Examples of interesting false proofs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:23:13Z http://mathoverflow.net/feeds/question/94742 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs Examples of interesting false proofs joro 2012-04-21T14:26:12Z 2013-03-21T15:18:16Z <p>According to Wikipedia <a href="https://en.wikipedia.org/wiki/False_proof" rel="nofollow">False proof</a></p> <blockquote> <p>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.</p> </blockquote> <p>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.</p> <blockquote> <p>What are some examples of interesting false proofs?</p> </blockquote> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94743#94743 Answer by Angelo for Examples of interesting false proofs Angelo 2012-04-21T14:38:34Z 2012-04-21T14:38:34Z <p>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 $p(A) = \det(AI-A) = 0$.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94744#94744 Answer by Karl Schwede for Examples of interesting false proofs Karl Schwede 2012-04-21T14:49:01Z 2012-04-21T14:49:01Z <p>Not so much of a proof but rather a computation.</p> <p>$$\frac{64}{16} = \frac{\not{6}4}{1\not{6}}= \frac{4}{1} = 4$$</p> <p>by canceling the $6$s.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94745#94745 Answer by Pietro Majer for Examples of interesting false proofs Pietro Majer 2012-04-21T14:55:13Z 2012-04-21T14:55:13Z <p>Let me recycle <a href="http://mathoverflow.net/questions/44716/counterexample-for-the-open-mapping-theorem" rel="nofollow">this</a>.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94746#94746 Answer by Steven Landsburg for Examples of interesting false proofs Steven Landsburg 2012-04-21T15:18:35Z 2012-04-21T15:18:35Z <p>Ethan Akin's <a href="http://www.sciencedirect.com/science/article/pii/0022404978900324" rel="nofollow">"proof"</a> that all vector bundles are stably trivial, and hence the $K$-theory of any space must vanish:</p> <p>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 $$V\oplus V=V\oplus T$$.</p> <p>Let $P$ be the principal bundle associated with $V$. Pull $P$ back over itself to get a bundle $Q$:</p> <p><img src="http://www.landsburg.org/akin.gif"></p> <p>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 $V\oplus T$. </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94786#94786 Answer by Daniele for Examples of interesting false proofs Daniele 2012-04-21T22:40:10Z 2012-04-21T22:40:10Z <p>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...</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94804#94804 Answer by Gerhard Paseman for Examples of interesting false proofs Gerhard Paseman 2012-04-22T03:32:13Z 2012-04-22T03:32:13Z <p>The Graham Pollak theorem is discussed at this link <a href="http://mathoverflow.net/questions/5449/combinatorial-results-without-known-combinatorial-proofs/5560#5560" rel="nofollow">http://mathoverflow.net/questions/5449/combinatorial-results-without-known-combinatorial-proofs/5560#5560</a> . 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.</p> <p>Gerhard "The Induction Looked So Pretty" Paseman, 2012.04.21 </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94820#94820 Answer by Filippo Alberto Edoardo for Examples of interesting false proofs Filippo Alberto Edoardo 2012-04-22T08:37:07Z 2012-04-22T10:47:22Z <p>One usual "proof" of <a href="http://en.wikipedia.org/wiki/Leopoldt%27s_conjecture" rel="nofollow"> Leopoldt Conjecture </a> 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 $\mathcal{O}^\times\otimes\mathbb{Z}_p$. </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94831#94831 Answer by Michael Renardy for Examples of interesting false proofs Michael Renardy 2012-04-22T10:45:36Z 2012-04-22T22:57:58Z <p>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.</p> <p>After considerable difficulty with implementing his solution he consults his lieutenant. The lieutenant checks the calculation by evaluating 13*6: </p> <p>3*6=18 1*6=6</p> <p>Add them: 24.</p> <p>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.</p> <p>13</p> <p>13</p> <p>13</p> <p>13</p> <p>13</p> <p>13</p> <p>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.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94833#94833 Answer by Piero D'Ancona for Examples of interesting false proofs Piero D'Ancona 2012-04-22T10:52:41Z 2012-04-22T10:52:41Z <p>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 $n$.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94842#94842 Answer by Gerald Edgar for Examples of interesting false proofs Gerald Edgar 2012-04-22T12:24:07Z 2012-04-22T12:24:07Z <p><strong>Ma &amp; Pa Kettle Math Lesson</strong><br> <a href="http://www.youtube.com/watch?v=QlAluML8TG4" rel="nofollow">YouTube</a></p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94866#94866 Answer by Timothy Chow for Examples of interesting false proofs Timothy Chow 2012-04-22T18:37:22Z 2012-04-22T18:37:22Z <p>$$e^i = (e^i)^{(2\pi/2\pi)} = (e^{2\pi i})^{1/2\pi} = 1^{1/2\pi} = 1.$$</p> <p>I first saw this one many years ago, written on the wall of a bathroom stall in the Princeton University math department.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94868#94868 Answer by Timothy Chow for Examples of interesting false proofs Timothy Chow 2012-04-22T18:40:17Z 2012-04-22T22:13:07Z <p>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) &amp;= {d\over dx}(xe^{-x^2})\cr &amp;= {d\over dx} \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy\cr &amp;= \int_0^1 {\partial \over \partial x} \biggl({x^3\over y^2} e^{-x^2/y}\biggr)\,dy\cr &amp;= \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 $\int_0^1 0\,dy = 0$.</p> <p>The main idea for this proof comes from an entry in Gelbaum and Olmstead's book <i>Counterexamples in Analysis</i>.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94887#94887 Answer by Joel David Hamkins for Examples of interesting false proofs Joel David Hamkins 2012-04-22T23:08:31Z 2012-05-19T15:31:50Z <p><b>True Theorem</b> The symmetric groups (consisting of all permutations) on infinite sets of different cardinalities are not isomorphic. </p> <p>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</p> <p>See the question: <a href="http://mathoverflow.net/questions/12943/can-the-symmetric-groups-on-sets-of-different-cardinalities-be-isomorphic" rel="nofollow">Can the symmetric groups on sets of differing infinite cardinalities be isomorphic?</a> for further information and a correct proof. </p> <p>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. </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94889#94889 Answer by Yannic for Examples of interesting false proofs Yannic 2012-04-22T23:39:04Z 2012-04-23T12:02:28Z <p>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).</p> <p><strong>Proposition</strong>: Let $(a_n)_{n\geq 0}$ and $(b_n)_{n \geq 0}$ be sequences. Then $$b_n=\sum_{k=0}^n\binom{n}{k} a_k \ \text{ for all } n \Longleftrightarrow a_n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b_k \ \text{ for all } n.$$</p> <p>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.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94899#94899 Answer by Gerry Myerson for Examples of interesting false proofs Gerry Myerson 2012-04-23T01:35:21Z 2012-04-23T01:35:21Z <p>Some years ago, I came up with this false proof of the irrationality of $\pi$. </p> <p>It suffices to prove that $x=\pi-3$ is irrational. </p> <p>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 $y$. </p> <p>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. </p> <p>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 $n$. </p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94909#94909 Answer by Hurkyl for Examples of interesting false proofs Hurkyl 2012-04-23T06:44:54Z 2012-04-23T06:44:54Z <p><strong>Theorem:</strong> All people have the same eye color.</p> <p>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.</p> <p>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.</p> <p>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. $\square$</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/95144#95144 Answer by Daniele for Examples of interesting false proofs Daniele 2012-04-25T10:32:42Z 2012-04-25T10:32:42Z <p>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 $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$ </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/95163#95163 Answer by joro for Examples of interesting false proofs joro 2012-04-25T14:48:12Z 2012-04-25T14:48:12Z <p>My favourites are "close" to <em>formal</em> false proofs in Coq.</p> <p>1) In reply to a challenge by coq developer</p> <blockquote> <p>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.</p> </blockquote> <p>Daniel Schepler solved it <a href="https://sympa-roc.inria.fr/wws/arc/coq-club/2011-06/msg00099.html?checked_cas=2" rel="nofollow">here</a>. 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.</p> <p>2) Damien Pous <a href="https://sympa-roc.inria.fr/wws/arc/coqdev/2011-07/msg00018.html" rel="nofollow">announced and gave link to code</a></p> <blockquote> <p>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.</p> </blockquote> <p>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).</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/95596#95596 Answer by Marc Palm for Examples of interesting false proofs Marc Palm 2012-04-30T18:04:52Z 2012-04-30T18:04:52Z <p>I think that the history of this wrong proof of the Riemann hypothesis is pretty interesting:</p> <p><a href="http://www.math.columbia.edu/~woit/wordpress/?p=707" rel="nofollow">http://www.math.columbia.edu/~woit/wordpress/?p=707</a></p> <p>In the end, it motivated a paper by Bombieri and Lagarias </p> <p><a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3791" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3791</a></p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/95613#95613 Answer by Feldmann Denis for Examples of interesting false proofs Feldmann Denis 2012-04-30T20:19:14Z 2012-04-30T20:19:14Z <p>$\pi$ 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). </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97393#97393 Answer by B D for Examples of interesting false proofs B D 2012-05-19T13:14:41Z 2012-05-19T13:14:41Z <p><strong>Theorem:</strong> Every totally disconnected set has the discrete topology.</p> <p><strong>Proof:</strong> 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.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97398#97398 Answer by James Cranch for Examples of interesting false proofs James Cranch 2012-05-19T14:05:31Z 2012-05-19T15:39:46Z <p>I can't remember where I first saw this: does anybody recognise it?</p> <p>Let $I$ be the operator, from $C^0(\mathbb{R})$ to itself, which takes $f(x)$ to $\int_0^xf(z)dz$.</p> <p>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 $$e(x) = \frac{1}{1-I}1 = (1 + I + I^2 + \cdots)1 = 1 + x + \frac{x^2}{2} + \cdots.$$</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97416#97416 Answer by John Engbers for Examples of interesting false proofs John Engbers 2012-05-19T18:06:12Z 2012-05-19T18:06:12Z <p>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.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97467#97467 Answer by Frank for Examples of interesting false proofs Frank 2012-05-20T07:25:23Z 2012-05-20T07:25:23Z <p>Claim: All positive integers are equal.</p> <p>Proof. Let $a$ and $b$ be positive integers. Let $n = M(a,b)$, where $M$ denotes the larger of $a$ and $b$. By induction on $n$ we get</p> <p>If $n=1$, we clearly have $a=b=1$.</p> <p>General case: Suppose $n=k$ implies $a=b$. Let now $n=k+1$. We have $M(a,b)=k+1$ thus $M(a-1,b-1)=k$, and from induction hypothesis $a-1=b-1$, so $a=b$ and the claim follows.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/99794#99794 Answer by Omnitic for Examples of interesting false proofs Omnitic 2012-06-16T14:46:25Z 2012-06-16T14:46:25Z <p>An excelent example is the iscosceles triangle fallacy. Here is a link to it in wikipedia <a href="http://en.wikipedia.org/wiki/Mathematical_fallacy#Fallacy_of_the_isosceles_triangle" rel="nofollow">http://en.wikipedia.org/wiki/Mathematical_fallacy#Fallacy_of_the_isosceles_triangle</a> </p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/99796#99796 Answer by Chandrasekhar for Examples of interesting false proofs Chandrasekhar 2012-06-16T16:08:04Z 2012-06-17T06:44:00Z <p>Here is an interesting false proof as to how to multiply $2 \cdot 2$. Taken from <a href="http://math-fail.com/2011/09/what-is-2-x-2.html" rel="nofollow">this link</a>.</p> <p><img src="http://img.math-fail.com/wp-content/uploads/joke.png" alt="alt text"></p> <hr> <p>$\Large\textbf{Another example}$:</p> <p><img src="http://skullsinthestars.files.wordpress.com/2008/12/neg1equals1.png?w=640" alt="alt text"></p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/118893#118893 Answer by Gro-Tsen for Examples of interesting false proofs Gro-Tsen 2013-01-14T16:10:52Z 2013-01-14T16:10:52Z <p>Here's a nice <strong>false proof of the continuum hypothesis</strong>.</p> <p>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 $\iota&lt;\omega_1$, 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.</p> <p>(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.)</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/118910#118910 Answer by Lior Bary-Soroker for Examples of interesting false proofs Lior Bary-Soroker 2013-01-14T19:00:06Z 2013-01-14T19:00:06Z <p><strong>Theorem</strong>: Every bounded differential function $f\colon \mathbb{R}\to \mathbb{R}$ is constant. </p> <p><strong>Proof</strong>. By assumption there exist real numbers $M,N$ such that<br> $$N\leq f(x) \leq M.$$ Taking derivatives we get $$0\leq f'(x)\leq 0.$$ Hence $f'(x)=0$ so $f$ is constant. QED</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/119006#119006 Answer by Noam D. Elkies for Examples of interesting false proofs Noam D. Elkies 2013-01-15T18:00:37Z 2013-01-15T18:00:37Z <p><strong>Theorem.</strong> $\int_0^\infty \sin x \phantom. dx/x = \pi/2$.</p> <p><strong>Poof.</strong> 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.</p> <p>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.</p> <p>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:</p> <p>$$\frac\pi2 = \int_0^\infty \sin^2 x \frac{dx}{x^2} = \int_0^\infty \sin^2 x \phantom. d(-1/x) = \int_0^\infty \frac1x d(\sin^2 x) = \int_0^\infty 2 \sin x \cos x \frac{dx}{x};$$ 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.</p> <p><strong>Exercise</strong> 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, <em>Nouvelles tables d'intégrales définies</em>, Amsterdam 1867; the general case is 3.827#1, from Gröbner and Hofreiter's <em>Integraltafel</em> II, Springer: Vienna and Innsbruck 1958.]</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/124986#124986 Answer by Denis Serre for Examples of interesting false proofs Denis Serre 2013-03-19T16:58:35Z 2013-03-21T10:35:49Z <p>I have known the following for 45 years: <em>in the Euclidian plane, every triangle is isosceles</em>.</p> <p>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 righttriangles $BIJ$ and $CIK$, we obtain that $BJ=CK$. We conclude that $$AB=AJ+JB=AK+KC=AC.$$</p> <p>The falsity is that one of $J$ or $K$ is <em>in</em> the triangle, and the other one is <em>out</em>. Therefore one of the sums above (and only one) should be a difference.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/125012#125012 Answer by Russ Woodroofe for Examples of interesting false proofs Russ Woodroofe 2013-03-19T19:38:10Z 2013-03-19T19:38:10Z <p>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 $$\operatorname{Var} ( \frac{1}{n} \sum_{i=1}^n X_i ) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n} \rightarrow 0 \textrm{ as } n\rightarrow\infty$$ 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.)</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/125071#125071 Answer by Joseph Van Name for Examples of interesting false proofs Joseph Van Name 2013-03-20T14:13:36Z 2013-03-20T14:13:36Z <p>e is irrational.</p> <p>Assume to the contrary that e were rational. Then e would be e-rational, a contradiction.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/125082#125082 Answer by Shahrooz for Examples of interesting false proofs Shahrooz 2013-03-20T15:38:34Z 2013-03-20T15:38:34Z <p>I think nobody point to these interesting false proof:</p> <p>Let $i=\sqrt{-1}$ be the complex number.</p> <p>$1)$ $1=\sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=i\times i=-1$.</p> <p>$2)$ 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: $$(\sqrt{(-1)^2})^\frac{1}{6}=(\sqrt{-1})^\frac{1}{3}\Rightarrow1=-1.$$</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/125084#125084 Answer by Torsten Schoeneberg for Examples of interesting false proofs Torsten Schoeneberg 2013-03-20T16:07:48Z 2013-03-20T16:07:48Z <p>In S. Bosch's <em>Algebra</em>, exercise 3.4.2 is to find an error in the following existence proof of an algebraic closure of a field $K$ (my translation):<br> "Consider all algebraic extensions of $K$. Since for a totally ordered (w.r.t. inclusion) family <code>$(K_i)_{i \in I}$</code> 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 $K$."</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/125177#125177 Answer by joro for Examples of interesting false proofs joro 2013-03-21T15:18:16Z 2013-03-21T15:18:16Z <p><a href="http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/pnp.pdf" rel="nofollow">Doron Zeilberger proved that P is equal to NP</a></p> <p>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).</p>