nyuuzyou/wikis huggingface dataset:
Measurements of Common Angles/Obtuse Angle

== Theorem ==
An obtuse angle measures $\theta$, where:
:$90 \degrees < \theta < 180 \degrees$
or:
:$\dfrac \pi 2 < \theta < \pi$
== Proof ==
An obtuse angle is defined to be an angle whose measure is between that of a right angle and a straight angle.
A right angle measures $90 \degrees$ or $\dfrac \pi 2$ and a straight angle measures $180 \degrees$ or $\pi$.
Hence the result.
== Sources ==
* 
* 
Category:Obtuse Angles
Category:Measurements of Common Angles
Measurements of Common Angles/Reflex Angle

== Theorem ==
A reflex angle measures $\theta$, where:
:$180 \degrees < \theta < 360 \degrees$
or:
:$\pi < \theta < 2 \pi$
== Proof ==
A reflex angle is defined to be an angle whose measure is between that of a straight angle and a full angle.
A straight angle measures $180 \degrees$ or $\pi$ and a full angle measures $360 \degrees$ or $2 \pi$.
Hence the result.
== Sources ==
* 
* 
Category:Reflex Angles
Category:Measurements of Common Angles
Measurements of Common Angles/Right Angle

== Theorem ==
The measurement of a right angle is $\dfrac {180 \degrees} 2 = 90 \degrees$ or $\dfrac \pi 2$.
== Proof ==
A right angle is equal to one half of a straight angle.
From Measurement of Straight Angle it follows that the measurement of a right angle is $\dfrac {180 \degrees} 2 = 90 \degrees$ or $\dfrac \pi 2$.
== Sources ==
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.1$ Unit Circle
* : $90$
* : $90$
* 
* 
* 
* 
* 
* 
* 
Category:Right Angles
Category:Measurements of Common Angles
Measurements of Common Angles/Straight Angle

== Theorem ==
The measurement of a straight angle is $180 \degrees$ or $\pi$ radians.
== Proof ==
From Measurement of Full Angle, a full rotation is defined to be $360 \degrees$ or $2 \pi$ radians.
Since lines are straight, it therefore follows that from any point on a line, the angle between one side of the line and the other is one half of a full rotation.
Therefore, the measurement of a straight angle is:
:$\dfrac {360 \degrees} 2 = 180 \degrees$
or:
:$\dfrac {2 \pi} 2 = \pi$
== Also see ==
* Two Angles making Two Right Angles make Straight Line
== Sources ==
* : $180$
* : $180$
* 
* 
* 
* 
* 
* 
* 
* 
Category:Straight Angles
Category:Measurements of Common Angles
Measures in Jordan Decomposition of Complex Measure are Finite

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$.
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$. 
Then:
:$\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ are finite.
== Proof ==
Let $\mu_R$ be the real part of $\mu$.
Let $\mu_I$ be the imaginary part of $\mu$.
Then:
:$\tuple {\mu_1, \mu_2}$ is the Jordan decomposition of $\mu_R$
and:
:$\tuple {\mu_3, \mu_4}$ is the Jordan decomposition of $\mu_I$.
From the definition of the real part and imaginary part, we have that both $\mu_R$ and $\mu_I$ are finite signed measures.
Now applying Jordan Decomposition of Finite Signed Measure to $\mu_R$, we have: 
:$\mu_1$ and $\mu_2$ are finite.
Applying Jordan Decomposition of Finite Signed Measure to $\mu_I$, we have: 
:$\mu_4$ and $\mu_4$ are finite
hence the result.
Category:Complex Measures
Mechanical Advantage of System of Ideal Pulleys

== Theorem ==
Let $S$ be a system of ideal pulleys, that is, without friction.
The mechanical advantage of $S$ is equal to the number $n$ of cords supporting the weight $W$ to be moved by the applied force $F$.
:
== Proof ==
== Sources ==
* 
* 
Category:Pulleys
Category:Mechanical Advantage
Mechanical Advantage of Type 3 Lever is Less than 1

== Theorem ==
Let $M$ be a type $3$ lever.
Then the mechanical advantage of $M$ is less than $1$.
Hence, instead of amplifying the force applied, a type $3$ lever amplifies the velocity ratio.
== Proof ==
From Principle of Lever:
:$\text {MA} = \dfrac b a$
where:
:$b$ is the distance of the effort from the fulcrum
:$a$ is the distance of the load from the fulcrum.
But in a type $3$ lever, the effort acts between the load and the fulcrum.
That is:
:$a > b$
and hence the result.
== Sources ==
* 
* 
Category:Type 3 Levers
Medial Area not greater than Medial Area by Rational Area

== Theorem ==
== Proof ==
Let $\rho \cdot \sqrt k \rho$ and $\rho \cdot \sqrt \lambda \rho$ be two medial areas which have been applied to the same rational number $\rho$.
Their difference is then:
:$\paren {\sqrt k - \sqrt \lambda} \rho^2$
Let $x = \sqrt k \rho$ and $y = \sqrt \lambda \rho$.
Let:
:$\rho \paren {x - y} = \rho z$
Suppose $\rho z$ is a rational area.
Then $z$ must be a rational straight line and:
:$(1): \quad z \frown \rho$
where $\frown$ denotes commensurability in length.
Since $\rho x$ and $\rho y$ are medial areas, it follows that $x$ and $y$ are rational and:
:$(2): \quad x \smile \rho$ and $y \smile \rho$
where $\smile$ denotes incommensurability in length.
Then from $(1)$ and $(2)$:
:$y \smile z$
Then:
But:
But $y^2 + z^2$ is rational.
Therefore $x^2$ must be irrational.
Therefore $x$ must be irrational.
But from $(2)$ $x$ is rational.
The result follows from this contradiction.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Medial is Irrational

== Theorem ==
== Proof ==
By definition, a medial is a mean proportional between two rational line segments which are commensurable in square only.
Let $\rho$ and $\rho \sqrt k$ be two rational line segments which are commensurable in square only.
Then:
:$\rho: \rho \sqrt k = \rho^2 : \rho^2 \sqrt k$
From Commensurability of Elements of Proportional Magnitudes, $\rho^2 \sqrt k$ is incommensurable in length with $\rho^2$.
From , $\rho^2 \sqrt k$ is an irrational area.
Thus $\sqrt {\rho^2 \sqrt k} = \rho \sqrt [4] k$ is by definition an irrational straight line.
By definition of mean proportional, $\rho \sqrt [4] k$ is the medial between $\rho$ and $\rho \sqrt k$
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Median Formula

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $CD$ be the median of $\triangle ABC$ which bisects $AB$.
:
The length $m_c$ of $CD$ is given by:
:${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Sources ==
* 
Category:Median Formula
Category:Medians of Triangles
Category:Named Theorems
Median Formula/Examples/(1, -2), (-3,4), (2,2)

== Example of Use of Median Formula ==
Consider the triangle $\triangle ABC$ whose vertices are:
:$A = \tuple {1, -2}, B = \tuple {-3, 4}, C = \tuple {2, 2}$
The length of the median of $\triangle ABC$ which which bisects $AB$ is $\sqrt {10}$.
== Proof ==
Let $\triangle ABC$ be embedded in a complex plane.
Let the position vectors of $A$, $B$ and $C$ be $z_1 = 1 - 2 i$, $z_2 = -3 + 4 i$, $z_3 = 2 + 2 i$ respectively.
:
Then:
Then:
Hence:
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $12$
Category:Median Formula
Median of Cauchy Distribution

== Theorem ==
Let $X$ be a continuous random variable with a Cauchy distribution:
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
for:
:$\lambda \in \R_{>0}$
:$\gamma \in \R$
The median of $X$ is $\gamma$.
== Proof ==
From the definition of the Cauchy distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
Note that $f_X$ is non-zero, sufficient to ensure a unique median.
By the definition of a median, to prove that $\gamma$ is the median of $X$ we must verify: 
:$\ds \map \Pr {X < \gamma} = \int_{-\infty}^\gamma \map {f_X} x \rd x = \frac 1 2$
We have:
== Sources == 
* 
* 
Category:Cauchy Distribution
Category:Medians
Category:Arctangent Function
Median of Continuous Uniform Distribution

== Theorem == 
Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$. 
Then the median $M$ of $X$ is given by: 
:$M = \dfrac {a + b} 2$
== Proof ==
From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {b - a}$
Note that $f_X$ is non-zero, so the median is unique. 
We have by the definition of a median: 
:$\ds \map \Pr {X < M} = \int_a^M \frac 1 {b - a} \rd x = \frac 1 2$
We have, by Primitive of Constant: 
:$\dfrac {M - a} {b - a} = \dfrac 1 2$
So: 
Category:Continuous Uniform Distribution
Category:Medians
Median of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$. 
Then the median of $X$ is equal to $\beta \ln 2$. 
== Proof ==
Let $M$ be the median of $X$.
From the definition of the exponential distribution, $X$ has probability density function: 
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
Note that $f_X$ is non-zero, so the median is unique.
We have by the definition of a median: 
:$\ds \map \Pr {X < M} = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x = \frac 1 2$
Evaluating this integral:
So: 
:$1 - e^{-\frac M \beta} = \dfrac 1 2$
Then: 
:$e^{-\frac M \beta} = \dfrac 1 2$
giving: 
:$-\dfrac M \beta = \map \ln {\dfrac 1 2}$
So, by Logarithm of Reciprocal: 
:$M = \beta \ln 2$
Category:Exponential Distribution
Category:Medians
Median of Logistic Distribution

== Theorem ==
Let $X$ be a continuous random variable with a logistic distribution:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
for $\mu \in \R, s \in \R_{>0}$.
The median of $X$ is $\mu$.
== Proof ==
From the definition of the logistic distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
Note that $f_X$ is non-zero, sufficient to ensure a unique median.
By the definition of a median, to prove that $\mu$ is the median of $X$ we must verify: 
:$\ds \map \Pr {X < \mu} = \int_{-\infty}^\mu \map {f_X} x \rd x = \frac 1 2$
We have:
Let:
and also:
Then:
Category:Logistic Distribution
Category:Medians
Median of Normal Distribution

== Theorem == 
Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then the median of $X$ is equal to $\mu$.
== Proof == 
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
Note that $f_X$ is non-zero, sufficient to ensure a unique median.
By the definition of a median, to prove that $\mu$ is the median of $X$ we must verify: 
:$\ds \map \Pr {X < \mu} = \int_{-\infty}^\mu \map {f_X} x \rd x = \frac 1 2$
We have:
Category:Normal Distribution
Category:Medians
Median of Trapezium is Parallel to Bases

== Theorem ==
Let $\Box ABCD$ be a trapezium such that $AB$ and $DC$ are the bases.
:
Let $E$ be the midpoint of $AD$.
Let $F$ lie on $BC$.
Then:
:$EF$ is parallel to both $AB$ and $DC$
:
:$F$ is the midpoint of $BC$.
That is, the median of $\Box ABCD$ is parallel to the bases of $\Box ABCD$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* 
Category:Median of Trapezium is Parallel to Bases
Category:Medians of Trapezia
Median of Trapezium is Parallel to Bases/Necessary Condition

== Theorem ==
Let $\Box ABCD$ be a trapezium such that $AB$ and $DC$ are the parallel sides.
:
Let $E$ be the midpoint of $AD$.
Let $F$ be the midpoint of $BC$.
Then $EF$ is parallel to both $AB$ and $DC$.
== Proof ==
:
 $EF$ is not parallel to $DC$.
By Playfair's axiom, there exists a unique straight line through $E$ which is parallel to $DC$.
Let $EF'$ be this line.
From Median of Trapezium is Parallel to Bases: Sufficient Condition, $F'$ is the midpoint of $BC$.
But $F$ is also the midpoint of $BC$.
That is:
:$F = F'$
So:
:$EF = EF'$
and so $EF$ is parallel to $DC$.
This contradicts our assertion that $EF$ is not parallel to $DC$.
Hence, by Proof by Contradiction, $EF$ is parallel to $DC$.
By definition of base of trapezium, $AB$ is parallel to $DC$.
By Parallelism is Transitive Relation, $EF$ is parallel to $AB$.
That is, $EF$ is parallel to both $AB$ and $DC$.
Category:Median of Trapezium is Parallel to Bases
Median of Trapezium is Parallel to Bases/Sufficient Condition

== Theorem ==
Let $\Box ABCD$ be a trapezium such that $AB$ and $DC$ are the parallel sides.
:
Let $E$ be the midpoint of $AD$.
Let $F$ lie on $BC$.
Let $EF$ be parallel to both $AB$ and $DC$.
Then $F$ is the midpoint of $BC$.
== Proof ==
:
Let $DH$ be constructed parallel to $BC$ to cut $AB$ at $H$.
From the Parallel Transversal Theorem:
:$DG : GH = DE : EA$
and so $G$ is the midpoint of $AH$.
That is:
:$(1): \quad DG = GH$
Then we have that:
:$DC$ is parallel to $GF$
and:
:$DG$ is parallel to $CF$
so, by definition, $\Box GFCD$ is a parallelogram.
Similarly, we have:
:$GF$ is parallel to $HB$
and:
:$GH$ is parallel to $FB$
so, by definition, $\Box GFBH$ is also a parallelogram.
By Opposite Sides and Angles of Parallelogram are Equal we have that:
:$CF = DG$
and:
:$GH = FB$
But from $(1)$:
:$DG = GH$
and so:
:$CF = FB$
and so $F$ is the midpoint of $CB$.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.25$: Corollary
Category:Median of Trapezium is Parallel to Bases
Median to Hypotenuse of Right Triangle equals Half Hypotenuse

== Theorem ==
Let $\triangle ABC$ be a right triangle such that $BC$ is the hypotenuse.
Let $AD$ be the median to $BC$.
Then $AD$ is half of $BC$.
== Proof ==
Construct $BE$ and $CE$ parallel to $AC$ and $AB$ respectively.
Then by definition $ABEC$ is a parallelogram.
By construction, $BC$ is a diagonal of $ABEC$ such that $AD$ is a bisector of it.
Thus by Quadrilateral is Parallelogram iff Diagonals Bisect each other, $AE$ is also a bisector of $ABEC$.
As $\angle BAC$ is a right angle it follows that $\angle BEC$ is also a right angle.
Thus by definition $ABEC$ is a rectangle.
From Diagonals of Rectangle are Equal:
:$AE = BC$
From above, $D$ is the midpoint of both $AE$ and $BC$.
Thus $AD = BD$ and hence the result.
== Also see ==
* Thales' Theorem, the converse of this.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.27$
Category:Right Triangles
Category:Medians of Triangles
Medians of Triangle Form Six Triangles of Equal Area

== Theorem ==
:
Let $\triangle ABC$ be an arbitrary triangle.
Draw $AL$, $BM$ and $CN$, the medians of $\triangle ABC$.
Then $\triangle ABC$ is divided into six small triangles:
* $\triangle GAN$
* $\triangle GBN$
* $\triangle GBL$
* $\triangle GCL$
* $\triangle GCM$
* $\triangle GAM$
Each one of these six triangles has the same area.
== Proof ==
:
:
* $AN = BN$
* $BL = CL$
* $CM = AM$
Each pair of triangles with bases on the same side of $\triangle ABC$ consists of two triangles:
:with a base of the same magnitude
:that share the same vertex, $G$
:so they have the same height
It follows by Triangles with Equal Base and Same Height have Equal Area:
:each member of the pair has the same area.
Now consider the pair $\triangle CAN$ and $\triangle CBN$:
:they have bases of equal magnitude
:they have the same vertex $C$
:so they have the same altitude and equal height.
By Triangles with Equal Base and Same Height have Equal Area:
:$\AA \paren { CAN } = \AA \paren { CBN }$
It follows that the component triangles of $\triangle CAN$ and $\triangle CBN$ add up to the same total area.
 
The result follows.
Category:Position of Centroid of Triangle on Median
Medians of Triangle Meet at Centroid

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then the medians of $\triangle ABC$ meet at a single point.
This point is called the centroid of $\triangle ABC$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Centroid of Triangle
* Position of Centroid of Triangle on Median, which shows that $G$ is $\dfrac 1 3$ the way along each median
== Sources ==
* 
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* 
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* 
Category:Medians of Triangle Meet at Centroid
Category:Centroids of Triangles
Category:Medians of Triangles
Medians of Triangle Meet at Centroid/Corollary

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $A'$ be the midpoint of $BC$.
Let $B'$ be the midpoint of $AC$.
Let $C'$ be the midpoint of $AB$.
Let $G$ be the centroid of $\triangle ABC$
Let $AA'$ be produced beyond $BC$ to $X$, where $A'X = A'G$.
Then the straight lines $BX$ and $CX$ are parallel to $CC'$ and $BB'$ respectively, and $\dfrac 2 3$ of their length.
== Proof ==
:
Consider the quadrilateral $\Box BGCX$.
Its diagonals are $GX$ and $BC$.
By construction, they bisect each other.
From Quadrilateral with Bisecting Diagonals is Parallelogram, $\Box BGCX$ is a parallelogram.
From Position of Centroid of Triangle on Median:
:$BG$ is $\dfrac 2 3$ the length of $BB'$
:$GC$ is $\dfrac 2 3$ the length of $CC'$
But as $\Box BGCX$ is a parallelogram:
:$BG = CX$
and:
:$GC = BX$
Hence the result.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The centre of gravity
Category:Medians of Triangle Meet at Centroid
Medians of Triangle Meet at Centroid/Proof 1

== Theorem ==
== Proof ==
:
Let $A'$ be the midpoint of $BC$.
Let $B'$ be the midpoint of $AC$.
Let $C'$ be the midpoint of $AB$.
Hence $AA'$, $BB'$ and $CC'$ are the medians of $\triangle ABC$.
Let $AA'$ and $BB'$ intersect at $G$.
Hence $A'B'$ is a midline of $\triangle ABC$.
By the Midline Theorem, $A'B'$ is parallel to $AB$ and half the length of $AB$.
We have that $\triangle AGB$ and $\triangle A'GB'$ are similar.
Hence:
Hence $BB'$ meets $AA'$ $\dfrac 1 3$ of the length along $AA'$ from $A'$.
That is:
:$A'G = \dfrac {AA'} 3$
Similarly, , it can be shown that $CC'$ also meets $AA'$ $\dfrac 1 3$ of the length along $AA'$ from $A'$.
Hence we have shown that $BB'$ and $CC'$ both meet at the same point on $AA'$
That is, the medians of $\triangle ABC$ meet at a single point.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The centre of gravity
Category:Medians of Triangle Meet at Centroid
Medians of Triangle Meet at Centroid/Proof 2

== Theorem ==
== Proof ==
Let $\vec a, \vec b, \vec c$ be $\vec{OA}, \vec{OB}, \vec{OC}$ respectively.
Let the midpoint of $BC, AC, AB$ be $\vec d, \vec e, \vec f$ respectively.
Then:
The three medians are $\vec{AD}, \vec{BE}, \vec{CF}$ respectively:
Their equations:
It can be verified that $x = y = z = \dfrac 2 3$ produce the same point:
When $x = \dfrac 2 3$, from $(1)$:
:$\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
When $y = \dfrac 2 3$, from $(2)$:
:$\vec r = \vec b + \dfrac 2 3 \paren {\dfrac {\vec a + \vec c - 2\vec b} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
When $z = \dfrac 2 3$, from $(3)$:
:$\vec r = \vec c + \dfrac 2 3 \paren {\dfrac {\vec a + \vec b - 2\vec c} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
Therefore, the three medians meet at a single point, namely $\dfrac {\vec a + \vec b + \vec c} 3$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $67$
Category:Medians of Triangle Meet at Centroid
Mediant is Between

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $r = \dfrac a b < \dfrac c d = s$.
Then:
:$r < \dfrac {a + c} {b + d} < s$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $r, s \in \R$ be such that:
:$r < s$
and:
:$r = \dfrac a b, s = \dfrac c d$
where $a, b, c, d$ are real numbers such that $b > 0, d > 0$.
Because $b, d > 0$, it follows from Real Number Ordering is Compatible with Multiplication that:
:$b d > 0$
Thus:
Then:
From Reciprocal of Strictly Positive Real Number is Strictly Positive:
:$\paren {a + c}^{-1} > 0$
and:
:$b^{-1} > 0$
It follows from Real Number Ordering is Compatible with Multiplication that:
:$\dfrac a b < \dfrac {a + c} {b + d}$
The other half is proved similarly.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (2)$
Category:Mediant is Between
Category:Real Analysis
Category:Mediants
Mediant is Between/Corollary 1

== Corollary to Mediant is Between ==
Let $a, b, c, d \in \Z$ be integers such that:
Then the mediant of $\dfrac a b$ and $\dfrac c d$ is between $\dfrac a b$ and $\dfrac c d$:
:$\dfrac a b < \dfrac {a + c} {b + d} < \dfrac c d$
== Proof ==
By definition, $\dfrac a b$ and $\dfrac c d$ are rational numbers.
From Rational Numbers form Subfield of Real Numbers, $\dfrac a b, \dfrac c d \in \R$.
Hence from Mediant is Between:
:$\dfrac a b < \dfrac {a + c} {b + d} < \dfrac c d$
Category:Mediant is Between
Mediant is Between/Corollary 2

== Corollary to Mediant is Between ==
Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $\dfrac a b = \dfrac c d$.
Then:
:$\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Mediant is Between
Mediant is Between/Corollary 2/Proof 1

== Corollary to Mediant is Between ==
== Proof ==
Let $p, q, r, s \in \R$ such that $q > 0, s > 0$.
Then from Mediant is Between:
:$\dfrac p q < \dfrac {p + r} {q + s} < \dfrac r s$
In order to present this in the form required by the Squeeze Theorem for Functions, we weaken the ordering:
:$(1): \quad \dfrac p q \le \dfrac {p + r} {q + s} \le \dfrac r s$
Let $f$, $g$ and $h$ be the constant real functions defined as:
From $(1)$:
:$\forall x \in \R: \map f x \le \map g x \le \map h x$
But :
:$\dfrac a b = \dfrac c d$
That is:
:$\forall x \in \R: \map f x = \map h x$
Hence from the Squeeze Theorem for Functions:
:$\forall x \in \R: \map f x = \map g x = \map h x$
That is:
:$\dfrac a b = \dfrac c d \implies \dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$
Category:Mediant is Between
Mediant is Between/Corollary 2/Proof 3

== Corollary to Mediant is Between ==
== Proof ==
Let:
:$t := \dfrac a b = \dfrac c d$
Then:
:$a = t c$
and:
:$c = t d$
Hence:
:$\dfrac {a + b}{c + d} = \dfrac {t \paren {c + d} }{c + d} = t$
Category:Mediant is Between
Mediant is Between/Corollary 2/Proof 4

== Corollary to Mediant is Between ==
== Proof ==
Let $\epsilon \in \R_{>0}$.
Then:
:$\dfrac a b < \dfrac {c + \epsilon} d$
By Mediant is Between:
:$\dfrac a b < \dfrac {a + c + \epsilon} {b + d} < \dfrac {c + \epsilon} d$
By Inequality of Sequences Preserved in Limit, letting $\epsilon \to 0$:
:$\dfrac a b \le \dfrac {a + c} {b + d} \le \dfrac c d$
Since :
:$\dfrac a b = \dfrac c d$
we have:
:$\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$
Category:Mediant is Between
Mediant is Dependent upon Representation

== Theorem ==
Let $r, s \in \Q$ be rational numbers.
Let $r$ and $s$ be expressed as:
where $a, b, c, d$ are integers such that $b > 0, d > 0$.
Then the mediant of $r$ and $s$ is dependent upon the specific integers chosen for $a, b, c, d$.
== Proof ==
;Proof by Counterexample
Let $r = \dfrac 1 2$ and $s = 1$.
We have:
:$r = \dfrac 1 2 = \dfrac 2 4 = \dfrac 3 6$
Then the mediant of $r = \dfrac 2 4$ and $s = \dfrac 1 1$ gives: 
:$\dfrac {2 + 1} {4 + 1} = \dfrac 3 5$
but the mediant of $r = \dfrac 1 2$ and $s = \dfrac 1 1$ gives: 
:$\dfrac {1 + 1} {2 + 1} = \dfrac 2 3$
which is not the same as $\dfrac {2 + 1} {4 + 1}$.
== Also see ==
* Applesellers' Problem
Category:Mediants