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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.
$$x^2=\underbrace{x+x+\cdots+x}_{(x\text{ times})}$$ $$\frac{d}{dx}x^2=\frac{d}{dx}[\underbrace{x+x+\cdots+x}_{(x\text{ times})}]$$ $$2x=1+1+\cdots+1=x$$ $$2=1$$
Endless chocolate bar (I do not know the author of this animation)
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
Simple one:
$1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = i\cdot i = -1$
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.
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.
$$\int \frac{1}{x}dx = x\cdot\frac{1}{x} - \int x\,d\frac{1}{x}$$ $$\int \frac{1}{x}dx = 1 + \int x\frac{1}{x^2}dx$$ $$\int \frac{1}{x}dx = 1 + \int \frac{1}{x}dx$$ $$0 = 1 $$
"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 $n=1$.
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
The scientist/engineer's proof that all odd numbers are prime...
Therefore, by extension (ignoring the scientific errors), all odd numbers are prime.
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...
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...
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.
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.
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.
The proof that $\pi = 4$, discussed here: Is value of $\pi = 4$?
Here's a "proof" of $e^{x} = 1$ for all $x$. $$\exp(x) = \exp(i2\pi\cdot\frac{x}{i2\pi}) = \exp(i2\pi)^{\frac{x}{i2\pi}} = 1 ^{\frac{x}{i2\pi}} = 1.$$
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 $\int_0^1 0\ dy=0$.
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.
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, $1 = 0$.
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:
$1\equiv 4$ $\quad \Rightarrow \quad$ $2^1\equiv 2^4$ $\quad \Rightarrow \quad$ $2\equiv 16$ $\quad \Rightarrow \quad$ $2\equiv 1$ $\quad \Rightarrow \quad$ $2-1\equiv 1-1$ $\quad \Rightarrow \quad$ $1\equiv 0$.
$$\begin{align*} \sum_{k\ge 1}\frac1k&=\sum_{k\ge 0}\left(\frac1{2k+1}+\frac1{2k+2}\right)\\ &=\sum_{k\ge 0}\frac{4k+3}{(2k+1)(2k+2)}\\ &=\sum_{k\ge 0}\left(\frac{2(2k+1)}{(2k+1)(2k+2)}+\frac1{(2k+1)(2k+2)}\right)\\ &=\sum_{k\ge 0}\left(\frac1{k+1}+\frac1{(2k+1)(2k+2)}\right)\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 0}\frac1{(2k+1)(2k+2)}\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 0}\left(\frac1{2k+1}-\frac1{2k+2}\right)\\ &=\sum_{k\ge 1}\frac1k+\sum_{k\ge 1}\frac{(-1)^{k+1}}k\\ &=\sum_{k\ge 1}\frac1k+\ln 2\;, \end{align*}$$
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 $\sum\limits_{k\ge 1}\frac{(-1)^{k+1}}k=\ln 2$.
$$(n+1)^2 = n^2+2n+1$$
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!
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!
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 $x=e^{\pi/2}$.
$1=2$: 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.
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!