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In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen.

I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.

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  • $\begingroup$ Are we looking to avoid oldies and elementary proofs? $\endgroup$
    – Git Gud
    Apr 1, 2013 at 15:01
  • $\begingroup$ A grad student I know has a wonderful proof involving Fermat's Last Theorem that $\sqrt2^n$ is irrational for all $n$. $\endgroup$
    – user47805
    Apr 1, 2013 at 15:06
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    $\begingroup$ math.stackexchange.com/questions/265853/too-simple-to-be-true/… $\endgroup$
    – tom
    Apr 1, 2013 at 15:10
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    $\begingroup$ One of the best ever fake, or at least apparently-fake, proofs in mathematics goes more or less as "I've found a truly wonderful proof but the margin is too $\rlap{-----------}{\text{...ejem,ejem...marginal...er...}}$ small to contain it..." $\endgroup$
    – DonAntonio
    Apr 1, 2013 at 15:27
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    $\begingroup$ There is this book amazon.com/Mathematical-Fallacies-Flaws-Flimflam-Spectrum/dp/… with one of the best collection I have seen. $\endgroup$ Apr 2, 2013 at 4:28

27 Answers 27

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$$x^2=\underbrace{x+x+\cdots+x}_{(x\text{ times})}$$ $$\frac{d}{dx}x^2=\frac{d}{dx}[\underbrace{x+x+\cdots+x}_{(x\text{ times})}]$$ $$2x=1+1+\cdots+1=x$$ $$2=1$$

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    $\begingroup$ Nice one. You took one $x$ as a variable and the other $x$ as constant!. $\endgroup$
    – hrkrshnn
    Apr 1, 2013 at 15:19
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    $\begingroup$ Also, the first statement isn't true unless $x$ is a positive integer, so it can hardly be used with differentiation by $x$ regardless. After all, $\sin \pi x=0$ is true for all integers $x$, but that doesn't mean we can differentiate both sides to get equality. @boywholived $\endgroup$ Apr 1, 2013 at 15:26
  • $\begingroup$ @Thomas Andrews.What you said is right and is a better reason too. $\endgroup$
    – hrkrshnn
    Apr 1, 2013 at 15:37
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    $\begingroup$ @boywholived No, your argument is good too - he is essentially differentiating $xy$ by $x$ and then substituting $y=x$. $\endgroup$ Apr 1, 2013 at 15:42
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    $\begingroup$ dougshaw.com/findtheerror/FTEdiff.html $\endgroup$
    – spin
    Apr 2, 2013 at 13:26
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enter image description here

Endless chocolate bar (I do not know the author of this animation)

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  • 5
    $\begingroup$ Also see Missing square puzzle and Dissection Fallacy (making it into a chocolate bar is pretty recent innovation intended to make the paradox accessible to non-math-enthusiasts that's been going around the memesphere lately). $\endgroup$
    – anon
    Apr 1, 2013 at 18:14
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    $\begingroup$ For those wondering... the two large pieces get taller as they move--watch their bottoms closely. $\endgroup$
    – ErikE
    Apr 1, 2013 at 19:25
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    $\begingroup$ In light of ErikE's comment, not really math. Still nice eye candy though. Pun intended. $\endgroup$ Apr 1, 2013 at 22:41
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    $\begingroup$ @Yoni, it is math. See the Expert's link for the missing square puzzle. It's connected to solving $ax+by = 1$ in integers. $\endgroup$
    – KCd
    Apr 1, 2013 at 23:10
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    $\begingroup$ @KCd If it was related to the missing square puzzle (as I originally thought), then sure. But it's not. The size of that piece is actually changing through the animation. $\endgroup$
    – Izkata
    Apr 2, 2013 at 14:35
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Proof that a dog has $9$ legs:

No dog has $5$ legs,

A dog has $4$ more legs than no dog.

A dog has $9$ legs

Source: Foolproof: A Sampling of Mathematical Folk Humour, P Renteln, A Dundes

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    $\begingroup$ Cries logical Bobby to Ned, will you dare // A bet, which has most legs, a mare, or no mare. // A mare, to be sure, replied Ned, with a grin, // And fifty I’ll lay, for I’m certain to win. // Quoth Bob, you have lost, sure as you are alive, // A mare has but four legs, and no mare has five. $\endgroup$
    – Pål GD
    Apr 1, 2013 at 17:44
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    $\begingroup$ By that theorem, a ham sandwich is better than eternal happiness. $\endgroup$
    – Joe Z.
    Apr 1, 2013 at 19:16
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    $\begingroup$ Actually, given that animals are occasionally born with two heads, I think the assumption that no dog has 5 legs is probably invalid. It's probably safer to use a larger number like 100. $\endgroup$
    – Tacroy
    Apr 1, 2013 at 23:03
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    $\begingroup$ ^ But what if genetic science finds a way to create a centipedal dog? $\endgroup$
    – Joe Z.
    Apr 17, 2013 at 17:06
  • $\begingroup$ I'll just leave this here... youtube.com/watch?v=FavUpD_IjVY $\endgroup$ Jul 31, 2013 at 13:35
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Simple one:

$1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = i\cdot i = -1$

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    $\begingroup$ $\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$. Is valid only when both $a$ and $b$ are positive. $\endgroup$
    – hrkrshnn
    Apr 1, 2013 at 16:03
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    $\begingroup$ @boywholived: yes, the idea is that these are fake proofs. $\endgroup$
    – robjohn
    Apr 1, 2013 at 16:04
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    $\begingroup$ A nice variant of this that does not require complex numbers is $-1=(-1)^{1/3}=(-1)^{2/6}=((-1)^2)^{1/6}=1^{1/6}=1$. $\endgroup$ Apr 1, 2013 at 16:30
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    $\begingroup$ @robjohn Yes...but boywholived is pointing out the problem with the proof...valuable to those who may not see the issue. After all, if the problem with the proof was obvious, it wouldn't qualify as much of a joke. $\endgroup$
    – Beska
    Apr 1, 2013 at 17:37
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    $\begingroup$ @boywholived It is valid only when at least one of them is non-negative. $\endgroup$
    – Alraxite
    Apr 2, 2013 at 7:41
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Let us prove the following (obviously) wrong proposition:
$\underline{\text{Proposition}}$
All positive natural numbers are definable in under eleven words.

$\underline{\text{Proof}}$
Suppose for the sake of contradiction(!) that not all positive natural numbers are definable in under eleven words. Then there is a smallest integer $n\in\mathbb N$ which is not definable in under eleven words. But this number is $$ \color{brown} { \text { the smallest positive integer not definable in under eleven words, } } $$ therefore, it is definable in ten words. Contradiction!

--

For more information see the wikipedia article Berry paradox.

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    $\begingroup$ So which number is "the largest positive integer definable in ten words or less"? $\endgroup$
    – Joe Z.
    Apr 1, 2013 at 19:17
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    $\begingroup$ @JoeZeng "Ackermann with (Ackermann with (Graham, Graham), Ackermann with (Graham, Graham))"? Does that count? $\endgroup$
    – Cole Tobin
    Apr 2, 2013 at 1:02
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    $\begingroup$ @JoeZeng why should there need to be one? $\endgroup$
    – crf
    Apr 2, 2013 at 3:12
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    $\begingroup$ I love that one :) The thing is, in the ZF axioms we have "the seperation schema", which says that if X is a set and phi is a formula, then there is a set Y such that for all a, (a /in Y) iff (a /in X and phi(a)). This is what you use here to get "the set of numbers not definable in 10 words or less". The thing is, the seperation schema demands that phi will be definable in first-order logic, which it isn't here :P $\endgroup$
    – Ekuurh
    Apr 10, 2013 at 18:40
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    $\begingroup$ Interestingly, one way to prove that the halting problem is unsolvable is to try to make this rigorous. If you can solve the halting problem, you can write a program that takes an input N, and then finds all outputs of programs of length $M$ with input $K$, with $M,K\leq N$, then find the least natural number not output in any of these cases. Then feed the length of this program to itself. With a solution to the halting problem, you can write this program easily. $\endgroup$ Dec 17, 2014 at 22:48
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A friend of mine came up with this proof that all perfect numbers are even. A little more advanced than your usual fake proof that $0=1$.

Assume $$2n = \sum_{d|n} d$$

Then by Möbius inversion:

$$n =\sum_{d|n} 2d\cdot\mu\left(\frac{n}{d}\right)$$

and therefore $n$ is even.

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    $\begingroup$ This is a great example! I heard it over 20 years ago, so I doubt your friend is the first person to make this "observation". $\endgroup$
    – KCd
    Apr 1, 2013 at 15:46
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    $\begingroup$ @KCd Hah, it was at math camp in the early 80s, so we're talking 20 years ago for me, too. $\endgroup$ Apr 1, 2013 at 15:47
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    $\begingroup$ It was at math camp for me too, in the late 80s. Did your camp happen to be located in Ohio? $\endgroup$
    – KCd
    Apr 1, 2013 at 15:52
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    $\begingroup$ Yep, the Ross program. :) Guess it was the same instance. $\endgroup$ Apr 1, 2013 at 15:53
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    $\begingroup$ @Thomas: You kids! :-) I did it in $1963$, when it was still at Notre Dame. $\endgroup$ Apr 1, 2013 at 20:37
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$$\int \frac{1}{x}dx = x\cdot\frac{1}{x} - \int x\,d\frac{1}{x}$$ $$\int \frac{1}{x}dx = 1 + \int x\frac{1}{x^2}dx$$ $$\int \frac{1}{x}dx = 1 + \int \frac{1}{x}dx$$ $$0 = 1 $$

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    $\begingroup$ From C to shining C! $\endgroup$ Apr 2, 2013 at 17:08
63
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"Proof" that $1$ is the largest natural number.

Let $n$ be the largest natural number. Then, $n^2$, being a natural number, is less than or equal to $n$. Therefore $n^2-n=n(n-1)\leq 0$. Hence, $0\leq n\leq 1$. Therefore $n=1$.

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    $\begingroup$ I know these are all joke proofs, but this is also a good reminder about the importance of proving the existence of a mathematical object before you go on to prove things about said object. Otherwise you can prove lots of interesting but nonsensical things. $\endgroup$
    – Josh Chen
    Jul 12, 2013 at 1:34
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    $\begingroup$ @JoshChen Sometimes it's good to prove some properties of a mathematical object first - if some of them give you an easy proof of nonexistence. $\endgroup$ Jul 27, 2013 at 21:01
  • $\begingroup$ @JoshChen that's the one I was searching for. Nice objection! $\endgroup$ May 8, 2015 at 10:00
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    $\begingroup$ It does prove that the largest natural number, if it exists, is equal to 1. $\endgroup$
    – gnasher729
    Apr 2, 2016 at 12:16
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    $\begingroup$ I'm sorry but I cannot seem to find the exact reference, but if I'm not mistaken, I learned of this fake proof from a book by Luc Tartar. His purpose was to illustrate (in the context of the calculus of variations) the danger in assuming the existence of an extremum without proving it. $\endgroup$ Jan 18, 2017 at 17:40
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Proof that Riemann hypothesis is true:

(1) At least one of the following statements is true

(2) The above statement is false

(3) Riemann hypothesis is true

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    $\begingroup$ Well I do not know the above theorem, but I know the fallacy is regarding that "Every mathematical statement is not necessarily either true or false". $\endgroup$
    – Grobber
    Apr 1, 2013 at 18:02
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    $\begingroup$ For example consider: "This statement is false", the cyclicity generated by its truth and falsehood makes the statement to be classifiable as none. $\endgroup$
    – Grobber
    Apr 1, 2013 at 18:04
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    $\begingroup$ @GregMartin: There is a difference: in your example, both (1) and (2) could be taken to be false, or both to be true, and there is no inconsistency in either case. In the above (fake) proof in the answer though, (1) is not a premise, but can be "proved" to be true: if (1) is false, then both (2) and (3) are false and in particular (2) is false, which means that (1) is true, which is a contradiction. Therefore (1) is true, and therefore (2) is false, and therefore as at least one of (2) and (3) are true, (3) must be true. $\endgroup$ Apr 2, 2013 at 6:28
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    $\begingroup$ @GregMartin: I don't understand what you mean by accepting the trio of statements as true: we did no such thing (and in fact "proved" statement (2) to be false). Self-reference is crucial to the argument (the way (1) refers to (2) which refers back to (1)), which is why your first example shows nothing, while your second example with "either this sentence is false..." does. If you mean accepting that each statement in the system is either true or false, yes that is the flaw as Grobber already pointed out above before you first commented — and self-reference is required for it to give anything. $\endgroup$ Apr 2, 2013 at 8:54
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    $\begingroup$ @HonzaBrabec: it’s not unrelated to Gödel’s incompleteness; but it’s most closely related, I think, to Tarski’s undefinability of truth, which essentially states that no reasonably strong logical system can talk about truth-values of its own statements. The two main ingredients of that are (a) some deep thought to formalize precisely what that self-reference means, and (b) having done this, a quick derivation of a contradiction. Part (b) can be done — among other ways — by the present false proof, with RH replaced by absurdity. $\endgroup$ Apr 2, 2013 at 14:54
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The scientist/engineer's proof that all odd numbers are prime...

  • 1 is prime
  • 3 is prime
  • 5 is prime
  • 7 is prime
  • 9 - experimental error
  • 11 is prime
  • 13 is prime
  • 15 - experimental error
  • 17 is prime
  • 19 is prime

Therefore, by extension (ignoring the scientific errors), all odd numbers are prime.

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    $\begingroup$ Someone with an art degree would say: all odd numbers are prime? Let's check: 1 is prime, 2 is prime, 3 is prime, 4 is prime, 5 is prime, ... $\endgroup$
    – SteeveDroz
    Apr 2, 2013 at 15:28
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    $\begingroup$ Isn't "$1$ is prime" some kind of error too? $\endgroup$
    – wchargin
    Jun 6, 2014 at 23:42
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    $\begingroup$ @WChargin: Experimenter error. $\endgroup$
    – user21820
    Jun 27, 2014 at 9:58
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    $\begingroup$ 3 isn't prime! It's third! :) $\endgroup$ Feb 22, 2017 at 19:33
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    $\begingroup$ This is so funny! $\endgroup$
    – jimm
    Jun 2, 2018 at 4:02
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All cats have the same color by induction on the number of cats.

Base case: If there is only one cat, then it doesn't matter the color. They all have the same color.

Inductive step: Fix $n$. Suppose for any set of $n$ cats they have the same color. Consider any set of $n+1$ cats. Then cats 1 through n have the same color by inductive hypothesis. There is only one remaining to check, but we also know that cats $2$ through $n+1$ have the same color by the inductive hypothesis. Thus they must all have the same color (for example, because they all have the same color as the second one in this proof). QED.

...But clearly not all cats have the same color...

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    $\begingroup$ There are a few typos: you typed group instead of set and cats instead of horses. $\endgroup$
    – Git Gud
    Apr 1, 2013 at 15:08
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    $\begingroup$ In the internet age, I think it is more appropriate to prove that "all cats on youtube have the same color" instead of "all horses have the same color." $\endgroup$
    – treble
    Apr 1, 2013 at 15:27
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    $\begingroup$ @GitGud I'm not sure how group/set actually matters here. Also, I intentionally didn't use "horses" so that people couldn't just google to figure out the error. $\endgroup$
    – Matt
    Apr 1, 2013 at 16:08
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    $\begingroup$ Since the above commenters seem to be unsure: The problem is that the proof assumes the two sets (1 thru n, and 2 thru n+1) intersect, which they do not for n = 1. $\endgroup$ Apr 1, 2013 at 18:43
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    $\begingroup$ The inductive step is valid for $n+1\ge 3$. So this leads to the "theorem" that if any two cats have the same color, then any finite set of cats consists of one color only. $\endgroup$ Apr 2, 2013 at 15:24
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Theorem: If we mark $n$ points on a circle and connect each point to every other point by a straight line, the number of regions that the interior of the circle is divided into is $2^{n-1}$.

Proof: First let's collect numerical evidence.

When $n = 1$ there is one region (the whole disc).

When $n = 2$ there are two regions (two half-discs).

When $n = 3$ there are 4 regions (three lune-like regions and one triangle in the middle).

When $n = 4$ there are 8 regions, and if you're still not convinced then try $n=5$ and you'll find 16 regions if you count carefully.

Our proof in general will be by induction on $n$. Assuming the theorem is true for $n$ points, consider a circle with $n+1$ points on it. Connecting $n$ of them together in pairs produces $2^{n-1}$ regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us $2 \cdot 2^{n-1} = 2^n$ regions.

Or does it...

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    $\begingroup$ A good lesson in the dangers of both making assumptions based on examples, and of lacking rigor in your proofs. $\endgroup$
    – Jack M
    Apr 1, 2013 at 16:41
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    $\begingroup$ It depends on the way you place the dots, but the maximum number of regions is given by the sequence $$\{1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196,\\ 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457, 41449, 46938,\\ 52956, 59536, 66712, 74519, 82993, 92171,\ldots\}$$. i.e. [A006522](oeis.org/A006522)$+n$ $\endgroup$
    – Spenser
    Apr 2, 2013 at 14:44
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    $\begingroup$ Actually it is in OEIS directly: A000127 $\endgroup$
    – Spenser
    Apr 2, 2013 at 14:52
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    $\begingroup$ It is interesting that there is actually a polynomial expression for this: $$a(n)=\frac{1}{24}n^4-\frac{1}{4}n^3+\frac{23}{24}n^2-\frac{3}{4}n+1$$ $\endgroup$
    – Spenser
    Apr 2, 2013 at 17:23
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    $\begingroup$ @Spenser: Yes, the small powers of 2 happen to match the polynomial sequence $1 + \binom{n}{2} + \binom{n}{4}$. Similarly, $1 + \binom{n}{2} + \binom{n}{4} + \cdots + \binom{n}{2r}$ equals $2^{n-1}$ for $1 \leq n \leq 2r+1$, and it equals $2^{n-1} - 1$ for $n = 2(r+1)$. $\endgroup$
    – KCd
    Apr 2, 2013 at 23:00
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Zero$^{th}$ fundamental theorem of calculus: $$\int\limits_{0}^{a}{f(x) dx}=0\space \space \quad \quad \forall a \in \mathbb{R} \text{ and } \forall f(x) \in\mathcal{R}\left(\left(0,a\right)\right)$$


Proof

It's well known that $$\int\limits_{0}^{0}{f(x) dx}=0\space \space \quad \quad \forall f(x) \space \tag{1} $$

Case I

When $a=0$ The theorem at this case follows directly from $(1) $

Case II

When $a\neq 0$

Let $\large I=\int\limits_{0}^{a}{f(x) dx}$ Let $\large \sin{\frac{\pi \cdot x}{a}}=t $. Then $\large \frac{\pi}{a}\cos{\frac{\pi \cdot x}{a}}dx={dt}$ $$\cos{\frac{\pi \cdot x}{a}}=\sqrt{1-\sin^{2}{\frac{\pi \cdot x}{a}}}=\sqrt{1-t^2}$$ $$dx=\frac{a}{\pi}\cdot\frac{dt}{\sqrt{1-t^2}}$$

As we have introduced a substitution we should change the variables of the limit. $$\begin{cases} x=0 & \text{then } t=0, \text{ (lower limit)} \\ x=a & \text{then } t=0, \text{ (upper limit)}\end{cases}$$

We have $$I=\int\limits_{0}^{a}{f(x) dx}=\large \int\limits_{0}^{0}{\left(\frac{f(\frac{a}{\pi}\cdot\arcsin{t})}{\sqrt{1-t^2}} dt\right)}=0$$

Q.E.D.

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  • 1
    $\begingroup$ There’s no reason to consider Case I separately — the current Case II covers it anyway (nowhere in Case II do you use the assumption that $a ≠ 0$). $\endgroup$ Apr 2, 2013 at 14:57
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    $\begingroup$ @Peter LeFanu Lumsdaine. I think we should consider case 1 $a=0$. Otherwise our initial substitution $\sin{\frac{\pi \cdot x}{\boxed{a}}}=t$ would . . . .. $\endgroup$
    – hrkrshnn
    Apr 2, 2013 at 15:22
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    $\begingroup$ @AGoogler: The first thing wrong I notice is that when $\dfrac{\pi}{2}<\theta\leq \pi$, $\cos(\theta)\neq \sqrt{1-\sin^2(\theta)}$. $\endgroup$ Apr 3, 2015 at 22:23
  • 1
    $\begingroup$ @AGoogler The inverse of sine is not well-defined in this domain. $\endgroup$ May 11, 2015 at 6:43
  • 1
    $\begingroup$ (Sorry for the necropost) But is it also true more generally that because $sin(\frac{\pi x}{a})$ is not a diffeomorphism between domains (it's not invertible), you can't define a proper pullback to integrate the form? $\endgroup$ Jan 4, 2018 at 19:40
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All numbers are interesting.

Assume there is at least one uninteresting number. Then there is a non-empty set of uninteresting numbers. There is also a smallest uninteresting number. That makes it an interesting number so we remove it from the set of uninteresting numbers and repeat until we have removed all uninteresting numbers from that set.

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  • 1
    $\begingroup$ I think you mean to say that the set of uninteresting numbers is non-empty. $\endgroup$
    – Git Gud
    Apr 1, 2013 at 17:18
  • $\begingroup$ I guess so. Of course if we did start with an empty set of uninteresting numbers there would be no need for the proof. $\endgroup$ Apr 1, 2013 at 17:29
  • $\begingroup$ I mean instead of "then there is a set of uninteresting numbers", so you can then conclude, by the well ordering principle that "there is also a smallest uninteresting number". $\endgroup$
    – Git Gud
    Apr 1, 2013 at 17:31
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    $\begingroup$ I have heard this proof with "There is a smallest uninteresting number. Hey, that's pretty interesting! Contradiction." $\endgroup$ Apr 1, 2013 at 18:24
  • $\begingroup$ Works for the natural numbers at least. I guess it would work for the reals by applying Zorn's lemma... $\endgroup$
    – usul
    Apr 1, 2013 at 18:52
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Here is a another proof that "$0=1"$. $$0=(1-1)+(1-1)+(1-1)+(1-1)+\cdots=1-1+1-1+1\cdots=1+(-1+1)+(-1+1)+(-1+1)+(-1+1)\cdots=1+0+0+\cdots=1$$ Of course, this proof fails because the series $\Sigma (-1)^n$ is divergent.

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    $\begingroup$ How about this one: Let S = 1+2+4+8+16+... Then 2S = 2+4+8+16+32+... Subtraction of first minus second gives -S = 1 ans so S = -1 therefore -1 = 1+2+4+8+16+..... $\endgroup$
    – imranfat
    Apr 1, 2013 at 16:49
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    $\begingroup$ @imranfat: That is actually true when working in the $2$-adics... $\endgroup$
    – TMM
    Apr 1, 2013 at 18:59
  • $\begingroup$ Am I not getting it or we just missed -1 after third equals sign just before the fourth one? $\endgroup$
    – Denys S.
    Apr 2, 2013 at 14:38
  • $\begingroup$ @TMM: I think proofs like these are canonical examples of proofs which are false in one setting (e.g. standard convergence over the reals), but which in other settings with the same formalism (e.g. p-adics, or formal power series, or even just other notions of convergence/summability over the reals) may turn out to be true, and quite useful. $\endgroup$ Apr 2, 2013 at 15:08
  • 2
    $\begingroup$ @Peter The original proof actually has applications in topology as the Mazur swindle where infinite sums are defined (but addition is not necessarily cancellative). $\endgroup$
    – Erick Wong
    Apr 2, 2013 at 15:25
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The proof that $\pi = 4$, discussed here: Is value of $\pi = 4$?

  1. Inscribe a unit circle inside a square with side $4$.
  2. Repeat the following: For every vertex of the polygon (at first the square) that does not touch the circle, "fold it" so it touches the circle (by creating two additional vertices, so that all angles are right angles).
  3. The length of the polygon is always $4$, but the polygon converges to the circle, so $\pi = 4$.
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    $\begingroup$ Fractals can be tricky when it comes to perimeter. $\endgroup$
    – Jon Purdy
    Apr 1, 2013 at 18:30
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    $\begingroup$ What does converge, however, is the area. $\endgroup$
    – obataku
    Apr 1, 2013 at 22:20
  • 2
    $\begingroup$ Very true; math is full of such convergent/divergent conundrums. The Gabriel's Horn, for instance, has infinite surface area (because it has infinite length/depth and at every point along its length a nonzero diameter), but a finite convergent volume. $\endgroup$
    – KeithS
    Apr 1, 2013 at 23:08
  • 5
    $\begingroup$ @KeithS: Careful, if one uses $\frac{1}{x^2}$ to make Gabriel's horn instead of $\frac{1}{x}$, then both the volume and surface area area finite, despite the fact that it "has infinite length/depth and at every point along its length a nonzero diamter." $\endgroup$ Apr 1, 2013 at 23:38
  • 6
    $\begingroup$ So, $\lim_{}{4} =\pi$? $\endgroup$
    – Cole Tobin
    Apr 2, 2013 at 1:17
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Here's a "proof" of $e^{x} = 1$ for all $x$. $$\exp(x) = \exp(i2\pi\cdot\frac{x}{i2\pi}) = \exp(i2\pi)^{\frac{x}{i2\pi}} = 1 ^{\frac{x}{i2\pi}} = 1.$$

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    $\begingroup$ This is a nice one, would you mind stating which = sign is incorrect? I assume the second one because of incorrect rules of exponents since there are complex numbers involved? $\endgroup$
    – imranfat
    Apr 1, 2013 at 20:09
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    $\begingroup$ Yes, it's the second equality :) $\endgroup$
    – Long
    Apr 1, 2013 at 20:13
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Timothy Chow has posted a proof that $1=0$ via an illicit differentiation under the integral sign on his website, here is a quote:

Proof that $1=0$

Given any $x$, we have (by using the substitution $u=\frac{x^2}{y}$) $$ \int_0^1 \frac{x^3}{y^2}e^{-\frac{x^2}{y}}dy=\left[xe^-{\frac{x^2}{y}}\right]^1_0=xe^{-x^2}. $$ Therefore, for all $x$, $$ \begin{array}{rcl} e^{-x^2}(1-2x^2) &=& \frac{d}{dx} (xe^{-x^2}) \\ &=& \frac{d}{dx} \int_0^1 \frac{x^3}{y^2} e^{-\frac{x^2}{y}} dy \\ &=& \int_0^1 \frac{\partial}{\partial x} \left(\frac{x^3}{y^2} e^{-\frac{x^2}{y}}\right) dy \\ &=& \int_0^1 e^{-\frac{x^2}{y}} \left(\frac{3x^2}{y^2} - \frac{2x^4}{y^3}\right) dy. \end{array} $$ Now set $x=0$; the left-hand side is $e^0(1-0)=1$, but the right-hand side is $\int_0^1 0\ dy=0$.

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    $\begingroup$ "Illicit differentiation" sounds so much more exciting than implicit differentiation. $\endgroup$
    – Adam Saltz
    Apr 1, 2013 at 18:56
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    $\begingroup$ There is no need for any kind of differentiation under the integral sign. The essence of the argument is this: for $a > 0$, $\int_1^\infty ae^{-ax}dx = e^{-a}$. Now let $a \rightarrow 0^{+}$. The left side has limit $\int_1^\infty 0dx = 0$ and the right side has limit $e^{-0} = 1$. Maybe not... $\endgroup$
    – KCd
    Apr 1, 2013 at 21:57
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Here's a fake proof that there are no natural numbers $p$ and $q$ such that $\sqrt{2} = \frac{p}{q}$. Suppose it were the case and further suppose, without loss of generality, that $p$ and $q$ are relatively prime. Then $\sqrt{2} = \frac{p}{q}$ and so $$2q^{2} = p^{2}, \tag{*}$$ which means that $p$ is even. Thus $p=2r$ for some natural number $r$. Plugging that into (*) gives $2q^{2} = 4r^{2}$, so that $q^{2} = 2r^{2}$, which means $q$ is also even, contrary to the assumption that $p$ and $q$ are relatively prime.

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    $\begingroup$ That proof is not fake. That is a proof by contradiction, a well established method for disproving conjectures $\endgroup$
    – imranfat
    Apr 1, 2013 at 18:48
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    $\begingroup$ @imranfat (I think that's his joke, considering it's April Fools day...) $\endgroup$ Apr 1, 2013 at 19:23
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Let $f(x) = 1$. It's easy to see that its Fourier transform is $0$ almost everywhere, so $\hat {\hat f}(x) = 0$. By the inversion theorem, $1 = 0$.

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$$\begin{align*} \sum_{k\ge 1}\frac1k&=\sum_{k\ge 0}\left(\frac1{2k+1}+\frac1{2k+2}\right)\\ &=\sum_{k\ge 0}\frac{4k+3}{(2k+1)(2k+2)}\\ &=\sum_{k\ge 0}\left(\frac{2(2k+1)}{(2k+1)(2k+2)}+\frac1{(2k+1)(2k+2)}\right)\\ &=\sum_{k\ge 0}\left(\frac1{k+1}+\frac1{(2k+1)(2k+2)}\right)\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 0}\frac1{(2k+1)(2k+2)}\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 0}\left(\frac1{2k+1}-\frac1{2k+2}\right)\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 1}\frac{(-1)^{k+1}}k\\ &=\sum_{k\ge 1}\frac1k+\ln 2\;, \end{align*}$$

so $\ln 2=0$, and $2=e^{\ln 2}=e^0=1$.


The legitimate version of this shows that $$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}k=H_{2n}-H_n\;,$$ where $H_n$ is the $n$-th harmonic number, and with the observation that

$$\ln\left(2-\frac1{n+1}\right)=\int_{n+1}^{2n+1}\frac{dx}x<H_{2n}-H_n<\frac1{n+1}+\int_n^{2n}\frac{dx}x=\frac1{n+1}+\ln 2$$

is one way to show that $\sum\limits_{k\ge 1}\frac{(-1)^{k+1}}k=\ln 2$.

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    $\begingroup$ This is also another proof that the harmonic series diverges. $\endgroup$ Apr 1, 2013 at 22:33
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    $\begingroup$ Or does it just prove the harmonic series doesn't converge absolutely? $\endgroup$ Apr 1, 2013 at 23:38
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    $\begingroup$ @Jason: Same thing, since it’s a positive series. $\endgroup$ Apr 1, 2013 at 23:40
  • $\begingroup$ @Brian: Oh, Duh! Thanks! $\endgroup$ Apr 1, 2013 at 23:40
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Let me prove that the number $1$ is a multiple of $3$.

To accomplish such a wonderful result we are going to use the symbol $\equiv$ to denote "congruent modulo $3$". Thus, what we need to prove is that $1 \equiv 0$. Next I give you the proof:

$1\equiv 4$ $\quad \Rightarrow \quad$ $2^1\equiv 2^4$ $\quad \Rightarrow \quad$ $2\equiv 16$ $\quad \Rightarrow \quad$ $2\equiv 1$ $\quad \Rightarrow \quad$ $2-1\equiv 1-1$ $\quad \Rightarrow \quad$ $1\equiv 0$.

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    $\begingroup$ How "does" $2 \equiv 16 \implies 2 \equiv 1$? Is it because $1 \equiv 4$ and $16 / 4 \in \mathbb{N}$? That doesn't seem to follow... what am I missing? $\endgroup$
    – wchargin
    Jun 6, 2014 at 23:45
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    $\begingroup$ @WChargin: That part's just legitimate transitivity, with $16 = 1 \text{ (mod 3)}$. I still can't think of anyone whom the exponentiation right at the start would fool, though. $\endgroup$ Oct 23, 2015 at 2:27
  • $\begingroup$ @boumol WHat is the mistake in this argument? $\endgroup$
    – Babai
    Mar 8, 2018 at 18:58
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    $\begingroup$ @Babai, $a \equiv b \Rightarrow a^k \equiv b^k$, but does not imply $k^a \equiv k^b$ $\endgroup$ Jul 25, 2018 at 6:39
  • $\begingroup$ @boumol $2^1 \equiv 2$ mod 3 but $2^4 \equiv 1$ mod 3. Multiplying two natural numbers that are the same mod 3 always gets you a number the same mod 3 but that's not the case for natural number exponentation. $\endgroup$
    – Timothy
    Jan 3, 2019 at 7:03
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$$(n+1)^2 = n^2+2n+1$$

Expansion: $$(n+1)^2-(2n+1) = n^2$$

Subtract from both sides: $$(n+1)^2-(2n+1)-n(2n+1) = n^2-n(2n+1)$$

Add to both sides: $$(n+1)^2-(n+1)(2n+1) = n^2-n(2n+1)$$

Factor: $$(n+1)^2-(n+1)(2n+1)+\frac{(2n+1)^2}{4} = n^2-n(2n+1)+\frac{(2n+1)^2}{4}$$

Add to both sides: $$\left[(n+1)-\frac{2n+1}{2}\right]^2 = \left[n-\frac{2n+1}{2}\right]^2$$

Square root: $$(n+1)-\frac{2n+1}{2} = n-\frac{2n+1}{2}$$

Subtract from both sides: $$n+1 = n$$

Subtract from both sides: $$1 = 0$$

Impossible!

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    $\begingroup$ Took me a while to figure this one out. I typed each expression in my TI and left hand side equals right hand side. Until I reache the square root part. The problem is the square root. If A² = B² that does not automatically imply A=B, it could also be A=-B That's an awesome example! $\endgroup$
    – imranfat
    Apr 2, 2013 at 19:48
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Here's a fun compilation of fake proofs in calculus. One of them is already here, but I'll give another example: \begin{align*} \int_{-1}^1\frac{dx}{1 + x^4} &=\int_{-1}^1\frac{\frac{1}{x^4}}{\frac{1}{x^4} + 1}dx\\ u &= \frac{1}{x^4}\\ du &= -\frac{4}{x^5}dx\\ \int_{-1}^1\frac{dx}{1 + x^4} &=\int_1^1\frac{u}{u+1}\left(-\frac{u^{-5/4}}{4}\right)\,du\\ &= -\frac{1}{4}\int_1^1\frac{du}{\sqrt[4]{u}\left(u + 1\right)}\\ &= 0. \end{align*} However, one look at the graph of $\frac{1}{1 + x^4}$ (or a quick estimate) tells us that the integral in question is certainly nonzero!

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  • $\begingroup$ Please look at the moderator's note at the beginning of the post. $\endgroup$
    – hrkrshnn
    Apr 1, 2013 at 16:47
  • $\begingroup$ @boywholived Right, I had posted that there because this post had linked there. I have since deleted the answer from the "duplicate" question. $\endgroup$
    – Stahl
    Apr 1, 2013 at 16:48
  • $\begingroup$ The answer to integral is not incorrect at all! It is just that you wanted to calculate the area under the curve and that didn't work. The answer to an integral itself does not always have to give you the area under the curve unless that is what the question states. $\endgroup$
    – imranfat
    Apr 1, 2013 at 16:49
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    $\begingroup$ @imranfat I agree that in general an integral does not have to represent the area under a curve. However, in this case, we can interpret the integral in question as the area under the curve of the graph of $1/(1 + x^4)$ without fear, and that's one way of seeing that this calculation cannot be correct. $\endgroup$
    – Stahl
    Apr 1, 2013 at 18:18
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    $\begingroup$ @Integral the substitution we consider is invalid: here's the theorem: Suppose that $f$ is continuous on an open interval $I$. Let $u$ and $u'$ be continuous on an open interval $J$, and assume that the range of $u$ is contained in $I$. If $a,b\in J$, then $\int_a^b f(u(x))u'(x)\,dx = \int_{u(a)}^{u(b)} f(u)\,du$. Certainly, the $u$ we use here is not continuous on the interval in question, $(-1 - \epsilon,1 + \epsilon)$. $\endgroup$
    – Stahl
    Apr 2, 2013 at 15:13
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We shall prove that $x=e^{\pi/2} \approx 4.8$ is a solution to $x^x=x$.

First, we compute $i^i = e^{i \log i} = e^{i(\pi/2i)} = e^{- \pi/2}$. Next we compute $i^{i^i} = e^{-pi/2 i} = -i$. Then $x=i^{i^{i^i}}=(-i)^i=e^{i\log (-i)}=e^{i(-\pi/2 i)}=e^{\pi/2}$.

Finally, $x^i=i^{i^{i^{i^i}}}=e^{i \pi/2}=i$.

Raise each side of the previous equation to the $i^{i^i}$th power. We get $$ x^{i^{i^{i^i}}}={i^{i^{i^i}}}$$ or $$x^x=x$$ where $x=e^{\pi/2}$.

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$1=2$: A Proof using Complex Numbers

This supposed proof uses complex numbers. If you're not familiar with them, there's a brief introduction to them given below. The Fallacious Proof:

Step 1: $-1/1 = 1/-1$
Step 2: Taking the square root of both sides: $\sqrt{-1/1} = \sqrt{1/-1}$.
Step 3: Simplifying: $\sqrt{-1} / \sqrt{1} = \sqrt{1} / \sqrt{-1}$
Step 4: In other words, $i/1 = 1/i$.
Step 5: Therefore, $i / 2 = 1 / (2i)$,
Step 6: $i/2 + 3/(2i) = 1/(2i) + 3/(2i)$,
Step 7: $i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) )$,
Step 8: $(i^2)/2 + (3i)/2i = i/(2i) + (3i)/(2i)$,
Step 9: $(-1)/2 + 3/2 = 1/2 + 3/2$,
Step 10: and this shows that $1=2$.

See if you can figure out in which step the fallacy lies.

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  • $\begingroup$ Hint : Its Steps 2 and 3 $\endgroup$
    – Ajo Koshy
    Apr 2, 2013 at 8:32
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Whatever goes up, must come down

Rules of Mathematics prove that this is a wrong rule and should be discarded

Proof:

Eq 1. Goes Up = Comes Down

Eq 2. (1 / Goes Up) = (1 / Comes Down)

Since Goes Up means inverse of Comes Down, hence

That means Goes Up = (1 / Comes Down)

From Eq2, we can say

Comes Down = (1 / Comes Down)

Which is Impossible!

Hence proved that this is wrong, and Whatever goes up must keep going up!

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    $\begingroup$ Well, what if Comes Down = 1? $\endgroup$
    – Joe Z.
    Apr 2, 2013 at 12:50
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    $\begingroup$ How would that be so? $\endgroup$ Apr 2, 2013 at 13:18
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    $\begingroup$ I dunno, you could arbitrarily set it to be that, I suppose. $\endgroup$
    – Joe Z.
    Apr 2, 2013 at 13:34
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    $\begingroup$ I would say mathematically, Goes up would be the additive inverse, and not multiplicative, of comes down. $\endgroup$
    – Guy
    Apr 7, 2014 at 6:15

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