OpenLLM-France/wikimedia huggingface dataset:
# Bilinear form/Symmetric/Definiteness/Minor criterion/Fact

Let $\left\langle -,-\right\rangle$ be a symmetric bilinear form on a finite-dimensional real vector space, and let $v_{1},\ldots ,v_{n}$ be a basis of $V$. Let $G$ be the Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis, and let $D_{k}$ denote the determinants of the square submatrices
    $M_{k}={\left(\left\langle v_{i},v_{j}\right\rangle \right)}_{1\leq i,j\leq k},\,k=1,\ldots ,n.$
* The form $\left\langle -,-\right\rangle$ is positive definite if and only if all $D_{k}$ are positive.
* The form $\left\langle -,-\right\rangle$ is negative definite if and only if the sign in the sequence $D_{0}=1,\,D_{1},\,D_{2},\ldots ,D_{n}$ changes at every step.
# Bilinear form/Symmetric/Definiteness/Minor criterion/Fact/Proof

(1). If the bilinear form is positive definite, then, because of exercise, the sign of the determinant of the Gram matrix equals $(-1)^{0}=1$, and is positive. Since the restriction of the form to the linear subspaces $U_{i}=\langle v_{1},\ldots ,v_{i}\rangle$ is also positive definite, the determinants of all relevant submatrices are positive. If, the other way round, all determinants are positive, then fact, implies that the bilinear form is positive definite.
(2) follows from (1) by considering the negative bilinear form, that is, $-\left\langle -,-\right\rangle$.
# Linear algebra (Osnabrück 2024-2025)/Bilinear form/Symmetric/Type/Minor criterion/Fact/Factreferencenumber

Theorem 39.
# Realvalued function/Partial derivatives/Hesse-Matrix/Definiteness and Extrema/Remark

For a function
    $f\colon \mathbb {R} ^{n}\longrightarrow \mathbb {R} ,$
one would like to know, like in the case $n=1$, where the function has local extrema, for example maxima, that is, points $P\in \mathbb {R} ^{n}$ with the property that in a small nighborhood of $P$, the function is bounded from above by $f(P)$. In case $n=2$, $f$ describes a mountain range over a plane and we are looking for the summits. If the function is twice continuously differentiable, then there exist, like in the one-dimensional situation, necessary and sufficient differential criteris for the existence of local maxima and minima. The necessary criterion is that $P$ is a critical point, that is, the partial derivatives ${\frac {\partial f}{\partial x_{i}}}(P)$ vanish for $i=1,\ldots ,n$. In this case, we consider the second partial derivatives, and form the so-called Hesse-matrix
    ${\left({\frac {\partial }{\partial x_{i}}}{\frac {\partial f}{\partial x_{j}}}(P)\right)}_{ij}.$
The corresponding symmetric bilinear form determines often whether we have a local maximum or a local minimum. If this form is positive definite, then we have an isolated local minimum; if it is negative definite, then we have an isolated local maximum; if it is indefinite, then we do not have a local extremum (see fact). In the remaining cases, e.g., in the case of the zero matrix, we need further considerations.
# Vector space/Perfect pairing/Definition

Let $K$ be a field, let $V$ and $W$ denote vector spaces over $K$, and let
    $\Psi \colon V\times W\longrightarrow K$
denote a bilinear form. We say that $\Psi$ defines a perfect pairing, if the mapping
    $V\longrightarrow {W}^{*},v\longmapsto {\left(w\mapsto \Psi (v,w)\right)},$
is bijective.
# Boolf prop/3-ary/nonmonotonicity

Number of blocks: 7 Integer partition: 1⋅5 + 1⋅20 + 1⋅29 + 1⋅36 + 1⋅46 + 2⋅60
| # | nonmonotonicity | block |
| 20 | 0 | [0, 128, 136, 160, 168, 170, 192, 200, 204, 224, 232, 234, 236, 238, 240, 248, 250, 252, 254, 255] |
| 60 | 3 | [1, 14, 24, 36, 44, 46, 50, 56, 58, 66, 74, 78, 84, 88, 92, 98, 100, 104, 106, 108, 110, 114, 116, 120, 122, 124, 126, 127, 129, 137, 139, 141, 143, 154, 156, 161, 163, 166, 169, 173, 177, 179, 180, 185, 189, 193, 197, 198, 201, 203, 209, 210, 213, 217, 219, 225, 227, 229, 231, 233] |
| 46 | 2 | [2, 4, 10, 12, 16, 17, 40, 42, 48, 51, 72, 76, 80, 96, 102, 112, 119, 130, 132, 142, 144, 152, 153, 164, 171, 175, 178, 188, 190, 191, 194, 205, 207, 212, 218, 222, 223, 230, 235, 237, 241, 243, 245, 246, 247, 249]File:Set of 3-ary Boolean functions 1318345726263181355624145172358117892981232755153868639571401447138503365652.svg |
| 60 | 4 | [3, 5, 6, 9, 11, 13, 15, 18, 20, 26, 28, 33, 35, 38, 43, 47, 49, 52, 59, 60, 62, 63, 65, 69, 70, 77, 79, 81, 82, 90, 93, 94, 95, 111, 113, 115, 117, 118, 123, 125, 131, 133, 134, 145, 146, 148, 155, 157, 158, 159, 165, 167, 181, 182, 183, 195, 199, 211, 214, 215] |
| 36 | 5 | [7, 19, 21, 25, 27, 29, 30, 31, 37, 39, 41, 45, 53, 54, 55, 57, 61, 67, 71, 73, 75, 83, 86, 87, 89, 91, 97, 99, 101, 103, 107, 109, 121, 135, 147, 149] |
| 29 | 1 | [8, 32, 34, 64, 68, 85, 138, 140, 162, 172, 174, 176, 184, 186, 187, 196, 202, 206, 208, 216, 220, 221, 226, 228, 239, 242, 244, 251, 253] |
| 5 | 6 | [22, 23, 105, 150, 151] |
# Tourism/Travel

This page is about mutual brainstorming, about what's the best way to travel and see the world at the lowest (possible) cost.

## Ways to travel and see the world

A lot of independent travelers travel the world by the backpacking method. Although this may be the cheapest way, it does not allow you to travel totally on your own and still locks you tied up to someone else which will transport you (eg airplane, subway, train, ...). Having your own method of transport (eg boat) may offer more advantages here (but might not be as cheap).

## By airplane

Several low cost airlines allow people to get to a destination at the lowest possible cost of all modes of transport. At this Wikivoyage article, you can find the low-cost airlines operating in each continent/area.
There has also been a surge in the creation of flight aggregator sites that allow you to cross-reference prices and find the cheapest flight possible.

## Airplane ticket

An airplane ticket is a paper or electronic document that contains passenger information as well as flight information. Origin and destination, flight time, flight number, passengers' names and arrival time are the most important features of the plane ticket.

## By train

Thomas Cook has a useful book on the destinations done by trains.

## By (sail)boat

Cruising is another way to see the world.
Sailing along on other people's sailboats or getting your own is another possibility. Cutters, as well as confiscated boats by the navy and fishing boats may be the cheapest ships to obtain (can be fitted afterwards with sails) or building one yourself may be an other option.

## The Spa Resort Experience

The spa at resorts is a holistic and immersive experience where the focus goes beyond traditional accommodations. Spa resorts prioritize well-being, offering a range of therapeutic treatments, wellness programs, and activities aimed at rejuvenating both the body and mind. Guests can enjoy personalized spa therapies, yoga and meditation sessions, and access to state-of-the-art fitness facilities.
Differences from Traditional Hotels
While hotels provide a place to stay, spa resorts are designed to create a complete retreat. The emphasis is on holistic well-being, with features like dedicated spa facilities, wellness consultations, healthy dining options, and mindfulness activities. The ambiance is often more tranquil, fostering a sense of relaxation and rejuvenation throughout your stay.
Options for Every Budget
Contrary to the misconception that spa resorts are exclusively for the affluent, there are options catering to various budgets. From boutique spa resorts to larger establishments, you can find accommodations that align with your financial plan. Many spa resorts also offer day spa packages, allowing you to enjoy their amenities without an overnight stay.

## Finding cheap hostels

Cheap hostels can be found at various Internet sites.

## Equipment

The Survival wallet may be very handy in making long voyages and a UMPC might also be handy. Further info may be obtained from Doug Lansky's Travel survival book.
# Computer Networks/Ping/Sweep

A ping sweep is a network reconnaissance using some type of ping (ICMP echo/reply) to locate hosts on a network. These activities will show you how to use the ping command to perform a ping sweep.

## Preparation

To prepare for this activity:
* Start Windows.
* Log in if necessary.

## Activity 1 - Identify the Host Network

To identify the host network:
* Open a command prompt.
* Use ipconfig to display the host IP address. Note the IPv4 Address displayed.
* Identify the host network. For this activity, the host network will be assumed to be a 24-bit network based on the first three octets of the host IP address. For example, if the IPv4 Address is 192.168.1.101, this activity will assume that the host network is 192.168.1.0, and has potential host addresses from 192.168.1.1 through 192.168.1.254.

## Activity 2 - Perform a Ping Sweep

To perform a Windows OS ping sweep:
* From a Microsoft OS command line, Type:
for /l %i in (1,1,254) do @ping -n 1 -w 100 <first three octets of host network>.%i.
For example, if the host network is 192.168.1.0, the command would be:
for /l %i in (1,1,254) do @ping -n 1 -w 100 192.168.%i.
To perform a Linux OS ping sweep:
This command instructs the computer to loop (for /l) counting from 1 by 1 to 254 (1,1,254) and execute (do) a ping command with a count (-n) of 1 and a wait (-w) time of 100 milliseconds and the given network address and changing host (%i) address. If using a bash command shell, on most Linux OS, a similar example would resemble for i in {1..25}; do ping -c 1 -W 1 10.1.1.$i ; done, where the count for ($i) would be from 1 to 25.
* Press Enter.
* Observe the results as the command sweeps the network searching for hosts.
Windows example to find live hosts(for /l %i in (1,1,254) do @ping -n 1 -w 100 192.168.0.%i | findstr "Reply")

## Activity 3 - List Responding Hosts

Most systems now have a firewall in place that prevents them from responding to the ping sweep. However, they will still acknowledge the underlying Address Resolution Protocol (ARP) request that first converts their IP address into a Media Access Control (MAC) address. To list hosts that responded to the ARP request:
* Use arp -a to view the ARP cache.
* Observe the list of hosts on the network.
* Close the command prompt to complete this activity.

## Readings

* Wikipedia: Ping (networking utility)
* Wikipedia: Ping sweep
* Wikipedia: Network Reconnaissance
* Wikipedia: Internet Control Message Protocol (ICMP)
* Wikipedia: IP Address
* Wikipedia: Media Access Control (MAC) Address
* Wikipedia: For loop
* fping for pinging multiple hosts https://fping.org/
# Minkowski-space/Type/Definition

A real vector space of dimension $n$, endowed with a bilinear form $\left\langle -,-\right\rangle$ of type $(n-1,1)$, is called a Minkowski-space.
# Minkowski-space/Several vectors/Definition

Let $V$ be a Minkowski-space, endowed with the Minkowski-form $\left\langle -,-\right\rangle$. A vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle =0\,$
is called lightlike, a vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle <0\,$
is called timelike, and a vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle >0\,$
is called spacelike.
# Minkowski-space/Motion/Realization by scaling/Example

In a four-dimensional standard-Minkowski-space, we want to move unformly from the point $P={\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}$ to the point $Q={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\\s\end{pmatrix}}$. In the classical framework, we would just take the vector
    ${\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}:={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\\s\end{pmatrix}}-{\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}\,.$
However, this is in general not an observer vector, and then the looked-for motion is not realizable. If $y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-t^{2}$ is negative, which means that we have a timelike vector, then we can rescale this vector to obtain an observer vector
    ${\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\\u\end{pmatrix}}={\frac {1}{\sqrt {-\left\langle {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}},{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\right\rangle }}}{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\,.$
Then, the mapping
    $x\longmapsto {\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}+x{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\\u\end{pmatrix}}$
describes a motion, which, for $x=0$, starts at the point $P$, and ends, for $x={\sqrt {-\left\langle {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}},{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\right\rangle }}$, at the point $Q$, and which is realizable in the physics world.
# Minkowski-space/Two observers/Relative velocity (vector)/Definition

Let $V$ denote a Minkowski-space, and let $B$ and $C$ denote observers with the four-velocities $v$ and $w$. The vector
    $v_{BC}=-v-{\frac {1}{\left\langle v,w\right\rangle }}w\,$
is called the velocity vector of $C$ relative to $B$. The number
    $\rho _{BC}={\sqrt {1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}}}\,$
is called the relative velocity of the two observers.
# Minkowski-space/Two observers/Relative velocity/Fact

Let $V$ be a Minkowski-space, and let $B$ and $C$ denote observers from the same time sheet with den four-velocities $v$ and $w$. Then the following statements hold.
* The relative velocity vector $v_{BC}$ is orthogonal to $v$.
* The relative velocity vector $v_{BC}$ is spacelike, and we have
    $\Vert {v_{BC}}\Vert =\Vert {v_{CB}}\Vert =\rho _{BC}=\rho \,.$
* We have
    $\left\langle v,w\right\rangle =-{\frac {1}{\sqrt {1-\rho ^{2}}}}\,.$
* We have and this is the decomposition of $w$ into the space component and the time component of $B$.
    $w={\frac {v}{\sqrt {1-\rho ^{2}}}}+{\frac {v_{BC}}{\sqrt {1-\rho ^{2}}}}\,,$
* The time coefficient of $w$ with respect to $B$ is ${\frac {1}{\sqrt {1-\rho ^{2}}}}$.
# Minkowski-space/Two observers/Relative velocity/Fact/Proof

* we have so these vectors are orthogonal to each other. Therefore, $v_{BC}$ belongs to the space component of $B$.
    $\left\langle v,v_{BC}\right\rangle =\left\langle v,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =\left\langle v,-v\right\rangle +\left\langle v,-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =1-1=0\,,$
* We have By part (1) (or because of fact (3)), this expression is non-negative. The square root of this number is the relative velocity $\rho$.
    ${\begin{aligned}\left\langle v_{BC},v_{BC}\right\rangle &=\left\langle -v-{\frac {1}{\left\langle v,w\right\rangle }}w,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v+{\frac {1}{\left\langle v,w\right\rangle }}w,v+{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v,v\right\rangle +2{\frac {\left\langle v,w\right\rangle }{\left\langle v,w\right\rangle }}+{\frac {1}{\left\langle v,w\right\rangle ^{2}}}\left\langle w,w\right\rangle \\&=1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}.\end{aligned}}$
* This follows directly from the definition using
    $\rho ={\sqrt {1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}}}\,,$
    $\left\langle v,w\right\rangle <0\,.$
* From and (3), we can infer Because of part (1), the vector $v_{BC}$ belongs to the space component of $B$.
    $v_{BC}=-v-{\frac {1}{\left\langle v,w\right\rangle }}w\,$
    $w=-\left\langle v,w\right\rangle v-\left\langle v,w\right\rangle v_{BC}={\frac {v}{\sqrt {1-\rho ^{2}}}}+{\frac {v_{BC}}{\sqrt {1-\rho ^{2}}}}\,.$
* From (4) we can see directly that the time cefficient of $w$ with respect to $B$ equals ${\frac {1}{\sqrt {1-\rho ^{2}}}}$.
# Minkowski-space/Simultaneity/Remark

For a four-velocity $v$ of an observers $B$ with the decomposition
    $V=V_{v}\oplus \mathbb {R} v\,,$
the set of points (events) of the form $sv+V_{v}$ with a fixed $s\in \mathbb {R}$ is called the space at the time $s$. The points from this set are called simultaneous for the observer $B$. For another observer $C$ with the four-velocity $w$, these points are not simultaneous. The concept of simultaneity of the observer $A$ depends on his decomposition of the Welt $V$ into space component and time component. If, for example, the four-velocity of the second observer is given, with respect to a Minkowski-basis of the first observer, by ${\begin{pmatrix}{\frac {3}{4}}\\0\\0\\{\frac {5}{4}}\end{pmatrix}}$, then
    ${\frac {15}{4}}{\begin{pmatrix}{\frac {1}{3}}\\0\\0\\{\frac {1}{5}}\end{pmatrix}},{\begin{pmatrix}0\\1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}$
is an orthonormal basis of the space component of the second observer. For the first observer, the events ${\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}$ and ${\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}$ are simultaneous, but they are not simultaneous for the second observer, because the first vector has the same description, and the second vector equals
    ${\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}={\frac {75}{16}}{\begin{pmatrix}{\frac {1}{3}}\\0\\0\\{\frac {1}{5}}\end{pmatrix}}-{\frac {3}{4}}{\begin{pmatrix}{\frac {3}{4}}\\0\\0\\{\frac {5}{4}}\end{pmatrix}}\,.$
The time component of the second event with respect to the second observer vector is $-{\frac {3}{4}}$.
# Permutation matrix/Cycle/Numbered/C/Eigenvalues/Fact

For a permutation matrix $M_{\rho }$ over $\mathbb {C}$ for a cycle $\rho \in S_{n}$ with $\rho :1\mapsto 2\mapsto \ldots \mapsto k\mapsto 1$
$k$
root of unity $\zeta$, the vectors
    $v_{\zeta }:=\zeta ^{k-1}e_{1}+\zeta ^{k-2}e_{2}+\cdots +\zeta e_{k-1}+e_{k}\,$
are eigenvectors of $M_{\rho }$ for the eigenvalue
$\zeta$
In particular, a permutation matrix of a cycle over $\mathbb {C}$ is diagonalizable.
# Permutation matrix/Cycle/Numbered/C/Eigenvalues/Fact/Proof

We have
    ${\begin{aligned}M_{\rho }(v_{\zeta })&=M_{\rho }(\zeta ^{k-1}e_{1}+\zeta ^{k-2}e_{2}+\cdots +\zeta e_{k-1}+e_{k})\\&=\zeta ^{k-1}M_{\rho }(e_{1})+\zeta ^{k-2}M_{\rho }(e_{2})+\cdots +\zeta M_{\rho }(e_{k-1})+M_{\rho }(e_{k})\\&=\zeta ^{k-1}e_{2}+\zeta ^{k-2}e_{3}+\cdots +\zeta e_{k}+e_{1}\\&=e_{1}+\zeta ^{k-1}e_{2}+\zeta ^{k-2}e_{3}+\cdots +\zeta e_{k}\\&=\zeta (\zeta ^{k-1}e_{1}+\zeta ^{k-2}e_{2}+\cdots +\zeta e_{k-1}+e_{k})\\&=\zeta v_{\zeta }.\end{aligned}}$
Since there are $k$ different $k$-th roots of unity in $\mathbb {C}$, these vectors are linearly independent due to fact, and they generate a $k$-dimensional linear subspace $U$ of $K^{n}$. In fact, we have
    $U=\langle e_{i},\,i=1,\ldots ,k\rangle \,.$
Since the vectors $e_{i}$, $i\geq k+1$, are fixed vectors, the $v_{\zeta }$ together with the $e_{i}$, $i\geq k+1$, form a basis consisting of eigenvectors of $M_{\rho }$. Hence, $M_{\rho }$ is diagonalizable.
# Permutation matrix/3-cycle/Invariant linear subspace/Minimal polynomial/Example

We consider the permutation matrix
    ${\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}.$
The line $K{\begin{pmatrix}1\\1\\1\end{pmatrix}}$ is the eigenspace for the eigenvalue $1$. Moreover,
    $U=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,$
is an invariant linear subspace (which, over $\mathbb {C}$, according to fact, can be decomposed further into smaller eigenspaces). With respect to the given basis, the restriction of the linear mapping to $U$ has the describing matrix
    ${\begin{pmatrix}0&-1\\1&-1\end{pmatrix}}.$
Therefore, the characteristic polynomial of this matrix is
    $X(X+1)+1=X^{2}+X+1\,.$
This is also the minimal polynomial of the restriction. The minimal polynomial of the permutation matrix is $X^{3}-1$, and indeed we have
    $X^{3}-1=(X-1)(X^{2}+X+1)\,$
in accordance with fact.
# Linear algebra (Osnabrück 2024-2025)/Permutation matrix/Cycle/Numbered/C/Eigenvalues/Fact/Factreferencenumber

Lemma 24.11
# President of the United States/Barack Obama

| 42nd President | 43rd President | 44th President | 45th President | 46th President |
| Bill Clinton 1993 - 2001 | George W. Bush 2001 - 2009 | Barack H. Obama 2009 - 2017 | Donald Trump 2017 - 2021 2025-2029 | Joe Biden 2021 - 2025 |
# What Matters/Community
## Community

A community is a group of people who have significant, positive social interactions. Citizen-to-citizen engagement, a focus on the well-being of the whole, and hospitality to all are the essential elements of a healthy community.
Extensive research strongly supports the conclusion that happiness is best predicted by the breadth and depth of one’s social connections.
Communities create social capital—trust, goodwill, and cooperation among people. Social capital provides many benefits including: improved education and children’s welfare, safe and productive neighborhoods, economic prosperity, health and happiness, and stronger democracy.
Engage with communities where you belong and can thrive and contribute.

### Assignment:

* Identify the communities you belong to that require meeting face-to-face with other people. Consider work you do for a political party, civic groups, religious groups, sports clubs, work or school groups, informal social connections, volunteer efforts and other such activities.
** List (privately) the people you have met through those communities whom: 1) you trust, 2) you would be comfortable asking for a favor, and 3) you would be happy to help out if they asked for a favor.
** Thinking about your neighbors, how many do you know by name? How many have you accumulated social capital with?
* Increase social capital—trust and cooperation among community members—through your participation in those communities.
** The Saguaro Seminar is a long-term research project aimed at significantly increasing Americans' connectedness to one another and to community institutions. You may wish to study their work and help to implement their suggestions.
* Engage with and strengthen your community.
* Strengthen and renew your friendships.

### Suggestions for further reading:

* Block, Peter (2009). Community: The Structure of Belonging. Berrett-Koehler Publishers. pp. 264. ISBN 978-1605092775.
* Putnam, Robert D. (2000). Bowling Alone. Touchstone Books by Simon & Schuster. pp. 544. ISBN 978-0743203043.
* (Evaluate the book: Better Together: Restoring the American Community. )
* (Evaluate the book: The Abundant Community: Awakening the Power of Families and Neighborhoods )
** and the associated website: http://www.abundantcommunity.com official

### References:

* ↑ Putnam, Robert D. (2000). Bowling Alone. Touchstone Books by Simon & Schuster. pp. 544. ISBN 978-0743203043. Page 332
* ↑ Putnam, Robert D. (2000). Bowling Alone. Touchstone Books by Simon & Schuster. pp. 544. ISBN 978-0743203043. Section IV