nyuuzyou/wikis huggingface dataset:
There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers

== Theorem ==
Let $n \in \N$ be a natural number.
Let $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ be the $n$th, $n + 1$th, $n + 2$th and $n + 3$th triangular numbers respectively.
Then it is not the case that all of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are sphenic numbers.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic: $T_{28}$, $T_{29}$ and $T_{30}$ that is, $406$, $435$ and $465$, are the smallest $3$ consecutive triangular numbers which are all sphenic.
* Sequences of 3 Consecutive Triangular Numbers which are Sphenic
Category:Triangular Numbers
Category:Sphenic Numbers
Category:There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers
There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers/Proof 1

== Theorem ==
== Proof ==
Let $\map \Omega n$ denote the number of prime factors of $n$ counted with multiplicity.
 there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Thus from Closed Form for Triangular Numbers:
Recall the definition of sphenic number:
Hence:
That is:
Suppose one of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.
Then at least one of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ is divisible by $4$, and so is not sphenic.
Hence none of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.
We have that:
and similarly:
That is:
:$\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$
Then we have:
There are three possibilities for $n$:
=== $\map \Omega n = 1$ ===
Suppose $\map \Omega n = 1$.
Then $n$ is prime by definition.
Then as $\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$ that means they are all prime. 
But $n$, $n + 2$ and $n + 4$ cannot be all primes unless $n = 3$.
But we have:
none of which are sphenic.
Thus:
:$\map \Omega n \ne 1$
=== $\map \Omega n = 2$ ===
Suppose $\map \Omega n = 2$.
We have:
At least one of $n$, $n + 1$, $n + 2$, $n + 3$ and $n + 4$ is divisible by $4$.
But because $\map \Omega 4 = 2$, the only natural number divisible by $4$ is $4$ itself.
Thus one of $n$, $n + 1$, $n + 2$, $n + 3$ and $n + 4$ has to be equal to $4$.
We have already noted that none of $T_3$ to $T_6$ is sphenic.
Thus:
:$\map \Omega n \ne 2$
=== $\map \Omega n = 3$ ===
Suppose $\map \Omega n = 3$.
From the above, we have:
:$\map \Omega {n + 1} = \map \Omega {n + 3} = 1$
and from $\map \Omega {n \paren {n + 1} } = 4$ we have:
:$\map \Omega {n + 1} = 1$
Thus $n + 1$ and $n + 3$ are twin primes.
Thus neither $n + 1$ nor $n + 3$ is divisible by $3$.
Thus $n + 2$ is divisible by $3$.
Suppose $n + 2$ is divisible by $12$.
Then as $\map \Omega {n + 2} = 3$, it follows that:
:$n + 2 = 12$
We investigate the case where $n + 2 = 12$:
So $n + 2$ is not $12$ and so cannot be divisible by $12$.
But $n + 2$ is divisible by $2$, but not $4$.
Thus $n$ and $n + 4$ must be divisible by $4$.
Hence either $n$ or $n + 4$ will be divisible by $8$.
This contradicts our deduction that neither $n$ nor $n + 4$ can be divisible by $8$.
Thus:
:$\map \Omega n \ne 3$
We have deduced that there exist no $n \in \N$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Hence the result by Proof by Contradiction.
Category:There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers
There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers/Proof 2

== Theorem ==
== Proof ==
 there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Observe from Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic that there are no such $n$ for $n < 12$.
Thus from Closed Form for Triangular Numbers:
Recall the definition of sphenic number:
Let $S = \set {n, n + 1, n + 2, n + 3, n + 4}$.
Then there exists $x \in S$ such that $4 \divides x$.
If $x = n$ or $x = n + 4$, then one of $n$ or $n + 4$ is divisible by $8$.
Then one of $T_n$ or $T_{n + 3}$ is divisible by $4$, so it cannot be sphenic.
Therefore $x \ne n$ and $x \ne n + 4$.
Hence $x - 1 \in S$ and $x + 1 \in S$.
Note that both $T_{x - 1}$ and $T_x$ are sphenic.
Write $x = 4 m$ for some integer $m$.
Since $x > n \ge 12$, we have $m > 3$.
Note that:
so as $T_x$ is sphenic, both $m$ and $4 m + 1$ must be prime.
Similarly, since $T_{x - 1}$ is sphenic, $4 m - 1$ must be prime.
One of $4 m - 1, 4 m, 4 m + 1$ is divisible by $3$.
As $4 m + 1 > 4 m - 1 > 3$, $4 m$ must be divisible by $3$.
By Euclid's Lemma, since $3 \perp 4$:
:$3 \divides m$
which contradicts the fact that $m$ is prime.
Hence the result by Proof by Contradiction.
Category:There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers
Think of a Number/Examples

== Examples of Think of a Number puzzles ==
=== Rhind Papyrus Problem $28$ ===
=== Rhind Papyrus Problem $30$ ===
=== Rhind Papyrus Problem $33$ ===
=== Bachet: $1$ ===
=== Bachet: $2$ ===
=== Bachet: $3$ ===
Category:Think of a Number Puzzles
Think of a Number/Examples/Bachet/1

== Classic Problem ==
:A person chooses secretly a number, and trebles it, telling you whether the product is odd or even.
:If it is even, he takes half of it,
:or if it is odd, he adds one and then takes one half.
:Next he multiplies the result by $3$,
:and tells you how many times $9$ will divide into the answer, ignoring the remainder.
:The number he chose is -- what?
== Solution ==
Let $n$ be the number given at the end.
If $n$ is stated as being even, then the number originally chosen was $2 n$.
If $n$ is stated as being odd, then the number originally chosen was $2 n + 1$.
== Proof ==
Let $x$ be the number chosen.
Let $x$ be either $2 n$ or $2 n + 1$, depending on whether it is odd or even.
If $x$ is even, the successive operations do the following:
:$2 n \to 6 n \to 3 n \to 9 n \to n$
If $x$ is odd, the successive operations do the following:
:$2 n + 1 \to 6 n + 3 \to 6 n + 4 \to 3 n + 2 = 9 n + 6 \to n$
== Sources ==
* 
* : Bachet: $105$
Category:Think of a Number Puzzles
Think of a Number/Examples/Bachet/2

== Classic Problem ==
:The subject chooses a number less than $60$
:and tells you the remainders when it is divided by $3$, $4$ and $5$, separately, not successively.
:The original number is -- what?
== Solution ==
Let $x$ be the number chosen.
Let $A$, $B$ and $C$ be the remainders on dividing $x$ by $3$, $4$ and $5$ respectively.
Then the original number is the remainder when dividing $40 A + 45 B + 36 C$ by $60$.
== Proof ==
We have that:
So $40 A + 45 B + 36 C$ has the same remainder on dividing by $60$ as does $x$.
Because $x$ has been chosen to be less than $60$, the result follows.
== Sources ==
* 
* : Bachet: $106$
Category:Think of a Number Puzzles
Think of a Number/Examples/Bachet/3

== Classic Problem ==
:The first person takes a number of counters greater than $5$.
:The second person takes $3$ times as many.
:The first person gives $5$ counters to the second.
:The second person then gives the first $3$ times as many as the first person holds in his hand.
:How many counters has the second person got in his hand?
== Solution ==
$20$
== Proof ==
Let $A$ and $B$ denote the first and second person respectively.
Let $x$ be the number taken by $A$.
Then $B$ takes $3 x$ from $A$.
Then $B$ gives $5$ counters to $A$.
At this point, $A$ has $x = 5$ and $B$ has $3 x + 5$.
Then $A$ gives $3 \paren {x - 5}$ to $B$.
Hence $B$ now has $3 x + 5 - 3 \paren {x - 5}$
Thus, doing the algebra, we see:
:$3 x + 5 - 3 x + 3 \times 5 = 20$
== Sources ==
* 
* : Bachet: $107$
Category:Think of a Number Puzzles
Think of a Number/Examples/Rhind Papyrus 28

== Example of Think of a Number puzzles ==
Problem $28$ of the is as follows:
:$\dfrac 2 3$ is to be added.
:$\dfrac 1 3$ is to be subtracted.
:There remains $10$.
== Solution ==
This can more clearly be expressed as:
:I think of a number.
:I add to it $\dfrac 2 3$ of the number.
:I then subtract $\dfrac 1 3$ of the sum.
:My answer is $10$.
:What number did I think of?
The number was $9$.
== Proof ==
Let $x$ be the number first thought of.
We have:
== Sources ==
* : Problem $28$
* : Problem $28$
* : Think of a Number: $7$
Category:Think of a Number Puzzles
Category:Rhind Papyrus
Think of a Number/Examples/Rhind Papyrus 28/Mistake

== Source Work ==
: 
:The Puzzles:
::Think of a Number: $7$
== Mistake ==
:Problem $29$ of the Rhind papyrus is not quite so clear, but it is plausibly the first ever 'Think of a Number' problem. It reads ...
== Correction ==
According to , which cites, this is actually Problem $28$.
It also needs to be pointed out that there are a number of such problems of a similar nature in this section of the , so this particular instance cannot really be considered to be "the first ever".
== Sources ==
* : Think of a Number: $7$
Category:Mistakes/Curious and Interesting Puzzles
Think of a Number/Examples/Rhind Papyrus 30

== Example of Think of a Number puzzles ==
Problem $30$ of the is as follows:
:If the scribe says to thee:
::$10$ has become $\dfrac 2 3 + \dfrac 1 {10}$ of what?
== Solution ==
This can more clearly be expressed as:
:I think of a number.
:$\dfrac 2 3$ of that number plus $\dfrac 1 {10}$ of it make $10$.
:What number did I think of?
The number was $13 + \dfrac 1 {23}$.
== Proof ==
Let $x$ be the number first thought of.
We have:
== Sources ==
* : Problem $30$
* : Problem $30$
* : Think of a Number: $8$
Category:Think of a Number Puzzles
Category:Rhind Papyrus
Think of a Number/Examples/Rhind Papyrus 33

== Example of Think of a Number puzzles ==
Problem $30$ of the is as follows:
:A number, plus its two-thirds, and plus its half, plus its seventh, makes $37$. What is the number?
It is to be expressed in Egyptian form.
== Solution ==
This can more clearly be expressed as:
:I think of a number.
:If you add $\dfrac 2 3$ of that number plus $\dfrac 1 2$ of that number and $\dfrac 1 7$ of that number, you get $37$.
:What number did I think of?
The number was $16 + \dfrac 1 {56} + \dfrac 1 {679} + \dfrac 1 {776}$.
== Proof ==
Let $x$ be the number first thought of.
We have:
== Sources ==
* : Problem $30$
* : Problem $30$
* : Think of a Number: $9$
Category:Think of a Number Puzzles
Category:Rhind Papyrus
Third Apotome/Example

== Example of Third Apotome==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a third apotome :
:$(1): \quad a \notin \Q$
:$(2): \quad b \notin \Q$
:$(3): \quad \dfrac {\sqrt {a^2 - b^2}} a \in \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = \sqrt {11}$ and $b = \sqrt {\frac {143} {49} }$.
Then:
Therefore $\sqrt {11} - \sqrt {\dfrac {143} {49}}$ is a third apotome.
Category:Examples of Apotome
Third Derivative of Inverse Function

== Theorem ==
Let $f$ be a real function which is of differentiability class $3$.
Let $f$ have an inverse $f^{-1}$, likewise of differentiability class $3$.
Then:
:$\dfrac {\d^3 x} {\d y^3} = -\paren {\dfrac {\d^3 y} {\d x^3} \dfrac {\d y} {\d x} - 3 \paren {\dfrac {\d^2 y} {\d x^2} }^2} \paren {\dfrac {\d y} {\d x} }^{-5}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Leibniz's Theorem for Differentiation of a Product: $3.3.11$
Category:Derivative of Inverse Function
Third Derivative of Natural Logarithm Function

== Theorem ==
Let $\ln x$ be the natural logarithm function.
Then:
:$\map {D^3_x} {\ln x} = \dfrac 2 {x^3}$
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Derivatives involving Logarithm Functions
Category:Natural Logarithms
Third Hyperoperation is Integer Power Operation

== Theorem ==
The $3$rd hyperoperation is the integer power operation restricted to the positive integers:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$
== Proof ==
By definition of the hyperoperation sequence:
:$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
Thus the $3$rd hyperoperation is defined as:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
H_2 \left({x, H_3 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$
From Second Hyperoperation is Multiplication Operation:
:$(1): \quad \forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
x \times H_3 \left({x, y - 1}\right) & : y > 0 \end{cases}$
The proof proceeds by induction.
For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
:$\forall x \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$
=== Basis for the Induction ===
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\forall x \in \Z_{\ge 0}: H_3 \left({x, k}\right) = x^k$
from which it is to be shown that:
:$\forall x \in \Z_{\ge 0}: H_3 \left({x, k + 1}\right) = x^{k + 1}$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$
Category:Hyperoperation
Third Isomorphism Theorem

== Theorem ==
=== Groups ===
=== Rings ===
== Also known as ==
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.
This particular result, for example, is also known both as the first isomorphism theorem and the second isomorphism theorem, according to source.
== Also see ==
* Isomorphism Theorems
Category:Isomorphisms (Abstract Algebra)
Category:Named Theorems
Category:Isomorphism Theorems
Third Isomorphism Theorem/Groups

== Theorem ==
Let $G$ be a group, and let:
:$H, N$ be normal subgroups of $G$
:$N$ be a subset of $H$.
Then:
:$(1): \quad N$ is a normal subgroup of $H$
:$(2): \quad H / N$ is a normal subgroup of $G / N$
:::where $H / N$ denotes the quotient group of $H$ by $N$
:$(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
:::where $\cong$ denotes group isomorphism.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
This result is also referred to by some sources as the first isomorphism theorem, and by others as the second isomorphism theorem.
== Also see ==
* Isomorphism Theorems
== Sources ==
* : $1$: Subgroups: $1.\text{T}.4$
Category:Group Isomorphisms
Category:Quotient Groups
Category:Normal Subgroups
Category:Isomorphism Theorems
Category:Third Isomorphism Theorem
Third Isomorphism Theorem/Groups/Corollary

== Corollary to Third Isomorphism Theorem for Groups ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $q: G \to \dfrac G N$ be the quotient epimorphism from $G$ to the quotient group $\dfrac G N$.
Let $K$ be the kernel of $q$.
Then:
:$\dfrac G N \cong \dfrac {G / K} {N / K}$
== Proof ==
From Kernel is Normal Subgroup of Domain we have that $K$ is a normal subgroup of $G$.
Thus the Third Isomorphism Theorem for Groups can be applied directly.
== Also known as ==
Some sources refer to this as the first isomorphism theorem.
== Sources ==
* : $\S 1.10$: Theorem $30$
* : Chapter $\text{II}$: Groups: Factoring Morphisms: Corollary
Category:Group Isomorphisms
Category:Quotient Groups
Category:Normal Subgroups
Category:Isomorphism Theorems
Third Isomorphism Theorem/Groups/Proof 1

== Theorem ==
== Proof ==
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
We define a mapping:
:$\phi: G / N \to G / H$ by $\map \phi {g N} = g H$
Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.
Suppose $x N = y N$.
Then:
:$y^{-1} x \in N$
Then:
:$N \le H \implies y^{-1} x \in H$
and so:
:$x H = y H$
So:
:$\map \phi {x N} = \map \phi {y N}$
and $\phi$ is indeed well-defined.
Now $\phi$ is a homomorphism, from:
Also, since $N \subseteq H$, it follows that:
:$\order N \le \order H$
So:
:$\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.
So:
The result follows from the First Isomorphism Theorem.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 68$. The First Isomorphism Theorem
* : Chapter $8$: The Homomorphism Theorem: Theorem $8.16$
Category:Third Isomorphism Theorem
Third Isomorphism Theorem/Groups/Proof 2

== Theorem ==
== Proof ==
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
Let $q_H$ and $q_N$ be the quotient mappings from $G$ to $\dfrac G H$ and $G$ to $\dfrac G N$ respectively.
Hence:
:$N \subseteq \map \ker {q_H}$
From Quotient Theorem for Group Homomorphisms: Corollary 2, it therefore follows that:
:there exists a group epimorphism $\psi: \dfrac G N \to \dfrac G H$ such that $\psi \circ q_N = q_H$
Then we have that :
:there exists a group epimorphism $\phi: \dfrac {G / N} {H / N} \to \dfrac G N$ such that $\phi \circ q_{H / N} = \psi$
:
:$H / N \subseteq \map \ker \psi$
Thus we form the composite:
:$\phi \circ q_{H / N} \circ q_N = q_H$
and it remains to be shown that $\phi$ is an isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.15$
Category:Third Isomorphism Theorem
Third Isomorphism Theorem/Groups/Proof 3

== Theorem ==
== Proof ==
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
Let $q_H$ denote the quotient mapping from $G$ to $\dfrac G H$.
Let $q_N$ denote the quotient mapping from $G$ to $\dfrac G N$.
Let $\RR$ be the congruence relation defined by $N$ in $G$.
Let $\TT$ be the congruence relation defined by $H$ in $G$.
Thus from Congruence Relation induces Normal Subgroup:
:$q_H = q_\TT$
and:
:$q_N = q_\RR$
where $q_\RR$ and $q_\TT$ denote the quotient epimorphisms induced by $\RR$ and $\TT$ respectively.
We have that:
That is:
:$\RR \subseteq \TT$
Let $\SS$ be the relation on the quotient group $G / N$ which satisfies:
:$\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x \mathrel \TT y$
That is, by definition of $\TT$:
:$\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x H = y H$
Then by Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
:$\SS$ is a congruence relation on $G / N$.
Hence by Congruence Relation on Group induces Normal Subgroup:
:$\SS$ induces a normal subgroup of $G / N$.
Again from Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
:there exists a unique isomorphism $\phi$ from $G / N$ to $G / H$ which satisfies:
::$\phi \circ q_\SS \circ q_\RR = q_\TT$
:where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.
That is:
:$\phi \circ q_{H / N} \circ q_N = q_H$
and the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.19 \ \text {(b)}$
Category:Third Isomorphism Theorem
Third Isomorphism Theorem/Rings

== Theorem ==
Let $R$ be a ring.
Let:
:$J, K$ be ideals of $R$
:$J$ be a subset of $K$.
Then:
:$(1): \quad K / J$ is an ideal of $R / J$
::where $K / J$ denotes the quotient ring of $K$ by $J$
:$(2): \quad \dfrac {R / J} {K / J} \cong \dfrac R K$
::where $\cong$ denotes ring isomorphism.
== Proof ==
By Ideal Contained in Larger Ideal is Ideal of Larger Ideal, $J$ is indeed an ideal of $K$.
Hence $K / J$ is adequately defined.
In Ring Homomorphism whose Kernel contains Ideal‎, take $\phi: R \to R / K$ to be the quotient epimorphism.
Then (from the same source) its kernel is $K$.
Thus we have that:
:$\phi = \psi \circ \nu$
where $\psi : R / J \to R / K$ is a homomorphism.
This can be illustrated by means of the following commutative diagram:
$\quad\quad \begin{xy}\xymatrix@L+2mu@+1em{
R \ar[dr]_*{\phi}
\ar[r]^*{\nu}
& R / J \ar@{-->}[d]^*{\psi}
\\ 
&
R / K 
}\end{xy}$
As $\phi$ is an epimorphism then from Surjection if Composite is Surjection we have that $\psi$ is a surjection.
So:
:$\Img \psi = \Img \phi = R / K$
and the First Ring Isomorphism Theorem applies.
== Also known as ==
== Also see ==
* Isomorphism Theorems
== Sources ==
* : $\S 2.2$: Homomorphisms: Theorem $2.11$
* 
Category:Third Ring Isomorphism Theorem
Category:Ring Isomorphisms
Category:Ideal Theory
Category:Isomorphism Theorems
Third Partial Derivative/Examples/u = ln (x^2 + y)

== Examples of Third Partial Derivatives ==
Let $u = \map \ln {x^2 + y}$ be a real function of $2$ variables such that $x^2 + y \in \R_{>0}$.
Then:
:$\dfrac {\partial^3 u} {\partial y^2 \partial x} = \dfrac {\partial^3 u} {\partial x \partial y^2} = \dfrac {\partial^3 u} {\partial x \partial y \partial x} = \dfrac {4 x} {\paren {x^2 + y}^3}$
== Proof ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives: Example $\text D$
Category:Examples of Partial Derivatives
Third Partial Derivatives of x^y

== Theorem ==
Let:
:$u = x^y$
Then:
:$\dfrac {\partial^3 u} {\partial x^2 \partial y} = \dfrac {\partial^3 u} {\partial x \partial y \partial x}$
== Proof ==
The two expressions are equal.
Hence the result.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $11$
Category:Examples of Partial Derivatives
Third Principle of Mathematical Induction

== Theorem ==
Let $\map P n$ be a propositional function depending on $n \in \N$.
If:
:$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$
:$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$
then $\map P n$ is true for all $n \in \N$.
== Proof ==
Let $A = \set {n \in \N: \map P n}$.
We show that $A$ is an inductive set.
By $(1)$:
:$\forall 1 \le i \le d: i \in A$
Let:
:$\forall x \ge d: \set {1, 2, \dotsc, x} \subset A$
Then by definition of $A$:
:$\forall k \in \N: x - \paren {d - 1} \le k < x + 1: \map P k$
Thus $\map P {x + 1} \implies x + 1 \in A$
Thus $A$ is an inductive set.
Thus by the fifth axiom of Peano:
:$\forall n \in \N: A = \N \implies \map P n$
== Sources ==
* : Exercise $2.3: \ 5$
Category:Number Theory
Category:Named Theorems
Category:Mathematical Induction
Category:Proof Techniques
Third Ring Isomorphism Theorem/Also known as

== Third Ring Isomorphism Theorem: Also known as == 
The Third Ring Isomorphism Theorem is also referred to by some sources as the First Ring Isomorphism Theorem, and by others as the Second Ring Isomorphism Theorem.
There is little consistency in the literature.
It can also be called the Third Isomorphism Theorem for Rings.
Category:Third Ring Isomorphism Theorem
Third Sylow Theorem

== Theorem ==
All the Sylow $p$-subgroups of a finite group are conjugate.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Some sources call this the fourth Sylow theorem, and merge it with what we call the Fifth Sylow Theorem.
Others call this the second Sylow theorem.
== Also see ==
* Sylow Theorems
== Historical Note ==
Category:Sylow Theorems
Category:Third Sylow Theorem
Third Sylow Theorem/Proof 1

== Theorem ==
== Proof ==
Suppose $P$ and $Q$ are Sylow $p$-subgroups of $G$.
By the Second Sylow Theorem, $Q$ is a subset of a conjugate of $P$.
But since $\order P = \order Q$, it follows that $Q$ must equal a conjugate of $P$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Corollary $11.11$
Category:Third Sylow Theorem