avewright/proofwiki-math huggingface dataset:
Theorem: Sum of Squares of Sum and Difference

:$\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$

Proof:
{{:Euclid:Proposition/II/9}}
:400px
That is, from the above diagram:
: $\left({AC + CD}\right)^2 + \left({AC - CD}\right)^2 = 2 \left({AC^2 + CD^2}\right)$
Let $AB$ be bisected at $C$, and let $D$ be another random point on $AB$.
Then the squares on $AD$ and $DB$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $DF$ parallel to $CE$ to intersect $BE$ at $F$, and join $AF$.
Construct $FG$ parallel to $AB$ to intersect $CE$ at $G$.
Since $AC = EC$ we have from Isosceles Triangle has Two Equal Angles that $\angle EAC = \angle AEC$.
Since $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles we have that $\angle EAC + \angle AEC$ also equals a right angle.
As they are equal, $\angle EAC$ and $\angle AEC$ are both equal to half a right angle.
For the same reason, $\angle EBC$ and $\angle BEC$ are also both equal to half a right angle.
We have that $\angle GEF = \angle BEC$, so $\angle GEF$ is also half a right angle.
From Parallelism implies Equal Corresponding Angles, $\angle EGF$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle EFG$ is half a right angle.
So $\angle GEF = \angle GFE$, and so from Triangle with Two Equal Angles is Isosceles, $EG = GF$.
From Parallelism implies Equal Corresponding Angles, $\angle FDB$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle BFD$ is half a right angle.
So $\angle BFD = \angle FBD$, and so from Triangle with Two Equal Angles is Isosceles, $FD = DB$.
Since $AC = CE$, the square on $AC$ equals the square on $CE$.
So the squares on $AC$ and $CE$ together are twice the square on $AC$.
But from Pythagoras's Theorem, the square on $AE$ equals the squares on $AC$ and $CE$ because $\angle ACE$ is a right angle.
So the square on $AE$ is double the square on $AC$.
By a similar argument, the square on $EF$ is double the square on $GF$.
But from Opposite Sides and Angles of Parallelogram are Equal, $GF = CD$.
So the square on $EF$ is double the square on $CD$.
So the squares on $AE$ and $EF$ are double the squares on $AC$ and $CD$.
We have that $\angle AEF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AE$ and $EF$.
So the square on $AF$ is double the squares on $AC$ and $CD$.
We have that $\angle ADF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AD$ and $DF$.
So the squares on $AD$ and $DF$ are double the squares on $AC$ and $CD$.
But $DF = DB$.
Hence the result.
{{Qed}}
Theorem: Sum of Squares of Sum and Difference

:$\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$

Proof:
{{:Euclid:Proposition/II/10}}
:400px
That is, from the above diagram:
:$\paren {AC + CD}^2 + \paren {CD - AC}^2 = 2 \paren {AC^2 + CD^2}$
Let $AB$ be bisected at $C$, and let $AB$ be produced to some point $D$.
Then the squares on $AD$ and $BD$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $EF$ parallel to $AB$, and Construct $DF$ parallel to $CE$.
From Parallelism implies Supplementary Interior Angles, $\angle CEF + \angle EFD$ equal two right angles.
Therefore $\angle FEB + \angle EFD$ is less than two right angles.
So by the Parallel Postulate, $EB$ and $FB$ must meet when produced in the direction $B$ and $D$.
Let this point be $G$, and let $AG$ be joined.
From Isosceles Triangle has Two Equal Angles, as $AC = AE$, $\angle EAC = \angle AEC$.
Because $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles, each of $\angle EAC$ and $\angle AEC$ are each half a right angle.
For the same reason, each of $\angle CEB$ and $\angle EBC$ are each half a right angle.
Since $\angle EBC$ is half a right angle, from Two Straight Lines make Equal Opposite Angles, $\angle DBG$ is also half a right angle.
From Parallelism implies Equal Alternate Angles, $\angle BDG = \angle DCE$, so $\angle BDG$ is also a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle DGB$ is half a right angle.
As $\angle DGB = \angle DBG$, from Triangle with Two Equal Angles is Isosceles, we have that $BD = BG$.
From Opposite Sides and Angles of Parallelogram are Equal, $\angle EFD = \angle ECD$ which is a right angle, 
Since $\angle EGF$ is half a right angle, and $\angle EFD$ is a right angle, $\angle FEG$ is half a right angle from Sum of Angles of Triangle Equals Two Right Angles.
So $\angle EGF = \angle FEG$, and so from Triangle with Two Equal Angles is Isosceles, $GF = EF$.
Since $EC = CA$, the square on $EC$ equals the square on $CA$.
So the squares on $EC$ and $CA$ are together twice the square on $CA$.
From Pythagoras's Theorem, the square on $EA$ is equal to the the squares on $EC$ and $CA$, and so twice the square on $AC$.
Again, since $FG = FE$, the square on $FG$ equals the square on $FE$.
So the squares on $FG$ and $EF$ are together twice the square on $FE$.
From Pythagoras's Theorem, the square on $EG$ is equal to the the squares on $GF$ and $FE$, and so twice the square on $EF$.
From Opposite Sides and Angles of Parallelogram are Equal, $EF = CD$.
But the square on $EA$ has been shown to be double the square on $AC$.
So the squares on $AE$ and $EG$ are twice the squares on $AC$ and $CD$.
From Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AE$ and $EG$.
So the square on $AG$is double the squares on $AC$ and $CD$.
But by Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AD$ and $DG$.
As $DG = DB$, the result follows.
{{qed}}
Theorem: Construction of Square equal to Given Polygon

A square can be constructed the same size as any given polygon.
{{:Euclid:Proposition/II/14}}

Proof:
:500px
Let $A$ be the given polygon.
Construct the rectangle $BCDE$ equal to the given polygon.
If it so happens that $BE = ED$, then $BCDE$ is a square, and the construction is complete.
Suppose $BE \ne ED$. Then WLOG suppose $BE > ED$.
Produce $BE$ from $E$ and construct on it $EF = ED$.
Bisect $BF$ at $G$.
Construct the semicircle $BHF$ with center $G$ and radius $GF$ (see diagram).
Produce $DE$ from $D$ to $H$.
From Difference of Two Squares, the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GF$.
But $GF = GH$.
So the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GH$.
From Pythagoras's Theorem, the square on $GH$ equals the squares on $GE$ and $EH$.
Then the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the squares on $GE$ and $EH$.
Subtract the square on $GE$ from each.
Then the rectangle contained by $BE$ and $EF$ is equal to the square on $EH$.
So the square on $EH$ is equal to the rectangle $BCDE$.
So the square on $EH$ is equal to the given polygon, as required.
{{qed}}
{{Euclid Note|14|II}}
Theorem: Finding Center of Circle

For any given circle, it is possible to find its center.
{{:Euclid:Proposition/III/1}}

Proof:
:300px
Draw any chord $AB$ on the circle in question.
Bisect $AB$ at $D$.
Construct $CE$ perpendicular to $AB$ at $D$, where $C$ and $E$ are where this perpendicular meets the circle.
Bisect $CE$ at $F$.
Then $F$ is the center of the circle.
The proof is as follows.
Suppose $F$ were not the center of the circle, but that $G$ were instead.
Join $GA, GB, GD$.
As $G$ is (as we have supposed) the center, then $GA = GB$.
Also, we have $DA = DB$ as $D$ bisects $AB$.
So from Triangle Side-Side-Side Congruence:
:$\triangle ADG = \triangle BDG$
Hence:
:$\angle ADG = \angle BDG$
But from {{EuclidDefLink|I|10|Right Angle}}:
:''{{:Definition:Euclid's Definitions - Book I/10 - Right Angle}}''
So $\angle ADG$ is a right angle.
But $\angle ADF$ is also a right angle.
So $\angle ADG = \angle ADF$, and this can happen only if $G$ lies on $CE$.
But then as $G$ is, as we suppose, at the center of the circle, then $GC = GE$.
Thus it follows that $G$ bisects $CE$.
But then $GC = FC$, and so $G = F$.
Hence the result.
{{qed}}
=== Porism ===
{{:Finding Center of Circle/Porism}}
Theorem: Finding Center of Circle

For any given circle, it is possible to find its center.
{{:Euclid:Proposition/III/1}}

Proof:
From Perpendicular Bisector of Chord Passes Through Center, $CE$ passes through the center of the circle.
The center must be the point $F$ such that $FE = FC$.
That is, $F$ is the bisector of $CE$.
Theorem: Conditions for Diameter to be Perpendicular Bisector

If in a circle a diameter bisects a chord (which is itself not a diameter), then it cuts it at right angles, and if it cuts it at right angles then it bisects it.
{{:Euclid:Proposition/III/3}}

Proof:
:300px
Let $ABC$ be a circle, in which $AB$ is a chord which is not a diameter (i.e. it does not pass through the center).
Theorem: Chords do not Bisect Each Other

If in a circle two chords (which are not diameters) cut one another, then they do not bisect one another.
{{:Euclid:Proposition/III/4}}

Proof:
:250px
Let $ABCD$ be a circle, in which $AC$ and $BD$ are chords which are not diameters (i.e. they do not pass through the center).
Let $AC$ and $BD$ intersect at $E$.
{{AimForCont}} they were able to bisect one another, such that $AE = EC$ and $BE = ED$.
Find the center $F$ of the circle, and join $FE$.
From Conditions for Diameter to be Perpendicular Bisector, as $FE$ bisects $AC$, then it cuts it at right angles.
So $\angle FEA$ is a right angle.
Similarly, $\angle FEB$ is a right angle.
So $\angle FEA = \angle FEB$, and they are clearly unequal.
From this contradiction, it follows that $AC$ and $BD$ can not bisect each other.
{{qed}}
{{Euclid Note|4|III}}
Theorem: Concentric Circles do not Intersect

If two circles are concentric, they share no points on their circumferences.
Alternatively, this can be worded:
:If two concentric circles share one point on their circumferences, then they share them all (that is, they are the same circle).

Proof:
This follows directly from:
:Intersecting Circles have Different Centers
:Touching Circles have Different Centers.
{{qed}}
Category:Concentric Circles
jrlz100khvixyaze2kcnsi6uaptwglk
Theorem: Existence of Real Logarithm

Let $b, y \in \R$ such that $b > 1$ and $y > 0$.
Then there exists a unique real $x \in \R$ such that $b^x = y$.
This $x$ is called the '''logarithm of $y$ to the base $b$'''.
Also see the definition of a (general) logarithm.

Proof:
We start by establishing a lemma:
Theorem: Baire Category Theorem/Complete Metric Space

Let $M = \struct {A, d}$ be a complete metric space.
Then $M = \struct {A, d}$ is also a Baire space.

Proof:
Let $U_n$ be a countable set of open sets of $M$ all of which are everywhere dense.
The strategy of this proof is to show that the intersection $\bigcap U_n$ is everywhere dense.
Let $W \subseteq A$ be a non-empty open set of $M$.
From Open Set Characterization of Denseness, since $U_1$ is everywhere dense:
:$W \cap U_1 \ne \O$
Thus there is a point $x_1 \in A$ and $\epsilon_1 \in \R_{>0}$ such that:
:$\map {\overline B} {x_1, \epsilon_1} \subset W \cap U_1$
where $\map B {x, \epsilon}$ and $\map {\overline B} {x, \epsilon}$ denote an open $\epsilon$-ball of $x$ and its closure, respectively.
Recall that the $U_n$ are everywhere dense.
Hence we can use {{Axiom-link|Dependent Choice}} in a recursive manner to find a pair of sequences $\sequence {x_n}$ and $\sequence {\epsilon_n} > 0$ such that:
:$\map {\overline B} {x_n, \epsilon_n} \subset \map B {x_{n - 1}, \epsilon_{n - 1} } \cap U_n$
as well as $\epsilon_n < 1/n $.
Since $x_n \in \map B {x_m, \epsilon_m}$ when $n > m$, we have that $x_n$ is a Cauchy Sequence.
Thus $x_n$ converges to some limit $x$ by definition of completeness.
For any $n$, by closedness:
:$x \in \map {\overline B} {x_{n + 1}, \epsilon_{n + 1} } \subset \map B {x_n, \epsilon_n}$
Thus:
:$\forall n \in \N: x \in W \cap U_n$
From Open Set Characterization of Denseness it follows that $\ds \bigcap U_n$ is (everywhere) dense.
So, by definition, $\struct {M, d}$ is a Baire space.
{{qed}}
{{ADC}}
{{Namedfor|René-Louis Baire|cat = Baire}}
Theorem: Baire Characterisation Theorem

A real-valued function $f$ defined on a Banach space $X$ is a Baire-$1$ function {{iff}}:
: for every non-empty closed subset $K$ of $X$, the restriction of $f$ to $K$ has a point of continuity relative to the topology of $K$.
{{explain|What is Baire-$1$? The linked definition does not have the numeration.}}

Proof:
{{ProofWanted}}
{{Namedfor|René-Louis Baire|cat = Baire}}
Category:Topology
4olj66tipmyrbw8qpve0xhj63f91bde
Theorem: Ramsey's Theorem

In any coloring of the edges of a sufficiently large complete graph, one will find monochromatic complete subgraphs.
For $2$ colors, Ramsey's theorem states that for any pair of positive integers $\tuple {r, s}$, there exists a least positive integer $\map R {r, s}$ such that for any complete graph on $\map R {r, s}$ vertices, whose edges are colored red or blue, there exists either a complete subgraph on $r$ vertices which is entirely red, or a complete subgraph on $s$ vertices which is entirely blue.
More generally, for any given number of colors $c$, and any given integers $n_1, \ldots, n_c$, there is a number $\map R {n_1, \ldots, n_c}$ such that:
:if the edges of a complete graph of order $\map R {n_1, \ldots, n_c}$ are colored with $c$ different colours, then for some $i$ between $1$ and $c$, it must contain a complete subgraph of order $n_i$ whose edges are all color $i$.
This number $\map R {n_1, \ldots, n_c}$ is called the Ramsey number for $n_1, \ldots, n_c$.
The special case above has $c = 2$ (and $n_1 = r$ and $n_2 = s$).
Here $\map R {r, s}$ signifies an integer that depends on both $r$ and $s$.
It is understood to represent the smallest integer for which '''Ramsey's Theorem''' holds.

Proof:
First we prove Ramsey's Theorem for the 2-color case, by induction on $r + s$.
It is clear from the definition that
:$\forall n \in \N: \map R {n, 1} = \map R {1, n} = 1$
because the complete graph on one node has no edges.
This is the basis for the induction.
We prove that $\map R {r, s}$ exists by finding an explicit bound for it.
By the induction hypothesis, $\map R {r - 1, s}$ and $\map R {r, s - 1}$ exist.
Theorem: Schur's Theorem (Ramsey Theory)

Let $r$ be a positive integer.
Then there exists a positive integer $S$ such that:
:for every partition of the integers $\set {1, \ldots, S}$ into $r$ parts, one of the parts contains integers $x$, $y$ and $z$ such that:
::$x + y = z$

Proof:
Let:
:$n = \map R {3, \ldots, 3}$
where $\map R {3, \ldots, 3}$ denotes the Ramsey number on $r$ colors.
Take $S$ to be $n$.
{{refactor|Extract the below process of "coloring" a partition into its own page|level = medium}}
partition the integers $\set {1, \ldots, n}$ into $r$ parts, which we denote by '''colors'''.
That is:
:the integers in the first part are said to be '''colored''' $c_1$
:the integers in the second part are said to be colored $c_2$ 
and so on till color $c_r$.
Thus $\set {1, \ldots, S}$ has been '''$r$-colored'''.
(This terminology is common in Ramsey theory.)
Now consider the complete graph $K_n$.
Now color the edges of $K_n$ as follows:
:An edge $xy$ is given color $c$ if $\size {x - y}$ was colored $c$ in the partitioning.
{{explain|When the page defining a "coloring" of a partition is written, make sure that the links are assigned appropriately from the two difference senses of "coloring" in the above.}}
From the definition of $\map R {3, \ldots, 3}$ and Ramsey's Theorem, $K_n$ will definitely contain a monochromatic triangle, say built out of the vertices $i > j > k$.
Let the triangle be colored $c_m$.
Now $i - j$, $i - k$ and $j - k$ will also be colored $c_m$.
That is, $i - j$, $i - k$ and $j - k$ will belong to the same part in the partition.
It only remains to take $x = i - j$, $y = j - k$ and $z = i - k$ to complete the proof.
{{qed}}
Theorem: Brahmagupta's Formula

The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:
:$\sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
where $s$ is the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$

Proof:
Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.
:300px
Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$
From Area of Triangle in Terms of Two Sides and Angle:
{{begin-eqn}}
{{eqn | l = \triangle ABC
      | r = \frac 1 2 a b \sin \angle ABC
      | c = 
}}
{{eqn | l = \triangle ADC
      | r = \frac 1 2 c d \sin \angle ADC
      | c = 
}}
{{end-eqn}}
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.
Hence we have:
{{begin-eqn}}
{{eqn | l = \sin \angle ABC
      | r = \sin \angle ADC
      | c = Sine and Cosine of Supplementary Angles
}}
{{eqn | l = \cos \angle ABC
      | r = -\cos \angle ADC
      | c = Sine and Cosine of Supplementary Angles
}}
{{end-eqn}}
This leads to:
{{begin-eqn}}
{{eqn | l = \Area
      | r = \frac 1 2 a b \sin \angle ABC + \frac 1 2 c d \sin \angle ABC
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\Area}^2
      | r = \frac 1 4 \paren {a b + c d}^2 \sin^2 \angle ABC
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 \paren {\Area}^2
      | r = \paren {a b + c d}^2 \paren {1 - \cos^2 \angle ABC}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \paren {a b + c d}^2 - \cos^2 \angle ABC \paren {a b + c d}^2
      | c = 
}}
{{end-eqn}}
Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$:
:$a^2 + b^2 - 2 a b \cos \angle ABC = c^2 + d^2 - 2 c d \cos \angle ADC$
From the above:
:$\cos \angle ABC = -\cos \angle ADC$
Hence:
:$2 \cos \angle ABC \paren {a b + c d} = a^2 + b^2 - c^2 - d^2$
Substituting this in the above equation for the area:
{{begin-eqn}}
{{eqn | l = 4 \paren {\Area}^2
      | r = \paren {a b + c d}^2 - \frac 1 4 \paren {a^2 + b^2 - c^2 - d^2}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 16 \paren {\Area}^2
      | r = 4 \paren {a b + c d}^2 - \paren {a^2 + b^2 - c^2 - d^2}^2
      | c = 
}}
{{end-eqn}}
This is of the form $x^2 - y^2$.
Hence, by Difference of Two Squares, it can be written in the form $\paren {x + y} \paren {x - y}$ as:
{{begin-eqn}}
{{eqn | o =
      | r = \paren {2 \paren {a b + c d} + a^2 + b^2 - c^2 - d^2} \paren {2 \paren {a b + c d} - a^2 - b^2 + c^2 + d^2}
      | c = 
}}
{{eqn | r = \paren {\paren {a + b}^2 - \paren {c - d}^2} \paren {\paren {c + d}^2 - \paren {a - b}^2}
      | c = 
}}
{{eqn | r = \paren {a + b + c - d} \paren {a + b + d - c} \paren {a + c + d - b} \paren {b + c + d - a}
      | c = 
}}
{{end-eqn}}
When we introduce the expression for the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
the above converts to:
:$16 \paren {\Area}^2 = 16 \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$
Taking the square root:
:$\Area = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
{{qed}}
Theorem: Euler-Binet Formula

The Fibonacci numbers have a closed-form solution:
:$F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5} = \dfrac {\phi^n - \paren {1 - \phi}^n} {\phi - \paren {1 - \phi}}$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
From {{Defof|Golden Mean|index = 2}}: $\phi = \dfrac {1 + \sqrt 5} 2$
Therefore, substituting $\sqrt 5 = 2\phi - 1 = \phi - \paren {1 - \phi} = \phi - \hat \phi$, the above can be written as:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\paren {\phi - \hat \phi}}$

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
=== Basis for the Induction ===
$\map P 0$ is true, as this just says:
:$\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \frac {\phi - \hat \phi} {\sqrt 5}
      | r = \frac {\paren {\frac {1 + \sqrt 5} 2} - \paren {\frac {1 - \sqrt 5} 2} } {\sqrt 5}
      | c = 
}}
{{eqn | r = \frac {\paren {1 - 1} + \paren {\sqrt 5 + \sqrt 5} } {2 \sqrt 5}
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = F_1
      | c = 
}}
{{end-eqn}}
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true.
So this is our induction hypothesis:
:$\forall 0 \le j \le k + 1: F_j = \dfrac {\phi^j - \hat \phi^j} {\sqrt 5}$
Then we need to show:
:$F_{k + 2} = \dfrac {\phi^{k + 2} - \hat \phi^{k + 2} } {\sqrt 5}$
=== Induction Step ===
This is our induction step:
We observe that we have the following two identities:
:$\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
:$\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$
This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.
Thus:
{{begin-eqn}}
{{eqn | l = \phi^{k + 2} - \hat \phi^{k + 2}
      | r = \paren {1 + \phi} \phi^k - \paren {1 + \hat \phi} \hat \phi^k
}}
{{eqn | r = \paren {\phi^k - \hat \phi^k} + \paren {\phi^{k + 1} - \hat \phi^{k + 1} }
}}
{{eqn | r = \sqrt 5 \paren {F_k + F_{k + 1} }
      | c = Induction Hypothesis
}}
{{eqn | r = \sqrt 5 F_{k + 2}
      | c = {{Defof|Fibonacci Numbers}}
}}
{{end-eqn}}
The result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
{{qed}}
Theorem: Euler-Binet Formula

The Fibonacci numbers have a closed-form solution:
:$F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5} = \dfrac {\phi^n - \paren {1 - \phi}^n} {\phi - \paren {1 - \phi}}$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
From {{Defof|Golden Mean|index = 2}}: $\phi = \dfrac {1 + \sqrt 5} 2$
Therefore, substituting $\sqrt 5 = 2\phi - 1 = \phi - \paren {1 - \phi} = \phi - \hat \phi$, the above can be written as:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\paren {\phi - \hat \phi}}$

Proof:
This follows as a direct application of the first Binet form:
:$U_n = m U_{n - 1} + U_{n - 2}$
where:
{{begin-eqn}}
{{eqn | l = U_0
      | r = 0
}}
{{eqn | l = U_1
      | r = 1
}}
{{end-eqn}}
has the closed-form solution:
:$U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$
where:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \sqrt {m^2 + 4}
}}
{{eqn | l = \alpha
      | r = \frac {m + \Delta} 2
}}
{{eqn | l = \beta
      | r = \frac {m - \Delta} 2
}}
{{end-eqn}}
where $m = 1$.
{{qed}}