsense. Look, most people want their ears tickled and do not care for the truth. That is why there are so many scams, not just in mathematics. Have you never heard of quantum bracelets and homeopathy? Thousands are selling such rubbish, but their reputation is only ruined in the eyes of truth-seekers, which are the minority. – user21820 Dec 29 '17 at 10:26
  • @user21820 but even if the interviewed mathematicians miss watching the problematic videos, at least some actual mathematicians have caught their mistakes (since the channel is popular) and report to them. The interviewed ones would be afraid of getting bad reputations, so they will at least correct the wrong ideas, and stop working with them. But as I see this is not the case. – Ooker Dec 29 '17 at 14:53
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    @Ooker: Indeed, it is hard to get people to fix their wrong statements, especially if they are popular personalities and stubborn. Think about it; why would those videos' creators want to take down the wrong video and upload a new one with the errors fixed? They wouldn't want to lose the number of views. Anyway this really isn't the right place to discuss such issues; you can come to the chat-rooms for that. – user21820 Dec 29 '17 at 15:02
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    [Dr. Padilla's response](http://www.nottingham.ac.uk/~ppzap4/response.html) to some objections raised against the 'sum of all integers' video. – pizzapants184 Dec 30 '17 at 02:38
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    @pizzapants184: Frankly, that linked post is utter rubbish. People who promote quantum bracelets and homeopathy have also generated lots of debate on the internet, and that **is definitely a bad thing**. Even till today, we cannot eradicate some diseases in certain parts of the world because the local witch doctors tell the locals to reject vaccines saying that they cause AIDS. What these video creators are doing is making mathematics seem nebulous and subjective. Just look at all the terrible mathematical misunderstandings due to the kind of shoddy reasoning that this video is promulgating! – user21820 Dec 30 '17 at 07:38
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    @pizzapants184: [Here](http://goodmath.scientopia.org/2014/01/17/bad-math-from-the-bad-astronomer) is a proper mathematical response. And the comments may help you grasp mathematicians' point of view, especially the second one by **non**-mathematician Ian Kaplan and the one by Casey Bowman. By the way, anyone who claims that the infinite series appears in physics is not doing proper physics. – user21820 Dec 30 '17 at 08:11
  • Found in the comments to the post @user21820 is linking to: "The way to reduce mistakes is to apply a cost to them. Public humiliation of someone with a doctorate in astronomy falls in that category." How aptly summarized! – Did Dec 30 '17 at 12:36

Let the natural number $x$ be the solution to $x=x+1$. Then (picks random contradiction out of a hat, as allowed by the principle of explosion (obligatory xkcd)), $4$ is prime. Proof: $3$ is known prime, since all smaller positive integers are either units or primes not dividing $3$. Also, since $3-x = 4-(x+1) = 4-(x)$, adding $x$ to both sides gives $3=4$. Since primeness is preserved by equality, $4$ is prime.

Eric Towers
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  • The nice thing about your example is that you do not even need the principle of explosion. All you need is an equational theory, namely one where you just perform substitutions to get from say "$4$ is even" to "$3 = 3+x-x = 3+(x+1)-x = 4$ is even". =) – user21820 Dec 30 '17 at 09:36
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    @user21820 : Yes, I was very lucky. The hat also contained a contradiction that was something about $0^\dagger$ ([zero dagger](https://en.wikipedia.org/wiki/Zero_dagger)) which I'm glad I didn't draw. – Eric Towers Dec 30 '17 at 11:00
  • Wow, that goes straight to my personal list of best xkcds! – Pedro A Dec 31 '17 at 21:39

In the general case, does not any proof that uses Reductio ad absurdum meet the criteria requested by the OP?

In particular, the case that I remember best is from high school, that being the proof that $\sqrt 2$ is irrational. This starts by assuming there exist some $a$ and $b$ that are both integers, have no common divisor, and that the following equation is true: $a / b = \sqrt 2$.

If I remember correctly (it's been over 40 years) the proof proceeds to show that if the above equation is true, $a$ and $b$ do have a common divisor, thus leading to a contradiction.

Hence in this case, assuming the existence of a solution to the problem of finding integers $a$ and $b$ that meet all the criteria of the second paragraph does indeed lead to a contradiction.

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A famous example is the existence of a solution for $$p^2 = 2q^2, p,q \in \mathbb Z$$

and $p,q$ share no common prime factors.

It is a rephrasing of a classical proof that $\sqrt{2}$ is irrational by assuming it can be written $\sqrt{2} = p/q$ for two such integers.

One comes to the conclusion that a square integer must have just 1 ( an odd number ) of the prime 2 in it's prime expansion, which is of course impossible, since a square must have each prime occurring an even number of times in it's prime factorization.

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Assume the existence of

$$S=\sum_{n=1}^{\infty}(-1)^nn = -1 +\sum_{n=2}^{\infty}(-1)^nn=-1 +\sum_{n=1}^{\infty}(-1)^{n+1}(n+1) \\=-1 -\sum_{n=1}^{\infty}\left[(-1)^{n}n+(-1)^{n}\right] = -1 -S -\sum_{n=1}^{\infty}(-1)^n$$

this leads to $$S = -\frac{1}{2} - \frac{1}{2}\sum_{n=1}^{\infty}(-1)^n =\color{red}{ -\frac{1}{2} - \frac{1}{2}K}$$

Hence the existence of $S$ entails the existence of $$ K=\sum_{n=1}^{\infty}(-1)^n = -1 +\sum_{n=2}^{\infty}(-1)^n = -1+\sum_{n=1}^{\infty}(-1)^{n+1} = -1-K$$

Then we have $K=-\frac{1}{2}$ and then $S= -\frac{1}{2} - \frac{1}{2}K = -\frac{1}{4} $

Finally we get $$ \color{blue}{ \sum_{n=1}^{\infty}(-1)^nn =-\frac{1}{4}~~~and ~~~\sum_{n=1}^{\infty}(-1)^n =-\frac{1}{2}}$$ But blatantly none of the above integrals sum converges in the usual sense.

Guy Fsone
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