Jekyll2023-01-15T10:21:20+01:00https://www.riemannslibrary.com/feed.xmlRiemann’s LibraryExploring the intersection of math, science, and culture. Follow us for educational and interesting content.Nadir SOUALEMProperties of Euler’s totient function2023-01-15T00:00:00+01:002023-01-15T00:00:00+01:00https://www.riemannslibrary.com/group-theory/euler-totient-function/2023/01/15/Properties-of-Euler's%20totient%20functionDefinition of Euler’s totient function $\varphi$

Euler’s totient function $\varphi$ counts the number of positive integers less than or equal to $n$ that are relatively prime (coprime) to $n$:

\begin{aligned} \forall n \in \mathbb{N}^*, \varphi(n) &=\operatorname{card}\{1\leqslant k\leqslant n \mid \operatorname{gcd}(k,n)=1\} \\ &=\sum_{k=1}^{n} [\gcd(k,n) = 1] \end{aligned}

where $\gcd(a,b)$ is the greatest common divisor of $a$ and $b$, and $[P]$ is the Iverson bracket, which is defined as:

$[P] = \begin{cases} 1 & \text{if P is true} \\ 0 & \text{if P is false} \end{cases}$

Euler’s totient function has a number of interesting properties, including the following

# Some properties of Euler’s totient function $\varphi$

Euler’s totient function $\varphi$ counts the number of positive integers less than or equal to $n$ that are relatively prime (coprime) to $n$:

$\forall n \in \mathbb{N}^*, \varphi(n)=\operatorname{card}\{1\leqslant k\leqslant n \mid \operatorname{gcd}(k,n)=1\}$

For example, $\varphi(9) = 6$ because ${1, 2, 4, 5, 7, 8}$ are all relatively prime to $9$.

Exercise 1. Show that if $n$ is prime, then $\varphi(n) = n-1$

Exercise 2. Invertible elements in $\mathbb{Z} / n \mathbb{Z}$ for a group under multiplication, denoted by $$(\mathbb{Z} / n \mathbb{Z})^{\times}=\{x \in \mathbb{Z} / n \mathbb{Z} \mid x \text { invertible }\} \text {. }$$

Show that: $\forall n \geq 2, \varphi(n)=\operatorname{card}{\left(\mathbb{Z} / n \mathbb{Z}^{\times})\right.}$

Exercise 3. Using the Chinese Remainder Theorem, if $m$ and $n$ are relatively prime, there exists a ring isomorphism:

$\mathbb{Z} / n m \mathbb{Z} \simeq \mathbb{Z} / n \mathbb{Z} \times \mathbb{Z} / m \mathbb{Z}$

Show that for $m$ and $n$ relatively prime, we have: $\varphi(mn) = \varphi(m)\varphi(n)$

In other words, $x, y \in(\mathbb{Z} / n \mathbb{Z})^{\times} \Longrightarrow x y \in(\mathbb{Z} / n \mathbb{Z})^{\times}$. Definition Euler’s totient function is $$\phi(n)=\#(\mathbb{Z} / n \mathbb{Z})^{\times}=\#\{0 \leqslant x<n \mid \operatorname{gcd}(x, n)=1\}$$

# Solutions

Exercise 1. Show that if $n$ is prime, then $\varphi(n) = n-1$

Solution. Euler’s totient function $\varphi$ counts the number of positive integers less than or equal to $n$ that are relatively prime (coprime) to $n$.

\begin{aligned} \forall n \in \mathbb{N}^*, \varphi(n) &=\operatorname{card}\{1\leqslant k\leqslant n \mid \operatorname{gcd}(k,n)=1\} \\ &=\sum_{k=1}^{n} [\gcd(k,n) = 1] \end{aligned}

where $\gcd(a,b)$ is the greatest common divisor of $a$ and $b$, and $[P]$ is the Iverson bracket, which is defined as:

$[P] = \begin{cases} 1 & \text{if P is true} \\ 0 & \text{if P is false} \end{cases}$

Consider the definition of a prime number, the only positive integer less than or equal to a prime $n$ which is not prime to $n$ is $n$ itself. If $n$ is a prime

$\underbrace{\operatorname{gcd}(k,n)=1, \quad \forall k \in \llbracket1,n \llbracket}_{n-1 \text{ numbers}}\text{ and } \operatorname{gcd}(n,n)=n$ $\color{green}{\varphi(n)=\operatorname{card}\{1\leqslant k\leqslant n \mid \operatorname{gcd}(k,n)=1\}=n-1}$

Exercise 2. Invertible elements in $\mathbb{Z} / n \mathbb{Z}$ for a group under multiplication, denoted by $$(\mathbb{Z} / n \mathbb{Z})^{\times}=\{\bar{x} \in \mathbb{Z} / n \mathbb{Z} \mid \bar{x} \text { invertible }\} \text {. }$$

Show that: $\forall n \geq 2, \varphi(n)=\operatorname{card}{(\mathbb{Z} / n \mathbb{Z})^{\times}}$

Solution. Let $\bar{x}\in(\mathbb{Z} / n \mathbb{Z})^{\times}, \bar{x} \text { invertible }$:

\begin{aligned} & \Longleftrightarrow \exists\bar{y}\in(\mathbb{Z} / n \mathbb{Z})^{\times} \mid \bar{x} \cdot\bar{y}=\overline{1} \\ & \Longleftrightarrow \exists\bar{y}\in(\mathbb{Z} / n \mathbb{Z})^{\times} \mid\overline{xy} =\overline{1} \text { since } (\mathbb{Z} / n \mathbb{Z})^{\times}\text { is a group for muliplication }\\ & \Longleftrightarrow \exists x, y \in \mathbb{Z} \mid x y \equiv 1 \bmod n \\ & \Longleftrightarrow \exists x, y \in \mathbb{Z} \mid x y=1+n k \text { for some } k \in \mathbb{Z} \\ & \Longleftrightarrow \exists x, y \in \mathbb{Z} \mid x y-n k=1 \text { for some } k \in \mathbb{Z} \\ & \Longleftrightarrow\operatorname{gcd}(x, n)=1 \text{ using Bézout's Identity} \end{aligned}

Bézout’s identity — For nonzero integers $a$ and $b$, let $d$ be the greatest common divisor $d=\operatorname{gcd⁡}(a,b)$. Then, there exist integers $u$ and $v$ such that $au+bv=d$

Here: $d=1=\operatorname{gcd}⁡(a=x,b=n)$, $⁡u=y$ and $v=-k$.

Euler’s totient function $\varphi$ counts the number of positive integers less than or equal to $n$ that are relatively prime (coprime) to $n$:

\begin{aligned} \forall n \in \mathbb{N}^*, \varphi(n) &=\operatorname{card}\{1\leqslant x\leqslant n \mid \operatorname{gcd}(x,n)=1\} \end{aligned}

Conclusion. $$\color{green}{\operatorname{card}\left(\mathbb{Z} / n \mathbb{Z})^{\times}\right.=\operatorname{card}\{\bar{x} \in \mathbb{Z} / n \mathbb{Z} \mid \bar{x} \text { invertible }\}=\operatorname{card}\{1\leqslant x\leqslant n \mid \operatorname{gcd}(x,n)=1\}=\varphi(n)}$$

Exercise 3. Using the Chinese Remainder Theorem, if $m$ and $n$ are relatively prime, there exists a ring isomorphism:

$\mathbb{Z} / mn \mathbb{Z} \simeq \mathbb{Z} / m \mathbb{Z} \times \mathbb{Z} / n \mathbb{Z}$

Show that for $m$ and $n$ relatively prime, we have: $\varphi(mn) = \varphi(m)\varphi(n)$, where the Euler’s totient function $\varphi$ counts the number of positive integers less than or equal to $n$ that are relatively prime (coprime) to $n$:

\begin{aligned} \forall n \in \mathbb{N}^*, \varphi(n) &=\operatorname{card}\{1\leqslant x\leqslant n \mid \operatorname{gcd}(x,n)=1\} \end{aligned}

Solution. We have seen in Exercise 2 that:

$\forall n \geq 2, \varphi(n)=\operatorname{card}\{(\mathbb{Z} / n \mathbb{Z})^{\times}\}$ \begin{aligned} \text { Let }\mathbb{A} \text { and } \mathbb{B} \text { be two rings. } & (a, b) \text { is an invertible element of } \mathbb{A} \times \mathbb{B} \\ & \Leftrightarrow \exists\left(a^{\prime}, b^{\prime}\right) \in \mathbb{A} \times \mathbb{B},(a, b) \times\left(a^{\prime}, b^{\prime}\right)=\left(1_{\mathbb{A}}, 1_{\mathbb{B}}\right) \\ & \Leftrightarrow \exists a^{\prime} \in \mathbb{A}, \exists b^{\prime} \in \mathbb{B}, a \times a^{\prime}=1_{\mathbb{A}} \text { and } b \times b^{\prime}=1_{\mathbb{B}}\\ & \Leftrightarrow a \text { is an invertible element of } \mathbb{A} \text { and } b \text { is an invertible element of } \mathbb{B}\\ & \Leftrightarrow (\mathbb{A} \times \mathbb{B})^{\times}=\mathbb{A}^{\times} \times \mathbb{B}^{\times} \end{aligned}

Then we have: $(\mathbb{Z} / mn \mathbb{Z})^{\times} \simeq (\mathbb{Z} / m \mathbb{Z} \times \mathbb{Z} / n \mathbb{Z})^{\times}\simeq (\mathbb{Z} / m \mathbb{Z})^{\times} \times (\mathbb{Z} / n \mathbb{Z})^{\times}$

Conclusion.

$\color{green}{\varphi(mn)=\operatorname{card}(\mathbb{Z} / mn \mathbb{Z})^{\times}=\operatorname{card}(\mathbb{Z} / m \mathbb{Z})^{\times} \times \operatorname{card}(\mathbb{Z} / n \mathbb{Z})^{\times}=\varphi(m)\times \varphi(n)}$]]>
Which-is-greater 1000^1001 or 1001^1000 ?2023-01-01T00:00:00+01:002023-01-01T00:00:00+01:00https://www.riemannslibrary.com/history/2023/01/01/Which-is-greater%201000%5E1001%20or%201001%5E1000Which-is-greater 1000^1001 or 1001^1000 ? Let’s simplify these number, to have a comparison criteria.

# Comparing $1000^{1001}$ or $1001^{1000}$

Since, $\ln x$ is strictly increasing, we can also compare

\begin{aligned} \ln(1000^{1001})&= \ln (\exp (1001 \cdot\ln 1000))=1001 \cdot\ln 1000 \\ \ln(1001^{1000})&= \ln (\exp (1000 \cdot\ln 1001))=1000 \cdot\ln 1001 \end{aligned}

Divided by $\color{blue}{1000\times1001}$, we can compare

\begin{aligned} (1001 \cdot\ln 1000)/(\color{blue}{1000\times1001})&=\ln 1000/1000=f(1000) \\ (1000 \cdot\ln 1001)/(\color{blue}{1000\times1001})&=\ln 1001/1001=f(1001) \end{aligned}

# Let’s study $f(x)=\ln (x)/x$

$\forall x \in \mathbb{R}_{+}^{*},f^{\prime}(x)=\frac{1-\log (x)}{x^2}; \quad 0 = f^{\prime}(x) \Longleftrightarrow 0 =\frac{1-\log (x)}{x^2} \Longleftrightarrow x=e \text{ and } x\neq 0 \\$ $\forall x \in \mathbb{R}_{+}^{*}, \quad 0 \lt f^{\prime}(x) \Longleftrightarrow 0 \lt \frac{1-\log (x)}{x^2} \Longleftrightarrow 0 \lt 1-\log (x)\Longleftrightarrow \log (x)\lt 1 \Longleftrightarrow x\lt e \\$ $\forall x \in \mathbb{R}_{+}^{*}, \quad f^{\prime}(x) \lt 0 \Longleftrightarrow \frac{1-\log (x)}{x^2} \lt 0\Longleftrightarrow 1-\log (x)\lt 0 \Longleftrightarrow 1 \lt\log (x) \Longleftrightarrow e\lt x$ $\color{green}{\forall x\in \left]e, +\infty \right[, f^{\prime}(x) \lt 0 \implies f \text{ is strictly decreasing}\implies \forall a, b \in \left]e, +\infty \right[, \quad a \lt b \implies f(a) \gt f(b)}$

# Taking $a=1000$ and $b=1001$

\begin{aligned} f(1000) &\gt f(1001) &\\ \ln (1000)/1000 &\gt \ln (1001)/1001 &\\ 1001 \cdot\ln 1000 &\gt 1000 \cdot\ln (1001) & \text{multiplied by }1000\times 1001\\ \exp (1001 \cdot\ln 1000) &\gt \exp (1000 \cdot\ln 1001)& \text{since } \exp \text{increasing} \end{aligned} $\color{purple}{\boxed{1000^{1001} \gt 1001^{1000}}}$]]>
Properties of number 20232022-12-30T00:00:00+01:002022-12-30T00:00:00+01:00https://www.riemannslibrary.com/news/2022/12/30/Properties-of-number-2023Happy New Year 2023! May this year bring you joy, prosperity, and all the things you wish for. Here are few properties of the number $2023$

## $2023$ Prime factorization

$2023 = (2+0+2+3)(2^2+0^2+2^2+3^2)^2 = 7\cdot17^2$

## $2023$ is Harshad number

It is an integer that is divisible by the sum of its digits $\sigma=2+0+2+3=7$

## $2023$ is a polite number

It can be written as the sum of two or more consecutive positive integers

$2023=111 +112+ \ldots +126+ 127$

## $2023$ and prime numbers around

The previous prime is $2017$. The next prime is $2027$, Follow us before:-)

## $2023$ Complete list of divisors

$1\space7\space 17\space 119\space 289\space 2023$]]>
The q-Pochhammer Symbol2022-12-28T00:00:00+01:002022-12-28T00:00:00+01:00https://www.riemannslibrary.com/history/2022/12/28/The%20q-Pochhammer%20SymbolThe Pochhammer symbol, also known as the rising factorial, is a notation that is used in combinatorics and special functions.

## Definition of Pochhammer Symbol

It is defined as follows:

$(a)_n = \begin{cases} 1 & \text{if } n = 0 \\ a(a+1)(a+2)\dotsm(a+n-1) & \text{if } n > 0 \end{cases}$

The Pochhammer symbol has a number of applications in various areas of mathematics, including combinatorics, special functions, and algebraic geometry. It is named after German mathematician Ernst Pochhammer, who introduced it in a paper published in 1877.

## Definition of q-Pochhammer Symbol

The q-Pochhammer symbol is a generalization of the Pochhammer symbol to the realm of q-series, which are series that involve powers of the variable q. The q-Pochhammer symbol was introduced by Leonard Carlitz in 1935. It has many of the same properties as the Pochhammer symbol, but is defined in terms of q-series rather than ordinary series.

The q-Pochhammer symbol is defined as follows:

$(a;q)_n = \begin{cases} 1 & \text{if } n = 0 \\ \displaystyle\prod_{i=0}^{n-1} (1-aq^i) = (1-a)(1-aq)(1-aq^2)\dotsm(1-aq^{n-1}) & \text{if } n > 0 \end{cases}$

## Properties

Here are some properties of the q-Pochhammer symbol:

• Recurrence relation: For all nonnegative integers $n$ and $m$, we have

$(a;q)_{n+m} = (a;q)_n (aq^n;q)_m$
• Infinity identity: For all complex numbers $a$ and $q$, we have

$(a;q)_n = \frac{(a;q)_\infty}{(aq^n;q)_\infty}$
• Binomial Equality: The Gaussian binomial coefficients are defined as follows:

$\binom{n}{k}_q = \frac{(1-q)(1-q^2)\dotsm(1-q^n)}{(1-q)(1-q^2)\dotsm(1-q^k)(1-q^{k+1})\dotsm(1-q^n)}$

where $n$ and $k$ are non-negative integers. If $k$ > $n$, this evaluates to $0$. For $k = 0$, the value is $1$ since both the numerator and denominator are empty products.

For all nonnegative integers $n$ and all complex numbers $a$ and $b$, we have

$\sum_{k=0}^n \binom{n}{k}_q a^k b^{n-k} = (a;q)_n (b;q)_n$

## Proofs

Here are the proofs for the above properties:

### Recurrence relation

We can prove the recurrence relation by induction on $m$:

• For the base case $m=0$, we have

\begin{aligned} (a;q)_n &= (a;q)_n (a;q)_0 \\ &= (a;q)_n \cdot 1 \end{aligned}

This holds, so the base case is proven.

• Inductive step: Assume that the formula holds for some arbitrary value of $\textcolor{purple}{m = k}$:

$\textcolor{purple}{(a;q)_{n+m} = (a;q)_{n+k} = (a;q)_n (aq^n;q)_k}$

We will now prove that it holds for $m = k+1$. We will prove the recurrence formula for the q-Pochhammer symbol using induction on $m$.

\begin{aligned} (a;q)_{n+m} &= (a;q)_{n+k+1} \\ &= (1-a)(1-aq)(1-aq^2)\dotsm(1-aq^{n+k-1})(1-aq^{n+k}) \\ &= \underbrace{(a;q)_{n+k}}_{\textcolor{purple}{\text{true for }m=k}}(1-aq^{n+k}) \\ &= \overbrace{(a;q)_n \textcolor{green}{(aq^n;q)_k}}(1-aq^{n+k}) \\ &= (a;q)_n\textcolor{green}{(1-aq^n)(1-aq^{n+1})(1-aq^{n+2})\dotsm(1-aq^{n+k-1})} (1-aq^{n+k})\\ &= (a;q)_n (aq^n;q)_{k+1} \\ &= (a;q)_n (aq^n;q)_{m} \end{aligned}

Thus, the formula holds for $m = k+1$. By induction, the formula holds for all values of $m$.

### Infinity identity

We can prove the infinity identity as follows:

\begin{aligned} \frac{(a;q)_\infty}{(aq^n;q)_\infty} &= \frac{\displaystyle\lim_{n \to +\infty}(a;q)_n }{\displaystyle\lim_{n \to +\infty}(aq^n;q)_n} = \frac{\displaystyle\prod_{i=0}^\infty (1-aq^{i})}{ \displaystyle \prod_{i=0}^\infty (1-aq^{i+n})}= \frac{\displaystyle\prod_{i=0}^{n-1} (1-aq^{i}) \prod_{i=n}^{\infty} (1-aq^{i})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}\\ &= \frac{\displaystyle\prod_{i=0}^{n-1} (1-aq^{i}) \prod_{k=0}^{\infty} (1-aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}= \prod_{i=0}^{n-1} (1-aq^{i}) \frac{\displaystyle\prod_{k=0}^{\infty} (1-aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}\\ &= \prod_{i=0}^{n-1} (1-aq^{i}) \cdot 1 \\ &= (a;q)_n \\ \end{aligned}

### Binomial Equality

We can prove the binomial equality by induction on $n$:

• For the base case $n=0$, we have
$\sum_{k=0}^0 \binom{0}{k}_q a^k b^{0-k} = \binom{0}{0}_q a^0 b^{0-0} = 1 = (a;q)_0 (b;q)_0$
• Inductive step: Assume that the formula holds for some arbitrary value of $\textcolor{purple}{n}$.We will now prove that it holds for $n+1$:
\begin{align*} \sum_{k=0}^{n+1} \binom{n+1}{k}_q a^k b^{n+1-k} &= \sum_{k=0}^n \binom{n+1}{k}_q a^k b^{n+1-k} + \binom{n+1}{n+1}_q a^{n+1} b^{n+1-(n+1)} \\ &= \textcolor{purple}{\sum_{k=0}^n \binom{n}{k}_q a^k b^{n-k}} + \frac{(1-q^{n+1})}{(1-q)} \cdot a^{n+1} b^0 \\ &= \textcolor{purple}{(a;q)_n (b;q)_n} + \frac{(1-q^{n+1})}{(1-q)} \cdot a^{n+1} b^0 \\ &= (a;q)_{n+1} (b;q)_{n+1} \end{align*}]]>
Bernhard Riemann - Biography2022-12-27T00:00:00+01:002022-12-27T00:00:00+01:00https://www.riemannslibrary.com/history/2022/12/27/Bernhard%20Riemann%20-%20Biography Georg Friedrich Bernhard Riemann 1826-1866

# Introduction

Bernhard Riemann was a German mathematician who made significant contributions to analysis, number theory, and differential geometry. He is best known for his work on the theory of Riemann surfaces, which helped to establish the modern field of complex analysis, and for his contributions to the study of multi-dimensional spaces.

# Early life

• Born: September 17, 1826 in Breselenz, Germany
• Showed an early aptitude for mathematics and was accepted into the University of Göttingen at the age of 19.

• Studied at the University of Göttingen under leading mathematicians including Carl Friedrich Gauss and Peter Gustav Lejeune Dirichlet.
• Became a professor at the University of Göttingen and later at the University of Berlin.

# Major contributions to mathematics

Bernhard Riemann made several significant contributions to mathematics throughout his career. Some of his most important theorems and concepts include:

• The Riemann integral, which allows for the definition of integrals of functions that are not necessarily continuous. The Riemann integral is defined as the limit of a sequence of Riemann sums, which are constructed by dividing the interval of integration into subintervals and approximating the function by a piecewise constant function. This construction allows for the integration of functions that may not have a well-defined value at every point in the interval, and is a key tool in the study of real analysis.

• The Riemann-Roch theorem, which relates the number of independent solutions to a system of algebraic equations to the topological properties of the solutions. The theorem is used in the study of algebraic curves and has important applications in geometry and topology.

Let $C$ be a compact Riemann surface of genus $g$, and let $D$ be a divisor on $C$. Then the Riemann-Roch theorem states that:

$\text{dim}(L(D)) - \text{dim}(L(-D)) = \text{deg}(D) + 1 - g$

where $L(D)$ denotes the space of meromorphic functions on $C$ with poles of at most order 1 at the points of $D$, and $\text{dim}(L(D))$ denotes the dimension of this space.

In this statement, the term $\text{deg}(D)$ denotes the degree of the divisor $D$, which is defined as the sum of the orders of the points in $D$. The term $g$ is the genus of the Riemann surface $C$, which is a topological invariant that measures the number of “holes” in the surface.

• The concept of a Riemannian manifold, which is a type of multi-dimensional space that is equipped with a metric tensor that allows for the definition of distance and angles between points. Riemannian manifolds are an important tool in the study of geometry and are used to model physical phenomena such as the curvature of space-time in general relativity.

A Riemannian manifold is a smooth manifold $M$ equipped with a smooth, symmetric, positive definite inner product $g_{ab}$ on each tangent space $T_pM$, where $p \in M$.

This inner product allows us to define the length of a tangent vector $v \in T_pM$ as:

$\left\|v\right\| = \sqrt{g_{ab}v^av^b}$

and the angle between two tangent vectors $u, v \in T_pM$ as:

$\cos \theta = \frac{g_{ab}u^av^b}{\left\|u\right\|\left\|v\right\|}$

where $u^a$ and $v^a$ are the components of $u$ and $v$ with respect to a local coordinate system on $M$.

The metric tensor $g_{ab}$ encodes information about the geometry of the manifold and can be used to define various geometric quantities, such as distances and angles between points, curvature, and volume. Riemannian manifolds are used in various areas of mathematics and physics to model the geometry of curved spaces and to study the behavior of physical phenomena in these spaces.

• The Riemann curvature tensor, which is a mathematical object that describes the curvature of a Riemannian manifold. The curvature tensor is used to measure the deviation of the manifold from flatness and is an important tool in the study of the geometry of curved spaces.

Let $(M, g_{ab})$ be a Riemannian manifold. The Riemann curvature tensor is a multilinear map

$R : T_pM \times T_pM \times T_pM \rightarrow T_pM$

defined at each point $p \in M$ by:

$R(u, v)w = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_{[u, v]}w$

where $u, v, w \in T_pM$ are tangent vectors, $\nabla$ is the Levi-Civita connection associated with the metric $g_{ab}$, and $[u, v]$ denotes the Lie bracket of $u$ and $v$.

The components of the Riemann curvature tensor in a local coordinate system $(x^1, \dots, x^n)$ on $M$ are given by:

$R_{ijkl} = g_{ia}R(e_j, e_k)e_l^a$

where $e_1, \dots, e_n$ is a local frame on $M$, and $e_i^a$ are the components of $e_i$ with respect to the frame.

The Riemann curvature tensor encodes information about the curvature of the Riemannian manifold and is used to study the geometry of curved spaces. It is a fundamental object in differential geometry and has applications in various areas of mathematics and physics.

Bernhard Riemann’s work on these and other concepts and theorems has had a lasting impact on mathematics and his ideas continue to be studied and developed by researchers today.

# Later life and legacy

• Died on July 20, 1866 at the age of 39 due to complications from tuberculosis.
• Despite his relatively short career, Riemann’s work has had a lasting impact on mathematics and his contributions continue to be studied and built upon by mathematicians today.
• Recognized as one of the greatest mathematicians in history and his name is associated with many important concepts and theorems in mathematics.

# Conclusion

Bernhard Riemann was a pioneering mathematician whose contributions to the fields of analysis, number theory, and differential geometry have had a lasting impact on the field. His work on the theory of Riemann surfaces and the concept of multi-dimensional spaces continue to be fundamental to modern mathematics and his ideas continue to be studied and developed by researchers today.

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Welcome to Riemann’s Library2022-12-22T17:56:15+01:002022-12-22T17:56:15+01:00https://www.riemannslibrary.com/2022/12/22/Welcome-to-Riemann's%20LibraryWelcome to our website Riemann’s Library, where we explore the intersection of mathematics, science, and culture!

Mathematics is a fundamental part of our world, playing a crucial role in fields ranging from physics and engineering to finance and computer science. It is a discipline that helps us understand and make sense of the world around us, and it has a rich and fascinating history. On our site, you will find a wide range of resources and articles on a variety of topics related to mathematics and its applications in science and culture. From the basics of arithmetic and algebra to more advanced concepts in geometry and topology, we cover it all.

We also delve into the cultural and historical context of mathematics, examining the lives and contributions of famous mathematicians and the ways in which math has shaped and influenced societies throughout the ages.

Whether you are a student looking to improve your math skills or simply someone with an interest in the subject, we hope you will find our site both educational and enjoyable. Thank you for visiting!

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