# Will assuming the existence of a solution ever lead to a contradiction?

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?

• 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 0:21
• I thought Proof By Contradiction was standard in Mathematics by now, though some would question the approach. – Dehbop Dec 28 '17 at 6:40
• – Ilmari Karonen Dec 28 '17 at 8:14
• 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 9:57
• @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

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.

• $\mathbb Q_2$ would like a word... – Artimis Fowl Dec 28 '17 at 6:31
• @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 6:58
• Why does $A = 1 + 2 \sum_{n=0}^{\infty}(2^n)$? – Brian J Dec 28 '17 at 17:02
• @AsafKaragila How do we go back to being farmers if all our fields are in universities? – Joonas Ilmavirta Dec 28 '17 at 20:06
• @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. – user491874 Dec 29 '17 at 19:09

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.

• maa.org/programs/maa-awards/writing-awards/… – Will Jagy Dec 28 '17 at 1:13
• 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 9:49
• @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
• @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. – Hurkyl Dec 28 '17 at 15:28
• 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

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.

• 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 2:36
• @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
• @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. – Hurkyl 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 $x=1$.)

• 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
• @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. – WetSavannaAnimal 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!

• this is my favorite – The Great Duck Dec 30 '17 at 4:26

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.

## 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)

• Please state clearly in your answer that the three links are rubbish. Otherwise it is misleading. – user21820 Dec 28 '17 at 15:16
• 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 - and physics.se also math.se – Mindwin Dec 28 '17 at 16:29
• @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 9:03
• ... 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 9:03
• @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

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.

• 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 9:36
• @user21820 : Yes, I was very lucky. The hat also contained a contradiction that was something about $0^\dagger$ (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.

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.

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.

## protected by J. M. is a poor mathematicianMar 13 '18 at 18:09

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