OpenLLM-France/wikimedia huggingface dataset:
# Polynomial ring/Field/1/Lemma of Bezout/Endomorphism/Application/Section
## Lemma

Let $\varphi \colon V\rightarrow V$ be a linear mapping on a finite-dimensional $K$-vector space $V$, and let
    $\chi _{\varphi }=P\cdot Q\,$
be a factorization of the characteristic polynomial in coprime polynomials $P,Q\in K[X]$. Then we have the direct sum decomposition
    $V=\operatorname {kern} P(\varphi )\oplus \operatorname {kern} Q(\varphi )\,,$
where these linear subspaces are $\varphi$-invariant.
$P(\varphi )$
$\operatorname {kern} Q(\varphi )$

### Proof

Due to the Lemma of Bezout, there exist polynomials $S,T\in K[T]$ such that
    $SP+TQ=1\,.$
Set $U=\operatorname {kern} P(\varphi )$ and $W=\operatorname {kern} Q(\varphi )$. Let $v\in V$. Due to the Theorem of Cayley-Hamilton , we have
    $0=\chi _{\varphi }(\varphi )=(P(\varphi )\circ Q(\varphi ))(v)=P(\varphi )(Q(\varphi )(v))\,.$
Therefore, the image of $Q(\varphi )$ belongs to the kernel of $P(\varphi )$ and vice versa. From
    ${\begin{aligned}v&=\operatorname {Id} _{V}(v)\\&=(SP+TQ)(\varphi )(v)\\&=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))\\&=P(\varphi )(S(\varphi )(v))+Q(\varphi )(T(\varphi )(v))\end{aligned}}$
we can read off that the left-hand summand belongs to $\operatorname {Im} P(\varphi )\subseteq \operatorname {kern} Q(\varphi )$ and the right-hand summand belongs to $\operatorname {Im} Q(\varphi )\subseteq \operatorname {kern} P(\varphi )$. Therefore, we have a sum decomposition, which is direct, since $P(\varphi )(v)=Q(\varphi )(v)=0$ implies $v=0$. For the $\varphi$-invariance of these spaces, see exercise. For $v\in \operatorname {kern} Q(\varphi )$, we have
    $v=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))=S(\varphi )(P(\varphi )(v))=P(\varphi )(S(\varphi )(v))\,,$
that is, we have $\operatorname {Im} P(\varphi )=\operatorname {kern} Q(\varphi )$. Therefore, the restriction of $P(\varphi )$ to the kernel of $Q(\varphi )$ is surjective, thus bijective.
$\Box$
# Permutation matrix/3-cycle/Characteristic polynomial/Coprime factorization/Direct sum/Example

We consider the permutation matrix
    ${\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}$
over $\mathbb {R}$, the characteristic polynomial is
    $\chi _{M}=X^{3}-1=(X-1){\left(X^{2}+X+1\right)}=P\cdot Q\,,$
where the two factors are coprime. We want to check fact in this example. We have
    $P(M)=M-E_{3}={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,$
with
    $\operatorname {kern} P(M)=\operatorname {Eig} _{1}{\left(M\right)}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\,,$
and
    $Q(M)={\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}}+{\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}+{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}\,$
with
    $\operatorname {kern} Q(M)=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$
We have
    $\mathbb {R} ^{3}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\oplus \langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$
Moreover, we have
    ${\begin{aligned}P(M){\begin{pmatrix}1\\-1\\0\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}1\\-1\\0\end{pmatrix}}\\&={\begin{pmatrix}-1\\2\\-1\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}+{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\end{aligned}}$
and
    ${\begin{aligned}P(M){\begin{pmatrix}0\\1\\-1\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\\&={\begin{pmatrix}-1\\-1\\2\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}-2{\begin{pmatrix}0\\1\\-1\end{pmatrix}}.\end{aligned}}$
From this, we can read off that the restriction of $P(M)$ to $\operatorname {kern} Q(M)$ is bijective. The representation of the $1$ from example yields the matrix equation
    ${\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\frac {1}{3}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}-{\frac {1}{3}}{\begin{pmatrix}2&0&1\\1&2&0\\0&1&2\end{pmatrix}}{\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,.$
# Linear algebra (Osnabrück 2024-2025)/Characteristic polynomial/Coprime factorization/Direct sum/Fact/Factreferencenumber

Lemma 26.9
# Commensurability/Q and R/Equivalence relation/Section
## Definition

Two line segments $s$ und $t$
$g$
$g$

## Definition

Real numbers $r,s\neq 0$ are called commensurable if ${\frac {r}{s}}$ is a
This is the equivalence relation for the subgroup
    $(\mathbb {Q} ^{\times },1,\cdot )\subseteq (\mathbb {R} ^{\times },1,\cdot )\,$
in the sense of definition.

## Remark

The subgroup relation
    $(\mathbb {Q} ,0,+)\subseteq (\mathbb {R} ,0,+)\,$
(which might be considered as a relation of a linear subspace of $\mathbb {Q}$-vector spaces) yields also an equivalence relation on the real numbers. Here, two real numbers are equivalent if their difference is a rational number.
# Commensurability/Historic/Motivation for real numbers/Remark

The question, whether there exist, beside the rational numbers, further useful numbers, has its roots in ancient Greece. The question was raised in the form whether two lengths, given in a natural way, are always commensurable to each other (whether there exists a third length such that the two given lengths are integer multiples of it). The Pythagorean school believed in the harmony of the universe, and this implied that all proportions between natural lengths must be expressed by a proportion of integer numbers. They found such integer proportions in music (music intervals), and thought that such proportions hold for the motions of the planets, and for geometry in general. Some people speculate that they knew the reasoning of fact showing the irrationality of ${\sqrt {2}}$ (the incommensurability of the side length and the length of the diagonal in a square), but that they concealed this knowledge. Anyhow, later in ancient times the knowledge was accepted that there are irrational numbers.
# Wikiphilosophers/Foundationalism

Foundationalism argues for sound beliefs that provide a secure basis from which more beliefs can be coherently generated. For example, René Descartes submitted "cogito ergo sum" or "I think, therefore I am" as a foundational belief.
Read users' philosophical views below.
* Micah believes that absolute certainty is currently impossible, which serves as a foundational belief: "nothing is certain, therefore anything is possible." Read more...
* S. Perquin thinks his own human mind is the only thing he can consider certain and he questions whether others have conscious experiences, but acknowledges the practical need to assume the existence of other minds to navigate life. Read more...
* User:Jaredscribe accepts as axiomatic that the natural world is real, that he and his species are rational animals - animals that think conceptually. He holds these truths to be self-evident, and denies that these are merely personal definitions of terms, inviting disputants to respond in this Wikidebate. Furthermore holds that rocks, plants, and animals are distinctively perceived by the common sense possessed by humans and most other animals alike, and that these natural genera and their species can be analytically known and conceptualized by homo sapiens like himself who have acquired the virtue of natural science and philosophy.
* User:lbeaumont describes how he chooses his beliefs based on knowing how he knows.
* User:Bobshimdt believes that all was made by mind, and all is illusory.
# Vector spaces/Linear mapping/Homomorphism theorem/Surjective and kernel/Fact

Let $K$ be a field, and let $V,Q,$ and $W$ denote vector spaces over $K$. Let $\varphi \colon V\rightarrow W$ be a linear mapping, and $\psi \colon V\rightarrow Q$ be a surjective linear mapping. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,$
holds.
    ${\tilde {\varphi }}\colon Q\longrightarrow W$
fulfilling
$\varphi ={\tilde {\varphi }}\circ \psi$
In other words: the diagram
    ${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&W&\\\!\!\!\!\!\,\,\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$
commutes.
# Applied Programming/Databases/Northwind SQL

SQL to create the Northwind database in SQLite. The Northwind database is available under a Microsoft Public License.
* ↑ Nucleon Software: Downloads
* ↑ CodePlex: Northwind License
# Vector spaces/Homomorphism theorem/Section
## Theorem

Let $K$ be a field, and let $V,Q,$ and $W$ denote vector spaces over $K$. Let $\varphi \colon V\rightarrow W$ be a linear mapping, and $\psi \colon V\rightarrow Q$ be a surjective linear mapping. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,$
holds. Then there exists a uniquely determined linear mapping
    ${\tilde {\varphi }}\colon Q\longrightarrow W$
fulfilling $\varphi ={\tilde {\varphi }}\circ \psi$. In other words: the diagram
    ${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&W&\\\!\!\!\!\!\,\,\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$

### Proof

$\Box$
The mapping ${\tilde {\varphi }}$ constructed in the preceding theorem is called the induced linear mapping; accordingly, the theorem is also called the theorem about the induced mapping.

## Corollary

Let $K$ be a field, and let
    $\varphi \colon V\longrightarrow W$
denote a surjective linear mapping. Then there exists a canonical linear isomorphism
    ${\tilde {\varphi }}\colon V/\operatorname {kern} \varphi \longrightarrow W.$

### Proof

$\Box$

## Theorem
### Proof

$\Box$
This statement can be briefly formuluated as:
Image $=$ preimage modulo kernel.
# Vector spaces/Homomorphism theorem/Group known/Section
## Theorem

Let $K$ be a field, and let $V,Q,$ and $W$ denote vector spaces over $K$. Let $\varphi \colon V\rightarrow W$ be a linear mapping, and $\psi \colon V\rightarrow Q$ be a surjective linear mapping. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,$
holds. Then there exists a uniquely determined linear mapping
    ${\tilde {\varphi }}\colon Q\longrightarrow W$
fulfilling $\varphi ={\tilde {\varphi }}\circ \psi$. In other words: the diagram
    ${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&W&\\\!\!\!\!\!\,\,\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$

### Proof

For every element $u\in Q$, there exists some $v\in V$ with $\psi (v)=u$. Due to the commutativity condition, we must have ${\tilde {\varphi }}(u)=\varphi (v)$. This means that there exists at most one ${\tilde {\varphi }}$. We have to show that, by this condition, a well-defined mapping is given. So let $v,v'\in V$ denote two preimages of $u$. Then
    $v'-v\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,;$
therefore, $\varphi (v)=\varphi (v')$. hence, the mapping is well-defined. Let $u,u'\in Q$ be gievn, with preimages $v,v'\in V$. Then $v+v'$ is a preimage of $u+u'$; therefore, we have
    ${\tilde {\varphi }}(u+u')=\varphi (v+v')=\varphi (v)+\varphi (v')={\tilde {\varphi }}(u)+{\tilde {\varphi }}(u')\,.$
This means that ${\tilde {\varphi }}$ is compatible with the addition. Let $u\in Q$ be given with a preimage $v\in V$, and let $\lambda \in K$. Then $\lambda v$ is a preimage of $\lambda u$; therefore,
    ${\tilde {\varphi }}(\lambda u)=\varphi (\lambda v)=\lambda \varphi (v)=\lambda {\tilde {\varphi }}(u)\,,$
and ${\tilde {\varphi }}$ is also compatible with the scalar multiplication.
$\Box$
The mapping ${\tilde {\varphi }}$ constructed in the preceding theorem is called the induced linear mapping; accordingly, the theorem is also called the theorem about the induced mapping.

## Corollary

Let $K$ be a field, and let
    $\varphi \colon V\longrightarrow W$
denote a surjective linear mapping. Then there exists a canonical linear isomorphism
    ${\tilde {\varphi }}\colon V/\operatorname {kern} \varphi \longrightarrow W.$

### Proof

$\Box$

## Theorem
### Proof

$\Box$
This statement can be briefly formuluated as:
Image $=$ preimage modulo kernel.
# Vector spaces/Linear mapping/Homomorphism theorem/Surjective and kernel/Fact/Proof2

For every element $u\in Q$, there exists some $v\in V$ with $\psi (v)=u$. Due to the commutativity condition, we must have ${\tilde {\varphi }}(u)=\varphi (v)$. This means that there exists at most one ${\tilde {\varphi }}$. We have to show that, by this condition, a well-defined mapping is given. So let $v,v'\in V$ denote two preimages of $u$. Then
    $v'-v\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,;$
therefore, $\varphi (v)=\varphi (v')$. hence, the mapping is well-defined. Let $u,u'\in Q$ be gievn, with preimages $v,v'\in V$. Then $v+v'$ is a preimage of $u+u'$; therefore, we have
    ${\tilde {\varphi }}(u+u')=\varphi (v+v')=\varphi (v)+\varphi (v')={\tilde {\varphi }}(u)+{\tilde {\varphi }}(u')\,.$
This means that ${\tilde {\varphi }}$ is compatible with the addition. Let $u\in Q$ be given with a preimage $v\in V$, and let $\lambda \in K$. Then $\lambda v$ is a preimage of $\lambda u$; therefore,
    ${\tilde {\varphi }}(\lambda u)=\varphi (\lambda v)=\lambda \varphi (v)=\lambda {\tilde {\varphi }}(u)\,,$
and ${\tilde {\varphi }}$ is also compatible with the scalar multiplication.
# Vector spaces/Linear mapping/Homomorphism theorem/Surjective and kernel/Fact/Proof

The statement about existence and uniqueness follows from fact. We have to show that the uniquly determined group homomorphism ${\tilde {\varphi }}$ is also compatible with scalar multiplication. Let $u\in Q$ be given with a preimage $v\in V$, and let $\lambda \in K$. Then $\lambda v$ is a preimage of $\lambda u$; therefore,
    ${\tilde {\varphi }}(\lambda u)=\varphi (\lambda v)=\lambda \varphi (v)=\lambda {\tilde {\varphi }}(u)\,,$
and ${\tilde {\varphi }}$ is also compatible with the scalar multiplication.
# Vector spaces/Linear mapping/Surjective and residue class space/Fact/Proof

This follows from fact and fact.
# Linear algebra (Osnabrück 2024-2025)/Group homomorphism/Homomorphism theorem/Surjective and kernel/Fact/Factreferencenumber

Theorem 47.1
# Linear algebra (Osnabrück 2024-2025)/Group homomorphism/Surjective and residue class group/Fact/Factreferencenumber

Corollary 47.3
# Linear algebra (Osnabrück 2024-2025)/Vector spaces/Linear mapping/Homomorphism theorem/Surjective and kernel/Fact/Factreferencenumber

Theorem 48.7
# Vector spaces/Linear mapping/Surjective and residue class space/Fact

Let $K$ be a field, and let
    $\varphi \colon V\longrightarrow W$
denote a surjective linear mapping between $K$-vector spaces.
linear isomorphism
    ${\tilde {\varphi }}\colon V/\operatorname {kern} \varphi \longrightarrow W.$
# Vector spaces/Linear mapping/Factorization/Fact

Let $K$ be a field, and let
    $\varphi \colon V\longrightarrow W$
be a linear mapping between $K$-vector spaces.
    $V{\stackrel {q}{\longrightarrow }}V/\operatorname {kern} \varphi {\stackrel {\theta }{\longrightarrow }}\operatorname {Im} \varphi {\stackrel {\iota }{\hookrightarrow }}W,$
where $q$ is the canonical projection, $\theta$ is a vector space isomorphism, and $i$ is the canonical inclusion of the image space
$W$
# Vector spaces/Linear mapping/Factorization/Fact/Proof

This follows from fact, applied to the surjective linear mapping
    $V\longrightarrow \operatorname {Im} \varphi .$
# Linear algebra (Osnabrück 2024-2025)/Vector spaces/Linear mapping/Surjective and residue class space/Fact/Factreferencenumber

Corollary 48.8
# Vector space/Linear subspace/Translation/Equivalence relation/Fact

Linear subspace/Situation
relation given by
    $v\sim w,{\text{ falls }}v-w\in U,$
is an equivalence relation
$V$
# Vector space/Linear subspace/Translation/Equivalence relation/Fact/Proof

We look at the conditions of an equivalence relationn The reflexivity follows from $v-v=0\in U$, the symmetry follows from $w-v=-(v-w)\in U$. The transitivity follows like this: from $u-v\in U$ and $v-w\in U$, we infer $u-w=(u-v)+(v-w)\in U$.
# Vector space/Linear subspace/Residue class space/Fact

Let $K$ be a field, $V$ be a $K$-vector space, and $U\subseteq V$ a linear subspace. Let $V/U$ be the set of all equivalence classes (the quotient set) of the equivalence relation on $V$ defined by $U$, and let
    $q\colon V\longrightarrow V/U,v\longmapsto [v],$
denote the canonical projection.
$K$
$V/U$
$q$
$K$
# Vector space/Linear subspace/Residue class space/Fact/Proof

The existence and uniqueness of the group structure on $V/U$, and the homomorphism property of the projection, follow from fact. For the projection to be linear, the scalar multiplication must satisfy
    $\lambda [v]=[\lambda v]\,.$
We have to show that by this condition, a well-defined scalar multiplication on $V/U$ is given. Let $v'=v+u$, with $u\in U$. Then we have
    $\lambda v'=\lambda (v+u)=\lambda v+\lambda u\,,$
and the is equivalent to $\lambda v$. The well-definedness of the scalar multiplication on $V/U$ and the surjectivity of the mapping imply that we have a vector space structure, and that the mapping is also compatible with the scalar multiplication.
# Vector space/Residue class space/Representative/Definition

Let $K$ be a field, let $V$ be a $K$-vector space, and let $U\subseteq V$ denote a linear subspace. Then the set $V/U$ of all equivalence classes, together with the structure of a vector space as proven in fact, is called the residue class space (or quotient space) of $V$ modulo $U$.
# Linear algebra (Osnabrück 2024-2025)/Vector space/Linear subspace/Residue class space/Fact/Factreferencenumber

Theorem 48.5
# Linear algebra (Osnabrück 2024-2025)/Group/Normal subgroup/Residue class group/Fact/Factreferencenumber

Theorem 46.14
# Vector space/Finite-dimensional/Linear subspace/Residue class space/Fact

Vector space/Finite-dimensional/Situation Let $U\subseteq V$ be a linear subspace.
    $\dim _{K}{\left(V/U\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,.$
# Vector space/Flag/Definition

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space of dimension $n=\dim _{K}{\left(V\right)}$. Then a chain of linear subspaces
    $0=V_{0}\subset V_{1}\subset \ldots \subset V_{n-1}\subset V_{n}=V\,$
is called a flag in $V$.
# Vector space/Finite-dimensional/Linear subspace/Residue class space/Fact/Proof

The canonical projection
    $V\longrightarrow V/U$
is surjective; its kernel is $U$. Because of the dimension formula, we obtain
    $\dim _{K}{\left(V/U\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,.$
# Vector space/Direct sum/Residue class space/Fact

Let $K$ be a field, and let $V$ denote a $K$-vector space, together with a direct sum decomposition
    $V=U\oplus W\,,$
with linear subspaces $U$ and $W$.
    $V/U\cong W\,.$
# Vector space/Direct sum/Residue class space/Fact/Proof

The kernel of the projection
    $V\longrightarrow W$
is $U$. Therefore, the statement follows from fact.
# Web Design/Produce basic client side script for dynamic web pages

Web Design → Produce basic client side script for dynamic web pages
This page is part of the Web Design project. :
This unit defines the competency required to produce a number of client side scripts for dynamic web pages, utilising a range of relevant features from different appropriate languages.
This competency can be viewed directly on the Australian training.gov.au website:
The competency is outlined here together with ideas for demonstrating your competence as well as related modules where you can learn more skills relevant to this unit.

## Learning Outcomes

The specific performance criteria for this unit can be viewed online on the training.gov.au site
In outline, this unit requires that you:
* Analyse the exact requirements for your specific project, including the dynamic functionality (e.g. form validation, fades, animations, interactivity etc.) and the client-side scripting language that you can use to achieve this functionality.
* Design and create your HTML (ensuring that you meet accessibility requirements and other relevant standards) with your initial client-side script embedded or linked in a separate file.
* Test, debug and retest your scripts until you achieve the functionality that you're aiming for.

## Ideas for demonstrating competence

A combination of the activities mentioned in the module An Introduction to Programming with Javascript may help to demonstrate competence in this unit.

## Related Web Design modules

* An Introduction to Programming with Javascript
# Computer Hardware

Computer hardware refers to the physical parts or components of a computer. Hardware includes components such as the monitor, keyboard, hard drive disk, mouse, printers, graphic cards, sound cards, memory, motherboard and chips, etc.

## Lessons

* Introduction
* IC3/Computer Hardware
* IT Fundamentals/Hardware
* IT Fundamentals/Peripherals
* Computer Support/Hardware
* Hardware
* Input Devices
* Output Devices
* Operating computer hardware
* Caring for computer hardware

## Readings
### Wikipedia

* Computer hardware
* List of open-source hardware
* History of computing hardware
* Timeline of computing hardware before 1950
* List of computer hardware manufacturers
* Glossary of computer hardware terms
* Lists of computer hardware (Category)
* Computer hardware (Category)

## Draft learning resources

You may try to learn from these, or develop them further. Please keep in mind that they are drafts though, if reading/studying them. If you want to contribute to open source education, please consider developing these resources.
* Draft:RAM
# Troubleshoot Machines
## Troubleshooting is an organized approach of observing and testing that narrows down possible causes to a specific fault.

This article will use industrial manufacturing machines as a template that can be transferred to automobiles,...

### Common machine faults are:

* loss of operating power; fluctuating electrical current or voltage, drop in air or hydraulic pressure,
* wear, loosening of bolts/nuts,
SAFETY!
* LOCK OUT / TAG OUT!
* stored energy!
* Make sure you are safe and coworkers are safe!
Primary Systems
* mechanical
** motion
** pneumatics
** hydraulics
** lubrication
* electrical power
** motors
** heaters
** power supplies for control circuits
* instrumentation
* code
Troubleshooting intermittently functioning machines
# Machine design

MACHINE DESIGN
Machine design is an area of study in which one learns to design machine parts by utilizing its whole knowledge and experience in field of mechanical engineering.
OBJECTIVES OF DESIGN
* DESIGN FOR STRENGTH- safety is first so part to be designed should strong enough to withstand forces in its service. It also required to be resist some probable happening during service due to increased load by some reason.
* DESIGN FOR FUNCTION- every part in a machine has its own function for which it is applied in machine. designer should ensure that the part to be designed should fulfill its function satisfactory.
* DESIGN FOR MANUFACTURING PROCESS- Designer have to modify its product in the light of available manufacturing processes and their limitations and strengths. Design of machined part is generally different from that of casting etc.
* DESIGN FOR SAFETY- designer should ensure that the part he is designing should safe in use. And any life or health hazard will not produce from its working.
* DESIGN FOR LOOK AND COMFORT- It is a secondary priority of designer to make a part with high level of comfort for user and also looks good.

## Learning projects and resources

* Machine design
* Search for Machine design at Wikiversity

## Wikimedia

* w: Machine design
# Wikiphilosophers/Time

User:Elías Fortaleza de la Fuerza Sánchez believes that time is the conceptual spectrum through which we exist and percieve our existence in various spectra. Understanding the...Read more