Examples of interesting false proofs According to Wikipedia False proof

For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.

The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics.

What are some examples of interesting false proofs?

 A: $\pi$ is irrational : if $\pi=a/b$ is irreducible, and $a$ is divisible by an odd prime $p$, the series for $\sin \pi =\pi-\pi^3/6+\pi^5/120-\dots$ converges in the $p$-adics, and the limit is obviously not zero, absurd (if $a=2^n$, $n>1$ and the convergence is assured in the 2-adics, with the same contradiction). 
A: A common mistake in using induction for statements concerning finite sets is the bad logic "prove it for 1-set, and if we have proved this for $n$-set, add an element and prove it for $(n+1)$-set". I like the following illustrative example proposed by Sergey Berlov:
Theorem. A simple undirected graph with $n$ vertices and $n$ edges contains a triangle.
Poof. For $n<3$ there are simply no such graphs. For $n=3$ a triangle exists. Now add a vertex and an edge. The triangle does not diappear, right?
A: This one is interesting also in the sense of having an interesting history: In his writings, Cantor used a principle "every set can be well-ordered." As far as I understand, he claimed that it was obvious.
In 1905, König proposed and published an alleged proof that $\mathbb{R}$ cannot be well-ordered:
Suppose that $\mathbb{R}$ is well-ordered with an ordering relation $\preceq$. Since there are uncountably many reals, there is an undefinable one (say, undefinable even in the language with the symbol $\preceq$). Since $\preceq$ is a well-ordering, there exists  the least one $x_0$ which is not definable. But we have just defined it. A contradiction.
In the same year, a paper by Zermelo ''A proof of the principle that every set can be well-ordered'' was published. The author reduced the controversial principle to several reasonably looking statements which became a basis of what is known as Zermelo--Fränkel set theory.
On the other hand, what became known as König paradox had to wait a bit to be resolved until better understanding of truth predicates was obtained. 
A: $$e^i = (e^i)^{(2\pi/2\pi)} = (e^{2\pi i})^{1/2\pi} = 1^{1/2\pi} = 1.$$
I first saw this one many years ago, written on the wall of a bathroom stall in the Princeton University math department.
A: I like this one, invented by T.Clausen in 1827: since $e^{2\pi i n}=1$ for all integers $n$, we have $e^{2\pi i n+1}=e$, which implies $e^{(2\pi i n+1)^2}=(e^{2\pi i n+1})^{2\pi i n+1}=e^{2\pi i n+1}=e$. Now expanding the square at the exponent gives
$$e^{1-4\pi^2n^2+4\pi n i}=e$$
and after simplifying
$$e^{-4\pi^2n^2}=1$$
for all $n$.
A: Given any $x$, we have (by using the substitution $u=x^2/y$)
$$ \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy
  = \biggl[x e^{-x^2/y}\biggr]_0^1 = x e^{-x^2}.$$
Therefore, for all $x$,
$$\eqalign{e^{-x^2}(1-2x^2) &= {d\over dx}(xe^{-x^2})\cr
 &= {d\over dx} \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy\cr
 &= \int_0^1 {\partial \over \partial x} \biggl({x^3\over y^2}
       e^{-x^2/y}\biggr)\,dy\cr
 &= \int_0^1 e^{-x^2/y} \biggl({3x^2\over y^2} -
       {2x^4\over y^3}\biggr)\,dy.\cr}
$$
Now set $x=0$; the left-hand side is $e^0(1-0) = 1$,
but the right-hand side is $\int_0^1 0\,dy = 0$.
The main idea for this proof comes from an entry in Gelbaum and Olmstead's book Counterexamples in Analysis.
A: I have known the following for 45 years: in the Euclidian plane, every triangle is isosceles.
The false proof needs a handmade picture; take your pen, it's easy. Start from a triangle $ABC$. Draw the perpendicular bisector of $BC$, and the angle bisector from $A$. Let $I$ be their intersection (if it is not unique, you are done). Let $J$ be the projection of $I$ over $AB$, $K$ that over $AC$. Considering the right triangles $AIJ$ and $AIK$, we see that (lengths) $AJ=AK$, and that $IJ=IK$. Then looking at right triangles $BIJ$ and $CIK$, we obtain that $BJ=CK$. We conclude that
$$AB=AJ+JB=AK+KC=AC.$$
The falsity is that one of $J$ or $K$ is in the triangle, and the other one is out. Therefore one of the sums above (and only one) should be a difference.
A: Let me recycle this.
$\phantom{*******}$
A: Josh Nichols-Barrer wrote a delightful proof of Fermat's Last Theorem (and much more) here:
https://groups.google.com/d/msg/rec.humor/wUZ9gBmMchM/V9OS_or6gIQJ
In a nutshell: if $x^n+y^n=z^n$ then by differentiating and dividing by $n$, we get $x^{n-1}+y^{n-1}=z^{n-1}$. There are no integer solutions to $x^0+y^0=z^0$, so by induction Fermat's Last Theorem holds. As corollaries, there are no Pythagorean triples, and also addition is a lie. (But this is just a summary of Josh's amusing post.)
A: In the definition of an equivalence relation $\sim$, the reflexivity of $\sim$ is redundant: Indeed, for any $x$, by the symmetric property we have $x \sim y$ implies $y \sim x$.  By transitivity we have $x \sim y$ and $y \sim x$ imply $x \sim x$.  Therefore, using only symmetry and transitivity, we obtain reflexivity.
A: Theorem: Every totally disconnected set has the discrete topology.
Proof: Let $X$ be a totally disconnected set. If $X$ has only one element, the conclusion clearly follows. Otherwise, for distinct points $a, b \in X$, we have that {$a, b$} $\subset X$ is not connected. Therefore, {$a, b$} admits a separation; but the only way to write this as a disjoint union of nonempty sets is {$a$} $\cup$ {$b$}. Since this gives a separation, each of {$a$} and {$b$} is open. In particular, {$a$} is open for any $a \in X$; so $X$ has the discrete topology. Q.E.D.
A: Many years ago, I came up with this stupid proof that all groups are abelian: $$ab^{-1}=a\cdot{1\over b}={a\over b}={1\over b}\cdot a=b^{-1}a$$ I called it The Passing Through Theorem.
A: Here is an interesting false proof as to how to multiply $2 \cdot  2$. Taken from this  link.


$\Large\textbf{Another example}$:

A: I'm fond of the following false proof of the Strong Law of Large Numbers.  Let $X$ be a random variable with expected value $\mu$ and  variance $\sigma^2$, and let $X_1, X_2, \dots$ be i.i.d. copies of $X$.  Then $$\operatorname{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n} \rightarrow 0 \textrm{ as } n\rightarrow\infty $$
and since a random variable with variance 0 takes on a single value with probability 1, we must have $$\lim_{n\rightarrow\infty} \frac{1}{n} \sum_{i=1}^n X_i = \mu \textrm{ almost surely.}$$
(It's a memorable heuristic reason to tell undergraduate probability students, even if not a true argument.)
A: This proof that $\pi=0$ may be of some interest in examinations.
The function $f(x)=\arctan(x)+\arctan(1/x)$ has derivative $f’(x)=\frac1{1+x^2}-\frac1{x^2} \frac{1}{1+\frac1{x^2}}=0$, hence it is constant. Therefore$\displaystyle \lim_{x\to+\infty}f(x)= \displaystyle \lim_{x\to-\infty}f(x)$, that is $\frac\pi2=-\frac\pi2$, whence $\pi=0$.$\quad\square$
A: Ethan Akin's "proof" that all vector bundles are stably trivial, and hence the $K$-theory of any space must vanish:
Let $V$ be a vector bundle over the base space $B$.  Let $T$ be a trivial bundle of the same rank as $V$.  To show that $V$ is stably trivial, it suffices to prove that $$V\oplus V=V\oplus T$$.
Let $P$ be the principal bundle associated with $V$.  Pull $P$ back over itself to get a bundle $Q$:

Then $Q$ (together with the map to $B$) is the principal bundle associated to $V\oplus V$.  But the bundle $Q\rightarrow P$ clearly has a section, namely the diagonal map (viewing $Q$ as a subspace of $P\times P$).  Thus $Q=P\times GL_n$, which (together with the same map to $B$) is the principal bundle associated to $V\oplus T$.
(Reference: Ethan Akin, K-theory doesn’t exist, JPAA 12 (1978) pp.177–179.)
A: Timothy Chow's answer has a nice application. Let $n,x,y,z$ be natural numbers such that $x^n+y^n-z^n=0$. It follows that $e^{x^n+y^n-z^n}=1=e^i$ and the absurd $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$ 
A: Theorem: Every bounded differentiable function $f\colon \mathbb{R}\to \mathbb{R}$ is constant. 
Proof. 
By assumption there exist real numbers $M,N$ such that
$$N\leq f(x) \leq M.$$
Taking derivatives we get 
$$0\leq f'(x)\leq 0.$$
Hence $f'(x)=0$ so $f$ is constant. QED
A: True Theorem The symmetric groups (consisting of all permutations) on infinite sets of different cardinalities are not isomorphic. 
False proof: The two groups have different cardinalities, since there are $2^\kappa$ many permutations of an infinite set of size $\kappa$, and $\kappa\lt\lambda$ implies $2^\kappa\lt 2^\lambda$. QED
See the question: Can the symmetric groups on sets of differing infinite cardinalities be isomorphic? for further information and a correct proof. 
I find the false proof illuminating, since it shows the limitation of a naive treatment of the continuum function $\kappa\mapsto 2^\kappa$. It simply isn't necessarily the case that the two groups have different cardinalities, even though it is necessarily the case that they are not isomorphic. 
A: Theorem: All people have the same eye color.
Proof by induction: we prove the statement "All members of any set of people have the same eye color". This is clearly true for any singleton set.
Now, assume we have a set $S$ of people, and the inductive hypothesis is true for all smaller sets. Choose an ordering on the set, and let $S_1$ be the set formed by removing the first person, and $S_2$ be the set formed by removing the last person.
All members of $S_1$ have the same eye color, and also for $S_2$. However, $S_1 \cap S_2$ has members from both sets, so all members of $S$ have the same eyecolor. $\square$
A: Theorem. $\int_0^\infty \sin x \phantom. dx/x = \pi/2$.
Poof. For $x>0$ write $1/x = \int_0^\infty e^{-xt} \phantom. dt$,
and deduce that $\int_0^\infty \sin x \phantom. dx/x$ is
$$
\int_0^\infty \sin x \int_0^\infty e^{-xt} \phantom. dt \phantom. dx
= \int_0^\infty
    \left( \int_0^\infty e^{-tx} \sin x \phantom. dx \right)
\phantom. dt
= \int_0^\infty \frac{dt}{t^2+1},
$$
which is the arctangent integral for $\pi/2$, QED.
The theorem is correct, and usually obtained as an application of
contour integration, or of Fourier inversion ($\sin x / x$ is a multiple of
the Fourier transform of the characteristic function of an interval).
The poof, which is the first one I saw
(given in a footnote in an introductory textbook on quantum physics),
is not correct, because the integral does not converge absolutely.
One can rescue it by writing $\int_0^M \sin x \phantom. dx/x$
as a double integral in the same way, obtaining
$$
\int_0^M \sin x \frac{dx}{x} =
\int_0^\infty \frac{dt}{t^2+1}
- \int_0^\infty e^{-Mt} (\cos M + t \cdot \sin M) \frac{dt}{t^2+1}
$$
and showing that the second integral approaches $0$ as $M \rightarrow \infty$;
but this detour makes for a much less appealing alternative to the usual
proof by complex or Fourier analysis.
Still the double-integral trick can be used legitimately to evaluate
$\int_0^\infty \sin^m x \phantom. dx/x^n$ for integers $m,n$ such that
the integral converges absolutely (that is, with $2 \leq n \leq m$;
NB unlike the contour or Fourier approach this technique applies
also when $m \not\equiv n \bmod 2$).
Write $(n-1)!/x^n = \int_0^\infty t^{n-1} e^{-xt} \phantom. dt$ to obtain
$$
\int_0^\infty \sin^m x \frac{dx}{x^n} = \frac1{(n-1)!}
 \int_0^\infty t^{n-1}
    \left( \int_0^\infty e^{-tx} \sin^m x \phantom. dx \right)
\phantom. dt,
$$
in which the inner integral is a rational function of $t$,
and then the integral with respect to $t$ is elementary.
For example, when $m=n=2$ we find
$$
\int_0^\infty \sin^2 x \frac{dx}{x^2}
= \int_0^\infty t \frac2{t^3+4t} dt
= 2 \int_0^\infty \frac{dt}{t^2+4} = \frac\pi2.
$$
As a bonus, we recover a correct proof of our starting theorem by
integration by parts:
$$
\frac\pi2 = \int_0^\infty \sin^2 x \frac{dx}{x^2}
 = \int_0^\infty \sin^2 x \phantom. d(-1/x)
 = \int_0^\infty \frac1x d(\sin^2 x)
 = \int_0^\infty 2 \sin x \cos x \frac{dx}{x};
$$
since $2 \sin x \cos x = \sin 2x$, the desired
$\int_0^\infty \sin x \phantom. dx/x = \pi/2$
follows by a linear change of variable.
Exercise Use this technique to prove that
$\int_0^\infty \sin^3 x \phantom. dx/x^2 = \frac34 \log 3$,
and more generally
$$
\int_0^\infty \sin^3 x \frac{dx}{x^\nu} =
 \frac{3-3^{\nu-1}}{4} \cos \frac{\nu\pi}{2} \Gamma(1-\nu)
$$
when the integral converges.  [Both are in Gradshteyn and Ryzhik,
page 449, formula 3.827; the $\nu=2$ case is 3.827#3, credited to
D. Bierens de Haan, Nouvelles tables d'intégrales définies,
Amsterdam 1867; the general case is 3.827#1, from Gröbner and
Hofreiter's Integraltafel II, Springer: Vienna and Innsbruck 1958.]
A: A cavalry sergeant has 24 horses available which he needs to put on 6 carriages. So he needs to divide 24 by 6. He figures that 6 will go into 24 at least once, so he puts down a 1. Subtracting 6 from 24, he gets 18, and he remembers that 18/6=3. So he comes up with the answer 13.
After considerable difficulty with implementing his solution he consults his lieutenant. The lieutenant checks the calculation by evaluating 13*6: 
3*6=18
1*6=6
Add them: 24.
Implementation of the result still remains elusive so they consult the colonel, who uses
a different method to check. Write down 13 six times and add.
13
13
13
13
13
13
In adding this up, the colonel arrives at the following sequence of intermediate results:
3,6,9,12,15,18,19,20,21,22,23,24.
A: An excelent example is the iscosceles triangle fallacy. Here is a link to it in wikipedia http://en.wikipedia.org/wiki/Mathematical_fallacy#Fallacy_of_the_isosceles_triangle 
A: In S. Bosch's Algebra, exercise 3.4.2 is to find an error in the following existence proof of an algebraic closure of a field $K$ (my translation):
"Consider all algebraic extensions of $K$. Since for a totally ordered (w.r.t. inclusion) family $(K_i)_{i \in I}$ of algebraic extensions of $K$, the union $\bigcup_{i \in I} K_i$ is an algebraic extension of $K$, Zorn's lemma shows the existence of a maximal algebraic extension, i.e. of an algebraic closure of $K$."
Added: Cf. https://math.stackexchange.com/q/621944/96384 for various discussions around, and actually working variants of, this flawed proof.
A: Mostly based on mlk's comments here
Lemma 1 $\lim_{x\to 0^+} x^0=1$, so $0^0=1$.
Lemma 2 $\lim_{x\to 0^+} 0^x=0$, so $0^0=0$.
Therefore $1=0$.
A: I have always found interesting, as a student as well as teacher, the "proof" that every derivative is continuous:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Fix any $x_0 \in \mathbb{R}$ and $h > 0$, by the mean value theorem we find $\xi \in (x_0,x_0+h)$ such that:
$$ f'(\xi) = \frac{f(x_0+h) - f(x_0)}{h} \implies \lim_{h \to 0} f'(\xi) = \lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} =f'(x_0),$$
where in the last equality we used that $f$ is differentiable. The conclusion follows since $h \to 0$ entails $\xi \to x_0$.
A: I came across this one in a book of false proofs, the name of which I can't remember. It stuck out because it's not the usual hidden division by $0$ or unestablished base case in an induction.
Theorem: Every implication or its converse must be true. 
Proof: 
Check the truth table for $(P\to Q)\vee (Q\to P)$ and note that it is a tautology. 
$\Box$
However we know that there are many cases where neither an implication nor its converse is true. For example take $P$ to be "$n$ is odd" and $Q$ to be "$n$ is prime."
A: Here's a nice false proof of the continuum hypothesis.
Consider the rational numbers $\mathbb{Q}$ as a totally ordered field.  We can add an indeterminate $T_0$ and make it positive but infinitely small (i.e., smaller than positive any element of $\mathbb{Q}$), that is, order $\mathbb{Q}(T_0)$ by lexicographic order of the Laurent series expansion at $0$.  Then we can add another indeterminate $T_1$ and make it positive but infinitely small (i.e., smaller than any positive element of $\mathbb{Q}(T_0)$).  This process can be iterated transfinitely and we can add $\aleph_1$ indeterminates $T_\iota$ for $\iota<\omega_1$, each infinitely smaller than all the previous ones.  The resulting field $K = \mathbb{Q}(T_\iota)$ has cardinality $\aleph_1$ as is easy to show.  Now any positive sequence converging to $0$ in $K$ must be eventually constant because it has to cross uncountably many $T_\iota$.  So any Cauchy sequence in $K$ is eventually constant.  So any Cauchy sequence in $K$ is convergent.  So $K$ is complete.  But since $K$ contains $\mathbb{Q}$, it contains $\mathbb{R}$.  So we have a set of cardinality $\aleph_1$ containing $\mathbb{R}$, which proves the continuum hypothesis.
(The error, of course, is simply that the notion of "completeness" is wrong and its use is nonsense.  But if you tell it quickly enough, many people will fall for it.)
A: Another subtle variant of the induction fallacy suggested by Fedor Petrov.
Theorem: every graph without isolated nodes is connected.
Proof Induction on the number of nodes. Clearly the result is true for graphs with 1 (void statement) and 2 nodes. Now, assume we have proved the statement for graphs with up to $n$ nodes. Take a graph with $n$ nodes; by induction hypothesis it must be connected. Let's add a non-isolated node to it. As this node is not isolated, it is connected to one of the other $n$ nodes. But then it's easy to conclude that the whole graph of $n+1$ nodes is connected!
A: The Graham Pollak theorem is discussed at this link Combinatorial results without known combinatorial proofs .  I came up with a nice short and incomplete proof of it.  The tricky part for me was to realize it was incomplete.  Follow the commentary if you want to see my "D'oh" moment.
The induction started by taking an a,b complete bipartite subgraph of an (a+b) complete graph.
Gerhard "The Induction Looked So Pretty" Paseman, 2012.04.21 
A: Some years ago, I came up with this false proof of the irrationality of $\pi$. 
It suffices to prove that $x=\pi-3$ is irrational. 
For real $y$ with $0\le y\lt1$,
and positive integer $j$, define $d_j(y)$ to be the $j$th digit in the
decimal expansion of $y$. 
Let $r_1,r_2,\dots$ be
an enumeration of the rationals in $[0,1)$. The $\it value$ of this
enumeration is $n$ if $d_n(r_n)=d_n(x)$ and $d_j(r_j)\ne d_j(x)$ for $j\lt n$.
If there is no such $n$, then the value of the enumeration is infinite. Note
that if there is an enumeration of infinite value, then $x$ is irrational; it
cannot equal any of the enumerated rationals, as it differs from the first
rational in (at least) the first decimal place, from the second in the second,
etc. 
Note also that there are enumerations of arbitrarily large value. For, given
any $n$, you can find $n$ rationals such that the first differs from $x$ in
the first decimal, the second differs from $x$ in the second decimal, and so
on, and then any enumeration that starts off with these $n$ rationals will
have value greater than $n$. 
Now, the set of all enumerations of the rationals can be partially ordered by
value; if $E_1$ and $E_2$ are enumerations, then $E_1>E_2$ if the value of
$E_1$ exceeds the value of $E_2$. By Zorn's Lemma, there is an enumeration
maximal with respect to this order. This maximal enumeration cannot have a finite
value --- as we have seen, there are enumerations of arbitrarily great finite
value. So, it must have infinite value. So, $x$ is irrational. 
An alternative use for this argument is to apply it to prove that $1/3$ is irrational, the contradiction with the known rationality of $1/3$ thereby establishing that Zorn's Lemma is false. 
A: I think that the history of this wrong proof of the Riemann hypothesis is pretty interesting:
http://www.math.columbia.edu/~woit/wordpress/?p=707
In the end, it motivated a paper by Bombieri and Lagarias 
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3791
A: I like to amuse calculus students with this trick: let us calculate by integrating by parts:
$$
\int \frac{dx}{x}=\int (x')\frac{1}{x}\,dx=x\cdot\frac{1}{x}-\int x\cdot \left(\frac{1}{x}\right)'\,dx=1+\int\frac{dx}{x}, 
$$
and we simplify to $0=1$.
A: I just found the following false proof of the (correct!) Skolem-Mahler-Lech theorem, which I think is interesting.
Statement of (correct) theorem: Suppose $f(z) := \sum_{n=0}^{+\infty} a_n z^n \in \mathbb{C}[[z]]$ is rational.  Let $b_n$ equal $1$ when $a_n\neq 0$ and $0$ when $a_n = 0$.  Then $g(z) := \sum_{n=0}^{+\infty} b_n z^n$ is also rational.
False proof: Since $f$ is defined by a linear recurrence relation, correcting for the uninteresting constant term, we can interpret it as the series recognized by a weighted finite automaton on the unary language (i.e., consisting of words over the single letter $z$; so the automaton is just a digraph with complex “multiplicities” associated to edges, and $a_n$ is the number of paths of length $n$, each counted with a multiplicity given by the product of the multiplicities of the edges, from an initial node to a final node: see, e.g., Bousquet-Mélou, “Rational and algebraic series in combinatorial enumeration”, §2).  Now make this automaton deterministic (or at least unambiguous) while forgetting multiplicities: in the new automaton, the number of paths of length $n$ from an initial node to a final node is simply $b_n$, i.e., $1$ or $0$ according as there is or isn't such a path in the original automaton.  But for the same reason (backwards), $g$ is now given by a linear recurrence relation, so it is rational.
Comment: The error is simply that when forgetting multiplicities we also forget possible cancellations between them: two paths could have multiplicities summing to zero.  But the proof does work, and generalize to more variables, when $f$ is $\mathbb{N}$-rational, because no cancellation is possible: see the correct statements in Berstel & Reutenauer, Noncommutative Rational Series with Applications, esp. chapter 3 lemma 1.4.  So the idea of the proof isn't stupid and gives related theorems, and the conclusion as stated is correct, yet the proof probably can't be fixed to yield that exact conclusion (because it would then work over any field, which isn't true), so I think this qualifies as an interesting proof.
A: Not so much of a proof but rather a computation.
$$\frac{64}{16} = \frac{\not{6}4}{1\not{6}}= \frac{4}{1} = 4$$
by canceling the $6$s.
A: One night I proved that every module is flat. Let $M$ be an $R$-module and let $\mathfrak{a}$ be any ideal of the ring $R$. Tensoring the natural inclusion $i:\mathfrak{a} \to R$ we obtain $i_\ast : M \otimes \mathfrak{a} \to M \otimes R$ such that $i_\ast(x\otimes y)=x\otimes i(y)=x\otimes y$, for every $x\in M$ and $y \in \mathfrak{a}$. So $i_\ast$ is injective and we conclude that $M$ is flat...
A: I always liked this proof, from the theory of Umbral Calculus developed by Rota (See "Combinatorics: The Rota Way", by Joseph Kung, Gian Carlo Rota and Catherin Yan, chapter 4.2).
Proposition: Let $(a_n)_{n\geq 0}$ and $(b_n)_{n \geq 0}$ be sequences. Then
$$b_n=\sum_{k=0}^n\binom{n}{k} a_k  \ \text{ for all } n \Longleftrightarrow a_n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b_k \ \text{ for all } n.$$
The heuristic proof use the notion of "raising and lowering subscripts and superscript". Raising subscripts at the left side we obtain
$$b^n=\sum_{k=0}^n\binom{n}{k}a^k=(a+1)^n.$$
Hence, for all $n$,
$$a^n=(b-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b^k.$$
Lowering exponents, we obtain the inverse relation.
A: My favourites are "close" to formal false proofs in Coq.
1) In reply to a challenge by coq developer

Who can address this challenge: find a "simple" statement $T$ (simple in the sense that anyone with a minimal background in logics can understand) such that you can prove both $T$ and $\neg T$ in Coq.

Daniel Schepler solved it here. Daniel's proof was valid and passed coqchk, though it was not enough to prove False in Coq - Coq gave an "Universe inconsistency".
AFAICT the proof encoded a paradox.
2) Damien Pous announced and gave link to code

There is a bug with vm_compute and values obtained from functors applications:
  using the attached code, I can produce an assumption-free proof of False, or Bus errors.

False proofs in Coq are difficult because Coq produces a "certificate" that can be checked for validity (if one doesn't check the certificate and is happy with the compiler as most people do, it is much easier).
A: My favorite example is the following proof of the Cayley-Hamilton theorem, which caused me some disconcertion when I was a student. Let $A$ be a square matrix, and call $p(t) = \det(tI - A)$ its characteristic polynomial. Then $p(A) = \det(AI-A) = 0$.
A: One usual "proof" of  Leopoldt Conjecture  is that $\mathbb{Z}_p$ is $\mathbb{Z}$-flat, hence the rank of the $p$-adic completion of the units of a number field has the same rank of the units themselves (which is Leopoldt Conjecture) because you can obtain the completion simply as $\mathcal{O}^\times\otimes\mathbb{Z}_p$. 
A: Ma & Pa Kettle Math Lesson
YouTube
A: I can't remember where I first saw this: does anybody recognise it?
Let $I$ be the operator, from $C^0(\mathbb{R})$ to itself, which takes $f(x)$ to $\int_0^xf(z)dz$.
Since the exponential function $e(x)$ is its own derivative, we integrate both sides to get
$e(x) = I(e(x)) + 1$. Regarding $1$ as the identity operator, we can rearrange to get
$$(1-I)e(x) = 1,$$ and hence
$$e(x) = \frac{1}{1-I}1 = (1 + I + I^2 + \cdots)1 = 1 + x + \frac{x^2}{2} + \cdots.$$
A: We have $$\int \text{sec}^2(x)\tan(x)dx=\int \text{sec}^2(x)\tan(x)dx$$ $$\int \tan(x)d(\tan(x))=\int \text{sec}(x)d(\text{sec}(x))$$ $$\frac{\tan^2(x)}{2}+C=\frac{\text{sec}^2(x)}{2}+C$$ $$\frac{\tan^2(x)}{2}=\frac{\text{sec}^2(x)}{2}$$ $$\tan^2(x)=\text{sec}^2(x)$$ for all $x\in\mathbb{R}$.
A: The limit of a function, if it exists, is unique. Indeed, from
$\lim_{x\to x_0} f(x)=L_1$ and $\lim_{x\to x_0} f(x)=L_2$, exploiting symmetry and transitivity of the equality you readily deduce $L_1=L_2$.
A: I think nobody point to these interesting false proof:
Let $i=\sqrt{-1}$ be the complex number.
$1)$   $1=\sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=i\times i=-1$.
$2)$ We know that $x^\frac{2}{6}=x^\frac{1}{3}\Rightarrow (\sqrt{x^2})^\frac{1}{6}=(\sqrt{x})^\frac{1}{3}$. Now, let $x=-1$ and so we have: $$(\sqrt{(-1)^2})^\frac{1}{6}=(\sqrt{-1})^\frac{1}{3}\Rightarrow1=-1.$$
A: Doron Zeilberger proved that P is equal to NP
Abstract: Using 3000 hours of CPU time on a CRAY machine, we settle the notorious P vs. NP problem in the affirmative, by presenting a “polynomial” time algorithm for the NP-complete subset sum problem. Alas the complexity of our algorithm is $O(n^{10^{10000}})$ (with the implied constant being larger than the Skewes number).
