nyuuzyou/wikis huggingface dataset:
Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Side Lies on Hypotenuse

== Theorem ==
:
Let $\triangle ABC$ be a right-angled triangle, where $\angle C = 90^\circ$.
Construct a perpedicular line from $C$ to side $AB$, and name the intersection $D$.
Construct the angle bisector of $\angle ADC$.
Let the point of intersection of this angle bisector and side $AC$ be $E$.
Construct a line parallel to $AB$ that passes through $E$ to meet $CD$ at $F$.
Extend $AF$ to side $BC$ at $G$.
Construct a line parallel to $AB$ that passes through $G$ to meet side $AC$ at $H$.
Construct perpedicular lines from $H$ and $G$ to side $AB$, and name the intersections $I$ and $J$ respectively.
Then $GHIJ$ is an inscribed square.
== Proof ==
Note that $HG \perp HI$, $HG \perp GJ$ and $HG \parallel IJ$ by construction.
Therefore $GHIJ$ is a rectangle.
By definition of an angle bisector, $\angle FDE = 45^\circ$.
Since $\angle EFD = 90^\circ$ by construction, by Sum of Angles of Triangle equals Two Right Angles:
:$\angle DEF + \angle EFD + \angle FDE = 180^\circ$
:$\therefore \angle DEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle FDE$
By Triangle with Two Equal Angles is Isosceles, $DF = FE$.
Since $EF \parallel HG$, $\triangle AEF \sim \triangle AHG$ by Equiangular Triangles are Similar.
Similarly, since $FD \parallel GJ$, $\triangle AFD \sim \triangle AGJ$ by Equiangular Triangles are Similar.
By definition of similar triangles:
:$\dfrac {EF} {HG} = \dfrac {AF} {AG} = \dfrac {FD} {GJ}$
As $EF = FD$, we must have $HG = GJ$.
Since $GHIJ$ is a rectangle with two adjacent equal sides, it must also be a square.
Category:Inscribed Squares in Right-Angled Triangle
Inscribed Squares in Right-Angled Triangle/Side Lengths

== Theorem ==
Let $a, b, c$ be the side lengths of the right-angled triangle, where $c$ is the length of the hypotenuse.
Then the side lengths $l$ of the inscribed squares are given by:
=== Shared right angle ===
=== Side lies on Hypotenuse ===
Category:Inscribed Squares in Right-Angled Triangle
Inscribed Squares in Right-Angled Triangle/Side Lengths/Shared Right Angle

== Theorem ==
Let $a, b, c$ be the side lengths of a right-angled triangle, where $c$ is the length of the hypotenuse.
Then the side length $l$ of the inscribed square that shares a right angle with the right-angled triangle is given by:
:$l = \dfrac {a b} {a + b}$
== Proof ==
:
In the figure above, let $BC = a$ and $AC = b$.
Note that $DE \parallel CF$.
Therefore $\triangle BDE \sim \triangle BCA$ by Equiangular Triangles are Similar.
Thus:
== Also see ==
* Haidao Suanjing/Examples/Example 1
Category:Inscribed Squares in Right-Angled Triangle
Inscribed Squares in Right-Angled Triangle/Side Lengths/Side Lies on Hypotenuse

== Theorem ==
Let $a, b, c$ be the side lengths of a right-angled triangle, where $c$ is the length of the hypotenuse.
Then the side length $l$ of the inscribed square that has a side lying on the hypotenuse of the right-angled triangle is given by:
:$l = \dfrac {a b c} {a b + c^2}$
== Proof ==
:
In the figure above, let $BC = a$ and $AC = b$.
$CD$ is drawn such that $AB \perp CD$.
Since $CD$ is the height of $\triangle ABC$:
:$CD = \dfrac {a b} c$
Note that $FH \parallel AB$.
Therefore $\triangle CFH \sim \triangle CAB$ by Equiangular Triangles are Similar.
Thus:
Category:Inscribed Squares in Right-Angled Triangle
Inscribing Circle in Regular Pentagon

== Theorem ==
In any given regular pentagon it is possible to inscribe a circle.
== Construction ==
:
Let $ABCDE$ be the given regular pentagon.
Let $\angle BCD$ and $\angle CDE$ be bisected by $CM$ and $DG$ respectively, and let these cross at $F$.
Then construct the circle whose center is $F$ and whose radius is $FM$ (or $FG$).
This circle is the one required.
== Proof ==
Join the straight lines $FB, FA, FE$.
We have that $BC = CD$, $CF$ is common and $\angle BCF = \angle DCF$.
So by Triangle Side-Angle-Side Congruence:
:$\triangle BCF = \triangle DCF$
and so:
:$BF = DF$
Thus $\angle CBF = \angle CDF$.
Since $\angle CDE = 2 \angle CDF$ and $\angle CDE = \angle CBF$, then $\angle CDF = \angle CBF$.
So $\angle ABF = \angle FBC$ and so $\angle ABC$ has been bisected by the straight line $BF$.
Similarly it can be shown that $\angle BAE, \angle AED$ have been bisected by the straight lines $FA, FE$ respectively.
Now join $FH, FK, FL$ from $F$ perpendicular to $BC, CD, DE$.
We have that:
:$\angle HCF = \angle KCF$
:$\angle FHC = \angle FKC$ (both are right angles)
:$FC$ is common and subtends one of the equal angles.
So from Triangle Side-Angle-Angle Congruence:
:$\triangle FHC = \triangle FKC$
and so:
:$FH = FK$
Similarly it is shown that $FL = FM = FG = FH = FK$.
Therefore the circle whose center is $F$ and radius is $FM$ (or $FG$) passes through all of the points $G, H, K, L, M$.
We have that the angles at each of those points is a right angle.
So from Line at Right Angles to Diameter of Circle, the circle $GHKLM$ is tangent to each of lines $AB, BC, CD, DE, EA$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Pentagons
Inscribing Circle in Square

== Theorem ==
== Construction ==
:
Let $\Box ABCD$ be the given square.
Let $AD, AB$ be bisected at $E$ and $F$ respectively.
Let $EH$ be drawn parallel to $AB$ or $CD$.
Let $FK$ be drawn parallel to $AD$ or $BC$.
Let $G$ be where $EH$ crosses $FK$.
Draw the circle $EFHK$ whose center is at $G$ and whose radius is $EG$.
Then $EFHK$ is the required circle.
== Proof ==
Each of the figures $AK, KB, AH, FD, AG, GC, BG, GD$ is a parallelogram from the method of construction.
The opposite sides of all of them is equal.
We have that $AD = AB$ and $2 AE = AD$, and $2 AF = AB$.
Therefore $AE = AF$.
Because the opposite sides are equal, $FG = GE$.
Similarly we can show that $GE = GF = GH = GK$.
So the circle whose center is at $G$ and whose radius is $EG$ passes through all of $E, F, H, K$.
The lines $AB, BC, CD, DA$ are all perpendicular to the diameters $FK$ and $EH$.
So from Line at Right Angles to Diameter of Circle, these lines are tangent to the circle.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Squares
Inscribing Circle in Triangle

== Theorem ==
== Construction ==
:
Let $\triangle ABC$ be the given triangle.
Let $\angle ABC$ and $\angle ACB$ be bisected by $BD$ and $CD$ and let these lines join at $D$.
From $D$ construct the perpendiculars $DE, DF, DG$ to $AB, BC, AC$ respectively.
Draw the circle with radius $DE$ and center $D$.
This is the required circle.
== Proof ==
We have that $\angle ABD = \angle CBD$ and $\angle BED = \angle BFD$, a right angle, and $BD$ is common.
So from Triangle Angle-Side-Angle Congruence:
:$\triangle EBD = \triangle FBD$
So $DE = DF$.
For the same reason $DG = DF$.
So $DE = DF = DG$.
So the circle drawn with radius $DE$ will pass through $E, F$ and $G$.
From Line at Right Angles to Diameter of Circle it follows that $AB, AC, BC$ are tangent to the circle $EFG$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
* : $\text {IV}$. Pure Geometry: Plane Geometry: The incentre
* 
* 
Category:Incircles of Triangles
Category:Circles
Category:Triangles
Inscribing Equilateral Triangle inside Square with a Coincident Vertex

== Theorem ==
Let $\Box ABCD$ be a square.
It is required that $\triangle DGH$ be an equilateral triangle inscribed within $\Box ABCD$ such that vertex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.
== Construction 1 ==
== Construction 2 ==
== Construction 3 ==
== Construction 4 ==
== Historical Note ==
== Sources ==
* : Abul Wafa (): $38$
Category:Squares
Category:Equilateral Triangles
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 1

== Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex ==
== Construction ==
:
By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$.
Let $AB$ be produced to $F$ such that $AB = BF$.
Draw an arc centred at $F$ with radius $FN$ to cut $AB$ at $G$.
Construct $H$ on $BC$ such that $DH = DG$.
Then $DGH$ is the required equilateral triangle.
== Proof ==
It is necessary only to note that $N$ passes through $DH$, which is demonstrated in the simpler Construction 4.
Then the proof for that construction can be applied.
== Sources ==
* 
* : Abul Wafa (): $38$
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 2

== Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex ==
== Construction ==
:
By Construction of Equilateral Triangle, let an equilateral triangle $\triangle DAI$ be constructed on $AD$ such that $I$ is inside $\Box ABCD$.
Bisect $\angle ADI$ and bisect it again towards $AD$ to cut $AB$ at $G$.
Construct $H$ on $BC$ such that $DH = DG$.
Then $DGH$ is the required equilateral triangle.
== Proof ==
First a lemma:
=== Lemma ===
Because $\triangle DAI$ is equilateral, we have that:
The result follows from the lemma.
== Sources ==
* 
* : Abul Wafa (): $38$
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 3

== Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex ==
== Construction ==
:
By Construction of Equilateral Triangle, let an equilateral triangle $\triangle CDN$ be constructed on $CD$ such that $N$ is inside $\Box ABCD$.
Bisect $\angle ADN$ to cut $AB$ at $G$.
Construct $H$ on $BC$ such that $DH = DG$.
Then $DGH$ is the required equilateral triangle.
== Proof ==
First a lemma:
=== Lemma ===
Because $\triangle CDN$ is equilateral, we have that:
The result follows from the lemma.
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 4

== Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex ==
== Construction ==
:
By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$.
Let $DN$ be produced to cut $BC$ at $H$.
Construct $G$ on $AB$ such that $DH = DG$.
Then $DGH$ is the required equilateral triangle.
== Proof ==
First a lemma:
=== Lemma ===
Because $\triangle ABN$ is equilateral:
:$AB = AN$
:$\angle BAN = 60 \degrees$
Thus $\triangle ADN$ is isosceles with apex at $A$.
Then $\angle DAN = 90 \degrees - 60 \degrees = 30 \degrees$.
From Sum of Angles of Triangle equals Two Right Angles:
:$\angle ADN + \angle AND = 180 \degrees - 30 \degrees = 150 \degrees$
From Isosceles Triangle has Two Equal Angles:
:$\angle ADN = \angle AND = \dfrac {150 \degrees} 2 = 75 \degrees$
Thus:
:$\angle CDH = 15 \degrees$
The result follows from the lemma.
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Lemma

== Lemma ==
Let $\Box ABCD$ be a square.
Let $\triangle DGH$ be an isosceles triangle inscribed within $\Box ABCD$ such that the apex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.
:
Then:
:$\triangle DGH$ is equilateral triangle
:
:$\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).
== Proof ==
First note that $\triangle DGH$ is isosceles.
First we note that:
Then:
So by Triangle Side-Side-Side Congruence:
:$\triangle ADG = \triangle CDH$
and in particular:
:$\angle ADG = \angle CDH$
=== Necessary Condition ===
Let $\angle ADG = \angle CDH = 15 \degrees$.
Then:
:$\angle GDH = 90 \degrees - 2 \times 15 \degrees = 60 \degrees$
We have that $\triangle DGH$ is isosceles.
Hence from Isosceles Triangle has Two Equal Angles:
:$\angle DGH = \angle DHG$
From Sum of Angles of Triangle equals Two Right Angles it follows that:
:$\angle DGH + \angle DHG = 180 \degrees - 60 \degrees = 120 \degrees$
from which:
:$\angle DGH = \angle DHG = 60 \degrees$
Hence all the vertices of $\triangle DGH$ equal $60 \degrees$.
It follows from Equiangular Triangle is Equilateral that $\triangle DGH$ is equilateral.
=== Sufficient Condition ===
Let $\triangle DGH$ be equilateral.
From Internal Angle of Equilateral Triangle:
:$\angle GDH = 60 \degrees$
Because $\Box ABCD$ is a square:
:$\angle ADC = 90 \degrees$
Thus:
But we have that:
:$\angle ADG = \angle CDH$
and so:
:$\angle ADG = \angle CDH = 15 \degrees$
Category:Inscribing Equilateral Triangle inside Square with a Coincident Vertex
Inscribing Regular 15-gon in Circle

== Theorem ==
In a given circle, it is possible to inscribe a regular 15-gon.
=== Corollary ===
== Construction ==
:
Let $ABCD$ be the given circle.
In $ABCD$ we inscribe an equilateral triangle, one of whose vertices is at $A$.
In $ABCD$ we also inscribe a regular pentagon, one of whose vertices is at $A$.
Thus we have that $AC$ is one of the sides of the equilateral triangle, and $AB$ is one of the sides of the regular pentagon.
Let $BC$ be bisected by a line which passes through the center of the circle.
Let this line intersect the circumference of the circle at $E$.
We fit as many copies of the straight line $BE$ around the circumference of $ABCD$, starting each one at the point the previous one ends.
The resulting polygon is the required regular 15-gon.
== Proof ==
Consider the circumference of circle $ABCD$ as divided into $15$ equal arcs.
Of these, there will be $5$ in the shorter arc $AC$, and $3$ in the shorter arc $AB$.
So there are $2$ in the shorter arc $BC$.
Once this has been bisected by the construction which produces $E$, we see that each of these parts is a copy of these $15$ equal arcs.
Then $BE$ and $EC$ are $\dfrac 1 {15}$ of the length of the circumference of $ABCD$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Inscribing Regular 15-gon in Circle
Category:Circles
Category:Regular Polygons
Inscribing Regular 15-gon in Circle/Corollary

== Corollary to Inscribing Regular 15-gon in Circle ==
== Proof ==
In the same way as for the regular pentagon, we can draw tangents to the circle at the vertices of the regular 15-gon.
This will draw a regular 15-gon which has been circumscribed about the circle.
Further, in a similar way to methods used for the regular pentagon, a circle can be inscribed in a regular 15-gon and circumscribed about a regular 15-gon.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Inscribing Regular 15-gon in Circle
Inscribing Regular Hexagon in Circle

== Theorem ==
In a given circle, it is possible to inscribe a regular hexagon.
=== Porism ===
== Construction ==
:
Let $ABCDEF$ be the given circle (although note that at this stage the positions relative to each other of the points $A, B, C, D, E, F$ have not been established).
Let $AD$ be a diameter of the circle $ABCDEF$.
Let the center $G$ be found.
Draw the circle $EGCH$ with center $D$ and radius $DG$.
Join $EC, CG$ and produce them to $B$ and $F$ on the circumference of the circle $ABCDEF$.
Join $AB, BC, CD, DE, EF, FA$.
This is the required regular hexagon.
== Proof ==
Since $G$ is the center of circle $ABCDEF$, it follows that $GE = GD$.
Since $D$ is the center of circle $EGCH$, it follows that $DE = GD$.
So $GE = GD = DE$ and so $\triangle EGD$ is equilateral and so equiangular.
By Sum of Angles of Triangle Equals Two Right Angles, $\angle EGD$ is one third of two right angles.
Similarly for $\angle DGC$.
Since the straight line $CG$ on $EB$ makes $\angle EGC + \angle CGB$ equal to two right angles, $\angle CGB$ is also equal to one third of two right angles.
So $\angle EGD = \angle DGC = \angle CGB$.
By the Vertical Angle Theorem, $\angle BGA = \angle AGF = \angle FGE = \angle EGD = \angle DGC = \angle CGB$.
From Equal Angles in Equal Circles, the six arcs $AB, BC, CD, DE, EF, FA$ are all equal.
So the six straight lines $AB, BC, CD, DE, EF, FA$ are all equal.
So the hexagon $ABCDEF$ is equilateral.
Now since the arc $FA$ equals the arc $ED$, let the arc $ABCD$ be added to each.
So arc $FABCD$ equals arc $ABCDE$.
Now $\angle FED$ stands on arc $FABCD$ and $\angle AFE$ stands on arc $ABCDE$.
So by Angles on Equal Arcs are Equal $\angle FED = \angle AFE$.
In the same way we show that all the angles around the hexagon $ABCDEF$ are equal.
Therefore $ABCDEF$ is a regular hexagon.
== Sources ==
* : Book $\text{IV}$. Propositions
* 
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Circles
Category:Hexagons
Inscribing Regular Hexagon in Circle/Porism

== Porism to Inscribing Regular Hexagon in Circle ==
== Proof ==
In the same way as for the regular pentagon, we can draw tangents to the circle at the vertices of the regular hexagon.
This will draw a regular hexagon which has been circumscribed about the circle.
Further, in a similar way to methods used for the regular pentagon, a circle can be inscribed in a regular hexagon and circumscribed about a regular hexagon.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Hexagons
Inscribing Regular Pentagon in Circle

== Theorem ==
In a given circle, it is possible to inscribe a regular pentagon.
== Construction ==
:
Let $ABCDE$ be the given circle (although note that at this stage the positions of the points $A, B, C, D, E$ have not been established).
Let $\triangle FGH$ be constructed such that $\angle FGH = \angle FHG = 2 \angle GFH$.
Let $ACD$ be inscribed in $ABCDE$ such that $\angle ACD = \angle FGH, \angle ADC = \angle FHG, \angle CAD = \angle GFH$.
Bisect $\angle ACD$ with $CE$ and bisect $\angle ADC$ with $DB$.
Then the pentagon $ABCDE$ is the required regular pentagon.
== Proof ==
We have that $\angle CDA = \angle DCA = 2 \angle CAD$.
As $\angle CDA$ and $ \angle DCA$ have been bisected, $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$.
From Equal Angles in Equal Circles it follows that the arcs $AB, BC, CD, DE, EA$ are all equal.
Hence from Equal Arcs of Circles Subtended by Equal Straight Lines, the straight lines $AB, BC, CD, DE, EA$ are all equal.
So the pentagon $ABCDE$ is equilateral.
Now since the arc $AB$ equals arc $DE$, we can add $BCD$ to each.
So the arc $ABCD$ equals arc $BCDE$.
So from Angles on Equal Arcs are Equal $\angle BAE = \angle AED$.
For the same reason, $\angle BAE = \angle AED = \angle ABC = \angle BCD = \angle CDE$.
So the pentagon $ABCDE$ is equiangular.
== Sources ==
* : Book $\text{IV}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Circles
Category:Pentagons
Inscribing Square in Circle

== Theorem ==
== Construction ==
:
Let $ABCD$ be the given circle.
Let two diameters be drawn at right angles to one another.
Join $AB, BC, CD, DA$.
Then $\Box ABCD$ is the required square.
== Proof ==
We have that $BE = ED$, $EA$ is common and $\angle BEA = \angle DEA$ are right angles.
So from Triangle Side-Angle-Side Congruence, $\triangle ABE = \triangle ADE$ and so $AB = AD$.
For the same reason $BC = CD = AD$ and so all four sides $AB, BC, CD, DA$ are equal.
So $\Box ABCD$ is equilateral.
Next we have that $BD$ is a diameter of circle $ABCD$.
So from Relative Sizes of Angles in Segments, $\angle BAD$ is a right angle.
For the same reason, $\angle ABC$, $\angle BCD$ and $\angle ADC$ are also all right angles.
So by definition, $\Box ABCD$ is a square.
== Sources ==
* : Book $\text{IV}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Circles
Category:Squares
Inscribing in Circle Triangle Equiangular with Given

== Theorem ==
== Construction ==
Let $ABC$ be the given circle and let $\triangle DEF$ be the given triangle.
:
Let $GH$ be drawn tangent to $ABC$ at $A$.
On the straight line $AH$ construct $\angle HAC$ equal to $\angle DEF$.
On the straight line $AG$ construct $\angle GAB$ equal to $\angle DFE$.
Join $BC$.
Then $\triangle ABC$ is the given triangle.
== Proof ==
We have that $AH$ touches the circle $ABC$, and $AC$ is a chord of circle $ABC$.
Then from the Tangent-Chord Theorem $\angle HAC = \angle ABC$.
But $\angle HAC = \angle DEF$.
Similarly, we have that $AG$ touches the circle $ABC$, and $AB$ is a chord of circle $ABC$.
Then from the Tangent-Chord Theorem $\angle GAB = \angle ACB$.
But $\angle GAB = \angle DFE$.
So from Sum of Angles of Triangle Equals Two Right Angles it follows that the remaining angles are equal also: $\angle BAC = \angle EDF$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Circles
Category:Triangles
Integer Addition Identity is Zero

== Theorem ==
The identity of integer addition is $0$:
:$\exists 0 \in \Z: \forall a \in \Z: a + 0 = a = 0 + a$
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.
From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $0 \in \N$.
So, we need to show that:
:$\forall a, b, c \in \N: \eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$
Thus:
So:
:$\eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {}$
The identity $\eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ is demonstrated similarly.
== Sources ==
* : Introduction $\S 5$: The system of integers
* : $\S 2.5$: Theorem $2.24$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 3$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization
* : $\S 1$
Category:Integer Addition
Integer Addition is Associative

== Theorem ==
The operation of addition on the set of integers $\Z$ is associative:
:$\forall x, y, z \in \Z: x + \paren {y + z} = \paren {x + y} + z$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Introduction $\S 5$: The system of integers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 1$
* : $\S 1$
Category:Integer Addition
Category:Examples of Associative Operations‎
Category:Integer Addition is Associative
Category:Associative Law of Addition
Integer Addition is Associative/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c, d, e, f \in \N$ such that:
:$x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$.
Then:
Category:Integer Addition is Associative
Integer Addition is Cancellable

== Theorem ==
The operation of addition on the set of integers $\Z$ is cancellable:
:$\forall x, y, z \in \Z: x + z = y + z \implies x = y$
== Proof ==
Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$ for some $x, y, z\in \Z$.
Then:
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Integer Addition