Pseudo Proofs that are intuitively reasonable

Solution 1:

Cayley-Hamilton Theorem. Let $A$ be an $n\times n$ matrix, and let $f(t)$ be its characteristic polynomial. Then $f(A)=0$.

"Proof." $f(t) = \det(A-tI)$. Therefore, $f(A) = \det(A-AI) = \det(A-A) = \det(0) = 0$.

Issues. One problem may not be obvious... the equation "$f(A)=0$" is really saying that the matrix we get via the evaluation (by identifying the underlying field with the subring of scalar matrices) is the zero matrix. However, the "proof" claims to prove that $f(A)$, which is supposed to be a matrix, is equal to the value of a determinant, which is a scalar.

As to why it "gives the right answers"... well, because the theorem is true. I am reminded of what Hendrik Lenstra once said in class after presenting an idea for a proof and explaining why it didn't quite work:

The problem with incorrect proofs to correct statements is that it is hard to come up with a counterexample.

Solution 2:

$$\begin{align*} x^{x^{x^{\scriptstyle\ldots}}} &= 2\\ x^{\left(x^{x^{x^{\scriptstyle\ldots}}}\right)} &= 2\\ x^2 &= 2\\ x &= \sqrt{2}\\ \end{align*}$$

Solution 3:

This is one that comes up on math.SE from time to time.

Proposition: $[0, 1]$ has the same cardinality as $[0, 1]^2$.

"Proof 1." Write $(x, y) \in [0, 1]^2$ in their binary expansions $x = 0.x_1 x_2 x_3 ..., y = 0.y_1 y_2 y_3 ...$, and consider the map which sends $(x, y)$ to $0.x_1 y_1 x_2 y_2 x_3 y_3 ...$.

Of course the problem with this proof is that the map is not well-defined since numbers do not have unique binary expansions, e.g. we have $0.1 = 0.0111...$.

"Proof 2." Repeat Proof 1, but this time don't allow any terminating binary expansions except $0.0$; instead, when a number has two binary expansions, pick the one which ends with infinitely many ones.

But now our map isn't surjective! The restriction on binary expansions above means, for example, that we can never hit $0.011010101...$.

A neat fix is just to show that $[0, 1]$ has the same cardinality as $\{ 0, 1 \}^{\mathbb{N}}$, which is straightforward to do by Cantor-Bernstein-Schroeder, and then the above argument actually works.

Solution 4:

Ito's Lemma: Assume that $x(t)$ has an SDE of the type $dx(t)=\mu(t)dt+\sigma(t)dW(t)$ for $\mu$ and $\sigma$ adapted processes. Also assume that $f\in C^{1,2}$. Define $z(t):=f(t,x(t))$, then $z(t)$ obeys the following SDE: $$ df(t,x(t))=\left\{\frac{\partial f}{\partial t}+\mu\frac{\partial f}{\partial x}+\frac{1}{2}\sigma^2\frac{\partial^2f}{\partial x^2}\right\}dt+\sigma\frac{\partial f}{\partial x}dW(t) $$ "Proof": Taylor expand to obtain $$ df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2f}{\partial x^2}dx^2+ \frac{1}{2}\frac{\partial^2f}{\partial t^2}dt^2+\frac{\partial^2f}{\partial t\partial x}dtdx. $$ Then use that $(dt)^2\approx 0$, $(dW)^2=dt$ and $dWdt\approx 0$, yielding the result.

Solution 5:

Let $X$ be an algebraic data type, i.e. an element of a semiring. Then the type $L$ of lists of $X$s is

$$L = 1 + XL$$

i.e. a list is either empty ($1$) or it is a pair consisting of an $X$ and another list ($XL$). Then we can write

$$(1-X)L = 1 \quad\Rightarrow\quad L = \frac{1}{1-X} \quad\Rightarrow\quad L = 1+X+X^2+X^3+\cdots$$

i.e. a list is either empty, or it has one element, or it has two elements, or it has three elements, etc. Which is true, despite that fact that we used subtraction and division in our "proof", neither of which are valid operations in a semiring.

The fact that this "proof" works is surprisingly deep, and has ties to category theory. This paper by Fiore and Leinster proves the result, showing that for some cases, proofs that 'pretend' the objects under consideration are complex numbers give valid results for semirings, monoids, rings and other classes of algebraic structure. Among other things they show that there is a bijection between the type of binary trees $T$, and the type of seven-tuples of binary trees, $T^7$, i.e. that

$$T=T^7$$

which follows trivially from the defining equation for binary trees holding elements of type $X$

$$T = 1 + XT^2$$

if you allow yourself free access to field operations.