Proofs without words 4

 Viviani's Theorem: In an equilateral triangle, the sum of the distances of an interior point to the three sides equals the altitude of the triangle.

Image from the web, attribution unknown

(Image from the web, attribution unknow

An overkill proof of the divergence of the harmonic series

 The following proof can be found in the MO thread [1].

Assume that the harmonic series $\sum_{n=1}^{+ \infty}\frac{1}{n}$ converges. Then the sequence of functions $\{f_n\}$ defined by $f_n = \frac{1}{n} \chi_{[0, \, n]}$ is dominated by the function $$g = \chi_{[0, \, 1]} + \frac{1}{2} \chi_{[1, \, 2]}+\frac{1}{3} \chi_{[2, \, 3]}+ \frac{1}{4} \chi_{[3, \, 4]}+\ldots,$$ which is by assumption absolutely integrable over $\mathbb{R}$. 

By applying Lebesgue's Dominate Convergence Theorem [2] we get: $$1 = \lim_n \int_{\mathbb{R}} f_n(x) dx = \int_{\mathbb{R}}  \lim_n f_n(x)dx = 0,$$ a contradiction. $\Box$

References

[1] https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts

[2] https://en.wikipedia.org/wiki/Dominated_convergence_theorem

Four colors suffice

The "Four colors suffice" postmark used by the University of Illinois at Urbana-Champaign (UIUC) after the celebrated proof of the Four Colors Problem by Kenneth Appel and Wolfgang Haken [1]. 

According to [2], the proof "was greeted with enthusiasm [...] but also with skepticism, great disappointment and outright rejection. Their extensive use of computers was widely criticized, and raised philosophical questions as to whether a proof is valid if it cannot be checked by hands."

References.

[1] K. Appel and W. Haken: Every planar map is four colorable I-II, Illinois J. Math. 21, No. 3 (1977).
[2] Wolfgang Haken (1928-2022), Memorial Tribute, Notices of the AMS 70, Number 9 (2023). DOI: https://doi.org/10.1090/noti2781

Automorphisms of the real numbers

Theorem. The only field automorphism of $\mathbb{R}$ is the identity.

Proof. If $f \colon \mathbb{R} \to \mathbb{R}$ is a field automorphism, we want to prove that $f=\operatorname{id}$. The proof will be divided into three steps, following the treatment given in Makoto Kato’s answer in [1].

Step 1. $f \colon \mathbb{R} \to \mathbb{R}$ is order-preserving. 

Let $x>0$ be a positive real number. Then there exists $y$ such that $x=y^2$, hence $f(x)=f(y)^2 > 0$. If $b > a$, then $b - a >0$ and so $f(b) - f(a)$ = $f(b - a) > 0$, i.e. $f(b) > f(a).$ 

Step 2. $f \colon \mathbb{R} \to \mathbb{R}$ fixes all elements of the subfield $\mathbb{Q}$. 

Since $f(1)=1$ and every natural number n can be written as $n=1+1+ \ldots +1,$ we infer $f(n)=n$. This shows that $f$ preserves all elements of $\mathbb{N}$, and so it preserves all elements of the subring $\mathbb{Z}$. Consequently, it preserves all elements of the fraction field of $\mathbb{ℤ},$ which is $\mathbb{Q}$.

Step 3. We can now conclude by using the density of $\mathbb{Q}$ in $\mathbb{R}$. In fact, take any real number $x$ and let $q, \, r$ be two rational numbers such that $q < x < r$. By the previous two steps, we get $q < f(x) < r.$ Since $q$ and $r$ can be chosen to be arbitrarily close to $x$, it follows that $f(x)=x. \quad \square$ 

Remark. The situation is entirely different for the field $\mathbb{C}$ of complex numbers. Using an argument similar to the one above, it is possible to show that the only continuous automorphisms of $\mathbb{C}$ are the identity and the complex conjugation. However, admitting the Axiom of Choice it is possible to prove that there are infinitely many non-continuous automorphisms, see [2]. In fact, the continuous automorphisms of $\mathbb{C}$ are precisely those fixing the subfield $\mathbb{R}$.

References.
[1] MSE449404-Is an automorphism of the field of real numbers the identity map?

[2] MSE412010-Wild automorphisms of the complex numbers

Trick or treat

A proof of the Fundamental Theorem of Algebra based on Differential Topology

The following proof of Fundamental Theorem of Algebra, which uses techniques of Differential Topology, can be found in [1, pp. 8-9].

Theorem. Every non-constant polynomial $P \in \mathbb{C}[z]$ must have a zero.

Proof. Using the stereographic projection from the point $\infty=(0, \, 0, \, 1)$, we can extend the polynomial function $P \colon \mathbb{C} \to \mathbb{C}$ to a function $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ such that $f(\infty)=\infty$. A straightforward use of the change of coordinates $z=1/w$ shows that $f$ is smooth on the whole of $\mathbb{S}^2$, including the point $\infty$.

Let us remark that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ has a finite number of critical points (and so a finite number of critical values), corresponding to the zeros of the derivative $P’$ and possibly to the point $\infty.$

If $0\in \mathbb{C}$ corresponds to a critical value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ then, by definition, it has a preimage via $P \colon \mathbb{C}^2 \to \mathbb{C}^2$, and there is nothing to prove.

We may therefore assume that $0\in \mathbb{C}$ corresponds to a regular value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$. The locus of these regular values is the complement of finitely many points on the sphere, so it is connected.

From the Implicit Function Theorem for differentiable manifolds, it follows that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ is a local diffeomorphism when restricted to the locus of regular values. So the number of points which are preimages of a fixed one is locally constant on such locus, hence constant (being the locus connected).

Since $\deg(P) \geq 1$, there are surely (infinitely many) points $y \in \mathbb{S}^2$ such that $f^{-1}(y) $ is non-empty. Thus, the same must be true for the point corresponding to $0 \in \mathbb{C}.$

In other words, $0$ is in the image of $P$. $\square$

References.

[1] J. W. Milnor: Topology from the Differentiable Viewpoint, Princeton University Press 1965.

An overkill way to recover the Taylor expansion of $e^x$

Let us consider the linear operator of Banach spaces $$F \colon C[0, \, 1/2] \to C[0, \, 1/2], \quad F(f(x)):= \int_0^x f(t)dt.$$ Then the identity $\int_0^x e^tdt=e^x-1$ can be rewritten as $$(I-F)(e^x)=1. \quad (\sharp)$$ The operator $F$ is bounded and we have $$||F|| = \sup_{f \in C[0, \, 1/2]} \frac{||F(f)||}{||f||} \leq \frac{1/2 \, ||f||}{||f||} = \frac{1}{2} <1.$$ This implies that $I-F$ is invertible as a bounded linear operator and  moreover $$(I-F)^{-1} = \sum_{k=0}^{+ \infty} F^k.$$ Substituting in $(\sharp)$, we get $$e^x=(I-F)^{-1}(1)=\left(\sum_{k=0}^{+ \infty} F^k \right) (1) = \sum_{k=0}^{+ \infty} F^k(1) = \sum_{k=0} ^{+ \infty}\frac{x^k}{k!},$$ which is the well-known Taylor series expansion of  $e^x$.

Fields

Hate is a lack of imagination.

Graham Green, "The power and the glory".

Il funerale di Renato Caccioppoli, Napoli 1959.

 

Photo copyright: Archivio fotografico Carbone - Napoli

Triangoli con lati razionali aventi stessa area e perimetro.

Supponiamo di avere un triangolo $T$ aventi lati di lunghezza razionale $a$, $b$, $c$ e area razionale $A$, e di volere determinare un altro triangolo $T'$ (non congruente a $T$) con lati di lunghezza razionale e avente lo stesso perimetro $2p=a+b+c$ e la stessa area $A$. 

Indicando con $$a'=a+x, \quad b'=b+y, \quad c'=c-x-y$$ le lunghezze dei lati del nuovo triangolo $T'$, applicando la formula di Erone otteniamo con semplici passaggi l'eguaglianza

\begin{equation} \label{erone} \big(x- \alpha\big) \big(y-\beta \big) \big(x+y+\gamma \big)- \frac{A^2}{p}= 0, \end{equation}

dove $$\alpha=p-a, \quad \beta = p-b, \quad \gamma = p-c.$$ Si può vedere \eqref{erone} come l'equazione di una curva ellittica $E$ definita su $\mathbb{Q}$, il cui luogo dei punti reali consiste di quattro rami: un ovale compatto contenuto nel triangolo definito nel piano $(x, \, y)$ dalle tre rette di equazione $$x= \alpha,  \quad y=\beta, \quad y=-x- \gamma$$ e tre rami illimitati aventi ciascuno due delle suddette rette come asintoti (si veda la Figura 1).

Figura 1

L'ovale compatto corrisponde esattamente ai valori $(x, \, y, \, z)$ per i quali esiste un triangolo  $T'$ che ha area $A$. Esso contiene in generale almeno sei punti a coordinate razionali: il punto $o=(0, \, 0, \, 0)$, corrispondente al triangolo di partenza $T$ avente lati $(a, \, b, \, c)$, e i cinque punti corrispondenti alle permutazioni  non banali di $(a, \, b, \, c)$. 

Il problema geometrico originale di determinare il triangolo $T'$ si traduce quindi nel seguente problema di Teoria dei Numeri:

Problema. Si determini un punto razionale non banale nell'ovale compatto della curva ellittica $E$.

Non intendiamo qui risolvere il problema nella sua generalità, ma ci limitiamo a trattare un importante esempio. Si consideri il famoso triangolo rettangolo $T$ di lati $(a, \, b, \, c)=(3, \, 4, \,5)$, che ha perimetro $6$ e area $12$. In questo caso, l'equazione di $E$ diventa \begin{equation} \label{erone2} \big(x- 3\big) \big(y- 2 \big) \big(x+y+1 \big)- 6= 0. \end{equation}

Per determinare un punto razionale non banale nell'ovale compatto, utilizziamo il classico metodo delle rette secanti. 

L'ovale compatto contiene il punto razionale $o=(0, \, 0, \, 0)$, corrispondente a $T$, il punto razionale $p_1=(1, \, 1)$, corrispondente al triangolo di lati $(a, \, b, \, c)=(4, \,5, \, 3)$ e il punto razionale $p_2=(1, \, -1)$, corrispondente al triangolo di lati $(a, \, b, \, c)=(4, \, 3, \, 5)$.

Le retta $y-x=0$ passa per $o$ e $p_1$ e interseca $E$ nell'ulteriore punto razionale $p_3=(7/2, \, 7/2)$; tale punto non è ancora una soluzione al nostro problema, in quanto appartiene ad uno dei rami illimitati. 

Possiamo però considerare la retta passante per $p_3$ e $p_2$, che interseca $E$ nell'ulteriore punto razionale $p_4=(38/21, \, 16/35)$. Tale punto sta nell'ovale limitato, quindi corrisponde effettivamente ad una soluzione  del problema (si veda la Figura 2, realizzata con il calcolatore grafico Desmos).

Figura 2

Infatti, $p_4$ corrisponde al triangolo $T'$ di lati razionali $$(a, \, b, \, c)=(101/21, \, 156/35, \, 41/15).$$ È semplice verificare che esso è un triangolo acutangolo di perimetro $12$ e area $6$, gli stessi valori del triangolo rettangolo $T$ di partenza (Figura 3).

Figura 3

Varianti di questo procedimento permettono di determinare ulteriori triangoli con lati razionali, perimetro $12$ e area $6$. Ad esempio, partendo dalla retta tangente ad $E$ nel punto $o=(0, \, 0, \, 0)$ si può ottenere il triangolo ottusangolo di lati $$(a, \, b, \, c)=(35380/10153, \, 81831/16159, \, 27689/8023),$$ si veda la Figura 4.

Figura 4