nyuuzyou/wikis huggingface dataset:
First Order ODE/y' = sin^2 (x - y + 1)

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} = \map {\sin^2} {x - y + 1}^2$
has the general solution:
:$\map \tan {x - y + 1} = x + C$
== Proof ==
Make the substitution:
:$z = x - y + 1$
Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = 1, b = - 1$:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $3 \ \text{(b)}$
Category:Examples of First Order ODEs
First Order ODE/y' = x

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} = x$
has the general solution:
:$y = \dfrac {x^2} 2 + C$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Linear First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
First Order ODE/y' ln (x - y) = 1 + ln (x - y)

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$
is an exact differential equation with solution:
:$\paren {x - y} \map \ln {x - y} = C - y$
== Proof ==
Let $(1)$ be expressed as:
:$\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$
Let:
:$\map M {x, y} = 1 + \map \ln {x - y}$
:$\map N {x, y} = -\map \ln {x - y}$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $18$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/y - x y' = y' y^2 exp y

== Theorem ==
The first order ODE:
:$(1): \quad y - x y' = y' y^2 e^y$
has the general solution:
:$x y^2 = e^y + C$
== Proof ==
Let $(1)$ be rearranged as:
:$\dfrac {\d y} {\d x} = \dfrac y {y^2 e^y + x}$
Hence:
:$(2): \quad \dfrac {\d x} {\d y} - \dfrac 1 y x = y e^y$
It can be seen that $(2)$ is a linear first order ODE in the form:
:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
where:
:$\map P y = -\dfrac 1 y$
:$\map Q y = y e^y$
Thus:
Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
:$\map {\dfrac {\d} {\d y} } {\dfrac x y} = e^y$
and the general solution is:
:$\dfrac x y = e^y + C$
or:
:$x = y e^y + C y$
== Sources ==
* : $\S 2.10$: Problem $4 \ \text{(b)}$
Category:Examples of First Order ODEs
First Order ODE/y dx + (x^2 y - x) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$
has the general solution:
:$\dfrac {y^2} 2 - \dfrac y x = C$
This can also be presented in the form:
:$\dfrac {\d y} {\d x} + \dfrac y {x^2 y - x}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of First Order ODEs
Category:First Order ODE/y dx + (x^2 y - x) dy = 0
First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 1

== Theorem ==
This can also be presented in the form:
:$\dfrac {\d y} {\d x} + \dfrac y {x^2 y - x}$
== Proof ==
We note that $(1)$ is in the form:
:$\map M {x, y} \d x + \map N {x, y} \d y = 0$
but that $(1)$ is not exact.
So, let:
:$\map M {x, y} = y$
:$\map N {x, y} = x^2 y - x$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
It can be observed that:
Thus $\dfrac {\map P {x, y} } {\map N {x, y} }$ is a function of $x$ only.
So Integrating Factor for First Order ODE: Function of One Variable can be used:
:$\map \mu x = e^{\int \map g x \rd x}$
Hence:
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 {x^2}$
which yields, when multiplying it throughout $(1)$:
:$\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
which is now exact.
By First Order ODE: $\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$, its solution is:
:$\dfrac {y^2} 2 - \dfrac y x = C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Example $1$
Category:First Order ODE/y dx + (x^2 y - x) dy = 0
First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 2

== Theorem ==
== Proof ==
Rearranging, we have:
:$x^2 y \rd y - \paren {x \rd y - y \rd x} = 0$
Aiming to use Quotient Rule for Differentials, divide by $x^2$:
:$y \rd y = \dfrac {x \rd y - y \rd x} {x^2}$
So from Quotient Rule for Differentials: Formulation 1
:$y \rd y = \map \d {\dfrac y x}$
from which the solution immediately drops:
:$\dfrac {y^2} 2 - \dfrac y x = C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Example $1$
Category:First Order ODE/y dx + (x^2 y - x) dy = 0
First Order ODE/y dx + x dy + 3 x^3 y^4 dy

== Theorem ==
The first order ODE:
:$(1): \quad y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$
has the general solution:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
This can also be presented in the form:
:$\dfrac {\rd y} {\rd x} + \dfrac y {3 x^3 y^4 + x} = 0$
== Proof ==
We note that $(1)$ is in the form:
:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
but that $(1)$ is not exact.
So, let:
:$\map M {x, y} = y$
:$\map N {x, y} = 3 x^3 y^4 + x = x \paren {3 x^2 y^4 + 1}$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
It can be observed that:
Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.
So Integrating Factor for First Order ODE: Function of Product of Variables can be used.
Let $z = x y$.
Then:
:$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$
Hence:
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 {x^3 y^3}$
which yields, when multiplying it throughout $(1)$:
:$\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$
which is now exact.
By First Order ODE: $\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$, its solution is:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $2 \ \text{(c)}$
Category:Examples of First Order ODEs
First Order ODE/y dx - x dy = x y^3 dy

== Theorem ==
The first order ODE:
:$(1): \quad y rd x - x \rd y = x y^3 \rd y$
has the general solution:
:$\ln \dfrac x y = \dfrac {y^3} 3 + C$
== Proof ==
Rearranging, we have:
:$\dfrac {y \rd x - x \rd y} {x y} = y^2 \rd y$
From Differential of Logarithm of Quotient:
:$\map \d {\ln \dfrac y x} = \dfrac {y \rd x - x \rd y} {x y}$
from which:
:$\map \d {\ln \dfrac x y} = y^2 \rd y$
Hence the result:
:$\ln \dfrac x y = \dfrac {y^3} 3 + C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $4 \ \text{(b)}$
Category:Examples of First Order ODEs
First Order ODE/y dy = k dx

== Theorem ==
Let $k \in \R$ be a real number.
The first order ODE:
:$y \, \dfrac {\d y} {\d x} = k$
has the general solution:
:$y^2 = 2 k x + C$
== Proof ==
Category:Examples of First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
First Order ODE/y dy = k x dx

== Theorem ==
Let $k \in \R$ be a real number.
The first order ODE:
:$y \rd y = k x \rd x$
has the general solution:
:$y^2 = k x^2 + C$
== Proof ==
Category:Examples of First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$ a e \ne b d$
can be solved by substituting:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
to obtain:
:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
which can be solved by the technique of Solution to Homogeneous Differential Equation.
== Proof ==
We have:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
Make the substitutions:
:$x := z - h$
:$y := w - k$
We have:
:$\dfrac {\d x} {\d z} = 1$
:$\dfrac {\d y} {\d w} = 1$
Thus:
In order to simplify this appropriately, we wish to reduce it to the form:
:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
by finding values of $h$ and $k$ such that:
:$a h + b k = c$
:$d h + e k = f$
So:
Similarly:
We note that the above works :
:$a e - b d \ne 0 \implies a e \ne b d$
Thus:
:$(1): \quad \dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
Letting:
:$\map M {z, w} = a z + b w$
:$\map N {z, w} = d z + e w$
we see:
:$\map M {t z, t w} = a t z + b t w = t \paren {a z + b w} = t \, \map M {z, w}$
:$\map N {t z, t w} = d t z + e t w = t \paren {d z + e w} = t \, \map N {z, w}$
Thus, by definition, $(1)$ is homogeneous.
== Examples ==
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $4 \text{(a)}$
* : : $(4)$ Equations reducible to homogeneous form
* : : $(4)$ Equations reducible to homogeneous form
Category:First Order ODEs
First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
== Examples ==
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $4 \text{(b)}$
Category:First Order ODEs
First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 1

== Theorem ==
The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$a e = b d$
can be solved by substituting:
:$z = a x + b y$
to obtain:
:$\dfrac {\d z} {\d x} = b \map F {\dfrac {a z + a c} {d z + f} } + a$
which can be solved by Solution to Separable Differential Equation.
== Proof ==
When $a e = b d$, it is not possible to make the substitutions:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
and so to use the technique of First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.
So, we consider what needs to be done to make $(1)$ separable.
Let us make the substitution:
:$z = x + r y$
Consider what, if any, value of $r$ would make $(1)$ separable.
We have:
To make $(1)$ separable: we make:
:$b = a r$
and:
:$d r = \dfrac {b d} a$
which comes to the same thing: that $r = \dfrac b a$.
So, we can make the substitution:
:$z = a x + b y$
so:
:$\dfrac {\d z} {\d x} = a + b \dfrac {\d y} {\d x}$
which leaves us with:
== Sources ==
* : : $(4)$ Equations reducible to homogeneous form
* : : $(4)$ Equations reducible to homogeneous form
Category:First Order ODEs
First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 2

== Theorem ==
The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$a e = b d$
can be solved by substituting:
:$z = d x + e y$
to obtain:
:$\dfrac {\d z} {\d x} = e \map F {\dfrac {b z + e c} {e z + e f} } + d$
which can be solved by Solution to Separable Differential Equation.
== Proof ==
When $a e = b d$, it is not possible to make the substitutions:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
and so to use the technique of First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.
So, we consider what needs to be done to make $(1)$ separable.
Let us make the substitution:
:$z = x + r y$
Consider what, if any, value of $r$ would make $(1)$ separable.
We have:
To make $(1)$ separable: we make:
:$e = d r$
and:
:$a r = \dfrac {a e} d$
which comes to the same thing: that $r = \dfrac e d$.
So, we can make the substitution:
:$z = d x + e y$
so:
:$\dfrac {\d z} {\d x} = d + e \dfrac {\d y} {\d x}$
which leaves us with:
Category:First Order ODEs
First Order ODE in form y' = f (a x + b y + c)

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$
can be solved by substituting:
:$z := a x + b y + c$
to obtain:
:$\ds x = \int \frac {\d z} {b \map f z + a}$
== Proof ==
We have:
:$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$
Put:
: $z := a x + b y + c$
Then:
This can be solved by Solution to Separable Differential Equation:
:$\ds x = \int \frac {\d z} {b \map f z + a}$
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $3$
Category:First Order ODEs
First Pappus-Guldinus Theorem

== Theorem ==
Let $C$ be a plane figure that lies entirely on one side of a straight line $\LL$.
Let $S$ be the solid of revolution generated by $C$ around $\LL$.
Then the volume of $S$ is equal to the area of $C$ multiplied by the distance travelled by the centroid of $C$ around $\LL$ when generating $S$.
== Proof ==
Let $V$ denote the volume of $S$
Let $\AA$ denote the area of $C$.
Let $C$ be embedded in a Cartesian plane such that $\LL$ coincides with the $x$-axis.
Let $\tuple {\overline x, \overline y}$ be the coordinates of the centroid of $C$.
Consider a rectangle $R$ bounded by the lines:
:$y = 0$
:$x = \xi$
:$x = \xi + \delta x$
:$y = \map f x$
The moment $M_y$ of $R$ about the $y$-axis is given by:
:$M_y = \map f x \xi \rdelta \xi$
Hence from Area under Curve:
:$\AA \overline x = \ds \int_a^b x \map f x \rd x$
The moment $M_x$ of $R$ about the $x$-axis is given by:
:$M_x = y \rdelta x \dfrac y 2$
that is, half way up.
Hence:
:$\AA \overline y = \dfrac 1 2 \ds \int_a^b y^2 \rd x$
It follows immediately that:
:$\dfrac V \AA \overline y = 2 \pi$
That is:
:$V = 2 \pi \AA \overline y$
== Also known as ==
== Also see ==
* Second Pappus-Guldinus Theorem
== Sources ==
* : $\text {II}$. Calculus: Area, Volume and Centre of Gravity: Centre of Gravity
* : Chapter $\text {A}.8$: Pappus (fourth century A.D.)
* 
* 
* 
Category:Pappus's Theorems
Category:Solids of Revolution
First Power of Element in B-Algebra

== Theorem ==
Let $\left({X, \circ}\right)$ be a $B$-algebra.
Then:
:$\forall x \in X: x^1 = x$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
== Proof ==
Hence the result.
Category:B-Algebras
First Principle of Transfinite Induction

== Theorem ==
Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $P$ be a property that satisfies the following condition:
:For all $x \in A$, if $P$ holds for every $y \prec x$, then $P$ holds for $x$.
Then $P$ holds for all $x \in A$.
== Proof ==
Let $s$ be the smallest element of $A$.
Then vacuously $P$ holds for every element $y \in A$ such that $y \prec s$.
Thus $P$ holds for $s$.
 $P$ fails to hold for some $z \in A$.
The class of elements of $A$ for which $P$ does not hold is therefore non-empty.
We have that $A$ is a well-ordered class.
Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.
This cannot be $s$ as we have seen above.
But for every $y \prec x$, we have that $P$ holds for $y$.
But then $P$ holds for $x$.
This contradicts our definition of $x$.
Hence there can be no such $z \in A$ for which $P$ fails to hold.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.8$ (Transfinite Induction Principle $1$)
Category:Well-Orderings
Category:Mathematical Induction
First Projection on Ordered Pair of Sets

== Theorem ==
Let $a$ and $b$ be sets.
Let $w = \tuple {a, b}$ denote the ordered pair of $a$ and $b$.
Let $\map {\pr_1} w$ denote the first projection on $w$.
Then:
:$\ds \map {\pr_1} w = \bigcup \bigcap w$
where $\ds \bigcup$ and $\ds \bigcap$ denote union and intersection respectively.
== Proof ==
We have by definition of first projection that:
:$\map {\pr_1} w = \map {\pr_1} {a, b} = a$
Then:
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.6 \ \text{(i)}$
Category:Ordered Pairs
Category:Projections
First Ring Isomorphism Theorem/Also known as

== First Ring Isomorphism Theorem: Also known as == 
The First Ring Isomorphism Theorem is also referred to as the First Fundamental Theorem on Ring Homomorphisms.
It can also be called the First Isomorphism Theorem for Rings.
== Sources ==
* : $\S 60.3$ Factor rings
Category:First Ring Isomorphism Theorem
First Socratic Paradox

== Paradox ==
 was supposed to have said:
:I know nothing at all.
In which case, how does Socrates know that he knows nothing?
=== Resolution ===
If he knows that he knows nothing, then it is false that he knows nothing, because there's (at least) one fact he has knowledge of: his lack of knowledge.
Hence I know nothing at all must always be false, whoever utters it.
== Also see ==
* Second Socratic Paradox
Category:Socratic Paradoxes
First Subsequence Rule

== Theorem ==
Let $T = \struct {A, \tau}$ be a Hausdorff space.
Let $\sequence {x_n}$ be a sequence in $T$.
Suppose $\sequence {x_n}$ has two convergent subsequences with different limits.
Then $\sequence {x_n}$ is divergent.
== Proof ==
As stated, let $T = \struct {A, \tau}$ be a Hausdorff space.
Let $\sequence {x_n}$ be a sequence in $T$.
Let $\sequence {y_n}$ and $\sequence {z_n}$ be convergent subsequences of $\sequence {x_n}$ with different limits.
 $\sequence {x_n}$ is convergent.
From Convergent Sequence in Hausdorff Space has Unique Limit, $\sequence {x_n}$ has exactly one limit.
From Limit of Subsequence equals Limit of Sequence, $\sequence {y_n}$ has the same limit as $\sequence {x_n}$.
This contradicts the assertion that $\sequence {y_n}$ and $\sequence {z_n}$ have different limits.
Hence by Proof by Contradiction it must be the case that $\sequence {x_n}$ is divergent.
== Also see ==
* Second Subsequence Rule
Category:Hausdorff Spaces
Category:Named Theorems
First Supplement to Law of Quadratic Reciprocity

== Theorem ==
$\quad \paren {\dfrac {-1} p} = \paren {-1}^{\paren {p - 1} / 2} = \begin {cases}
+1 & : p \equiv 1 \pmod 4 \\
-1 & : p \equiv 3 \pmod 4
\end {cases}$
where $\paren {\dfrac {-1} p}$ is defined as the Legendre symbol.
== Proof ==
From Euler's Criterion for Quadratic Residue, and the definition of the Legendre symbol, we have that:
:$\paren {\dfrac a p} \equiv a^{\paren {p - 1} / 2} \pmod p$
The result follows by putting $a = -1$.
== Examples ==
== Also see ==
* Law of Quadratic Reciprocity
* Second Supplement to Law of Quadratic Reciprocity
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
* 
* 
Category:Legendre Symbol
Category:Named Theorems
Category:Law of Quadratic Reciprocity
First Supplement to Law of Quadratic Reciprocity/Examples

== Examples of First Supplement to Law of Quadratic Reciprocity ==
=== $-1$ is not Quadratic Residue of $3$ ===
=== $-1$ is Quadratic Residue of $5$ ===
=== $-1$ is not Quadratic Residue of $7$ ===
=== $-1$ is not Quadratic Residue of $11$ ===
=== $-1$ is Quadratic Residue of $13$ ===
=== $-1$ is Quadratic Residue of $17$ ===
=== $-1$ is not Quadratic Residue of $19$ ===
Category:Law of Quadratic Reciprocity
First Supplement to Law of Quadratic Reciprocity/Examples/11

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic non-residue of $11$.
== Proof ==
From Quadratic Residues of 11:
:the set of quadratic residues modulo $11$ is:
::$\set {1, 3, 4, 5, 9}$
:but not $10 \equiv -1 \pmod {11}$.
We also have that:
:$11 = 4 \times 3 - 1$
and so is not of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:11
First Supplement to Law of Quadratic Reciprocity/Examples/13

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic residue of $13$.
== Proof ==
We have that:
:$5^2 = 25 \equiv -1 \pmod 13$
and so $-1$ is therefore a quadratic residue of $13$.
We also have that:
:$13 = 4 \times 3 + 1$
and so is of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:13
First Supplement to Law of Quadratic Reciprocity/Examples/17

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic residue of $17$.
== Proof ==
From Quadratic Residues of 17:
:the set of quadratic residues modulo $13$ is:
:$\set {1, 2, 4, 8, 9, 13, 15, 16}$
and so $-1 \equiv 16 \pmod {17}$ is therefore a quadratic residue of $17$.
We also have that:
:$17 = 4 \times 4 + 1$
and so is of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:17
First Supplement to Law of Quadratic Reciprocity/Examples/19

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic non-residue of $19$.
== Proof ==
We have that:
:$19 = 4 \times 5 - 1$
and so is not of the form $4 n + 1$.
The result follows from the First Supplement to Law of Quadratic Reciprocity.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:19
First Supplement to Law of Quadratic Reciprocity/Examples/3

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic non-residue of $3$.
== Proof ==
From Quadratic Residues of 3:
:there exists exactly $1$ quadratic residue modulo $3$, and that is $1$.
We also have that:
:$3 = 4 \times 1 - 1$
and so is not of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:3
First Supplement to Law of Quadratic Reciprocity/Examples/5

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic residue of $5$.
== Proof ==
From Quadratic Residues of 5:
:the set of quadratic residues modulo $5$ is:
::$\set {1, 4}$
We have that:
:$4 \equiv -1 \pmod 5$
and so $-1$ is therefore a quadratic residue of $5$.
We also have that:
:$5 = 4 \times 1 + 1$
and so is of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:5
First Supplement to Law of Quadratic Reciprocity/Examples/7

== Examples of First Supplement to Law of Quadratic Reciprocity ==
:$-1$ is a quadratic non-residue of $7$.
== Proof ==
From Quadratic Residues of 7:
:the set of quadratic residues modulo $7$ is:
::$\set {1, 2, 4}$
:but not $6 \equiv -1 \pmod 7$.
We also have that:
:$7 = 4 \times 2 - 1$
and so is not of the form $4 n + 1$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
Category:Law of Quadratic Reciprocity
Category:7
First Sylow Theorem

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also see ==
* Sylow Theorems
== Note ==
By Orbits of Group Action on Sets with Power of Prime Size, it was clear that $k \divides \size {\Orb S}$.
However here, since it is established that $\size {\Stab S} = p^n$, and by Orbit-Stabilizer Theorem, we also have $k = \size {\Orb S}$.
== Historical Note ==
=== First Sylow Theorem ===
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.18$
Category:Sylow Theorems
Category:First Sylow Theorem
First Sylow Theorem/Corollary

== Corollary to First Sylow Theorem ==
Let $p$ be a prime number.
Let $G$ be a group.
Let:
:$p^n \divides \order G$
where:
:$\order G$ denotes the order of $G$
:$n$ is a positive integer.
Then $G$ has at least one subgroup of order $p$.
== Proof 1 ==
== Proof 2 ==
This result can also be proved directly:
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem
Category:Sylow Theorems
Category:First Sylow Theorem
First Sylow Theorem/Corollary/Proof 1

== Corollary to First Sylow Theorem ==
== Proof ==
Let $\order G = k p^r$ where $p \nmid k$.
From the First Sylow Theorem, $G$ has a subgroup $S$ of order $p^r$.
From (need to find it), $S$ itself has subgroups of order $p^n$ for all $n \in \set {1, 2, \ldots, r}$.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 56$. First Sylow Theorem
Category:First Sylow Theorem