nyuuzyou/wikis huggingface dataset:
Difference of Positive and Negative Parts

== Theorem ==
Let $X$ be a set, and let $f: X \to \overline{\R}$ be an extended real-valued function.
Let $f^+$, $f^-: X \to \overline{\R}$ be the positive and negative parts of $f$, respectively.
Then $f = f^+ - f^-$.
== Proof ==
Let $x \in X$. 
Consider the following four cases for the value of $\map f x$ in $\overline{\R}$:
$(1): \quad \map f x = -\infty$
: By ordering on extended reals:
::$\map {f^+} x = \map \max {0, \map f x} = \map \max {0, -\infty} = 0$
::$\map {f^-} x = - \map \min {0, \map f x} = - \map \min {0, -\infty} = +\infty$
: By extended real subtraction, it thus follows that:
::$\map {f^+} x - \map {f^-} x = 0 - \paren {+\infty} = - \infty = \map f x$
$(2): \quad \map f x \in \openint {-\infty} 0$
: Since $\map f x < 0$, it follows that:
::$\map {f^+} x = \map \max {0, \map f x} = 0$
::$\map {f^-} x = - \map \min {0, \map f x} = - \map f x$
: Thence it immediately follows that:
::$\map {f^+} x - \map {f^-} x = 0 - \paren {- \map f x} = \map f x$
$(3): \quad \map f x \in \hointr 0 {+\infty}$
: Since $\map f x \ge 0$, we obtain:
::$\map {f^+} x = \map \max {0, \map f x} = \map f x$
::$\map {f^-} x = - \min {0, \map f x} = 0$
: Then, these immediately imply:
::$\map {f^+} x - \map {f^-} x = \map f x - 0 = \map f x$
$(4): \quad \map f x = +\infty$
: By ordering on extended reals:
::$\map {f^+} x = \map \max {0, \map f x} = \map \max {0, +\infty} = +\infty$
::$\map {f^-} x = - \map \min {0, \map f x} = - \map \min {0, +\infty} = 0$
: By extended real subtraction, it now follows that:
::$\map {f^+} x - \map {f^-} x = +\infty - 0 = +\infty = \map f x$
In all cases, $\map {f^+} x - \map {f^-} x = \map f x$.
As $x \in X$ was arbitrary, we may conclude that:
:$\forall x \in X: \map {f^+} x - \map {f^-} x = \map f x$
That is, we have shown:
:$f = f^+ - f^-$
== Sources ==
* : $8.7 \ \text{(v)}$, $\S 8$: Problem $6$
Category:Mapping Theory
Category:Positive Parts
Category:Negative Parts
Difference of Projections

== Theorem ==
Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
Then the following are equivalent:
:$(1): \quad P - Q$ is a projection
:$(2): \quad P Q = Q$
:$(3): \quad Q P = Q$
== Proof ==
First it is shown that $(2)$ is equivalent to $(3)$.
Then, equivalence to $(1)$ is shown.
=== $(2)$ iff $(3)$ ===
Suppose $P Q = Q$.
Then $P = P + Q - P Q$, and by Product of Projections, necessarily $P Q = Q P$.
That is, $Q P = P Q = Q$.
The converse is obtained by swapping the rôles of $P$ and $Q$ in Product of Projections.
=== $(2), (3)$ imply $(1)$ ===
Suppose $P Q = Q P = Q$. Then as $P, Q$ are projections:
:$\paren {P - Q}^2 = P^2 + Q^2 - P Q - Q P = P - Q$
That is, $P - Q$ is idempotent.
From Adjoining is Linear, $\paren {P - Q}^* = P^* - Q^* = P - Q$.
An application of Characterization of Projections, statement $(4)$ shows that $P - Q$ is a projection.
=== $(1)$ implies $(2)$ ===
Let $P - Q$ be a projection.
Then by Characterization of Projections, statement $(6)$, one has:
:$\innerprod {P h} h_H - \innerprod {Q h}, h_H \ge 0$
Applying this statement on $P, Q$ also, one obtains that:
:$P h = 0 \implies Q h = 0$
that is:
:$\ker P \subseteq \ker Q$
Orthocomplement Reverses Subset and that $P, Q$ are projections combine to state $\Rng Q \subseteq \Rng P$.
So for every $h \in H$, there is a $p \in H$ with $Q h = P p$.
It follows that:
:$P Q h = P P p = P p = Q h$
Hence $PQ = Q$.
== Also see ==
* Sum of Projections
* Product of Projections
== Sources ==
* $II.3 \text{ Exercise } 6$
Category:Linear Transformations on Hilbert Spaces
Difference of Squares of Cosecant and Cotangent/Also presented as

== Difference of Squares of Cosecant and Cotangent: Also presented as == 
Difference of Squares of Cosecant and Cotangent can also be presented as:
:$\csc^2 x = 1 + \cot^2 x \quad \text{(when $\sin x \ne 0$)}$
or:
:$\cot^2 x = \csc^2 x - 1 \quad \text{(when $\sin x \ne 0$)}$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(3)$
* : $\text V$. Trigonometry: Definitions of the ratios
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Difference of Squares of Cosecant and Cotangent
Difference of Squares of Hyperbolic Cosine and Sine

== Theorem ==
:$\cosh^2 x - \sinh^2 x = 1$
where $\cosh$ and $\sinh$ are hyperbolic cosine and hyperbolic sine.
== Proof ==
== Also presented as ==
== Also see ==
* Sum of Squares of Sine and Cosine
== Sources ==
* : $8.11$: Relationships among Hyperbolic Functions
* : Chapter $15$: Differentiation of Hyperbolic Functions: Solved Problems: $1$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $5$
* 
* 
::in which a mistake occurs
* 
* 
Category:Difference of Squares of Hyperbolic Cosine and Sine
Category:Hyperbolic Cosine Function
Category:Hyperbolic Sine Function
Difference of Squares of Hyperbolic Cosine and Sine/Also presented as

== Difference of Squares of Hyperbolic Cosine and Sine: Also known as == 
Difference of Squares of Hyperbolic Cosine and Sine can also be presented as:
:$\cosh^2 x = 1 + \sinh^2 x$
or:
:$\sinh^2 x = \cosh^2 x - 1$
== Sources ==
* 
Category:Difference of Squares of Hyperbolic Cosine and Sine
Difference of Squares of Hyperbolic Cosine and Sine/Mistake

== Source Work ==
:
:hyperbolic function
== Mistake ==
:The hyperbolic functions satisfy the identity:
::$\sinh^2 \alpha - \cosh^2 \alpha = 1$
== Correction ==
This should read:
:The hyperbolic functions satisfy the identity:
::$\cosh^2 \alpha - \sinh^2 \alpha = 1$
This is demonstrated in Difference of Squares of Hyperbolic Cosine and Sine.
== Sources ==
* 
Category:Mistakes/Dictionary of Mathematics
Difference of Squares of Hyperbolic Cotangent and Cosecant

== Theorem ==
:$\coth^2 x - \csch^2 x = 1$
where $\coth$ and $\csch$ are hyperbolic cotangent and hyperbolic cosecant.
== Proof ==
== Also presented as ==
== Also see ==
* Sum of Squares of Hyperbolic Secant and Tangent
== Sources ==
* : $8.13$: Relationships among Hyperbolic Functions
* 
* 
Category:Difference of Squares of Hyperbolic Cotangent and Cosecant
Category:Hyperbolic Cotangent Function
Category:Hyperbolic Cosecant Function
Difference of Squares of Hyperbolic Cotangent and Cosecant/Also presented as

== Difference of Squares of Hyperbolic Cotangent and Cosecant: Also known as == 
Difference of Squares of Hyperbolic Cotangent and Cosecant can also be presented as:
:$\csch^2 x = \coth^2 x - 1$
or:
:$\coth^2 x = 1 + \csch^2 x$
== Sources ==
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $5$
Category:Difference of Squares of Hyperbolic Cotangent and Cosecant
Difference of Squares of Secant and Tangent/Also presented as

== Difference of Squares of Secant and Tangent: Also presented as == 
Difference of Squares of Secant and Tangent can also be presented as:
:$\sec^2 x = 1 + \tan^2 x \quad \text {(when $\cos x \ne 0$)}$
or:
:$\tan^2 x = \sec^2 x - 1 \quad \text {(when $\cos x \ne 0$)}$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(2)$
* : $\text V$. Trigonometry: Definitions of the ratios
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Difference of Squares of Secant and Tangent
Difference of Two Even-Times Odd Powers

== Theorem ==
Let $\F$ be one of the standard number systems, that is $\Z, \Q, \R$ and so on.
Let $n \in \Z_{> 0}$ be a (strictly) positive odd integer.
Then:
== Proof ==
whence the result.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.22$
* : $\S 2$: Special Products and Factors: $2.22.$
Category:Difference of Two Powers
Difference of Two Even Powers

== Theorem ==
Let $\GF$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then for all $a, b \in \GF$:
== Proof ==
whence the result.
== Also see ==
* Factors of Difference of Two Even Powers
* Difference of Two Odd Powers
Category:Difference of Two Powers
Difference of Two Fifth Powers

== Theorem ==
:$x^5 - y^5 = \paren {x - y} \paren {x^4 + x^3 y + x^2 y^2 + x y^3 + y^4}$
== Proof ==
From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
The result follows directly by setting $n = 5$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.15$
* : $\S 2$: Special Products and Factors: $2.15.$
Category:Fifth Powers
Category:Examples of Use of Difference of Two Powers
Difference of Two Fourth Powers

== Theorem ==
:$x^4 - y^4 = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2$: Special Products and Factors: $2.14$
* : $\S 2$: Special Products and Factors: $2.14.$
Category:Difference of Two Fourth Powers
Category:Fourth Powers
Category:Examples of Use of Difference of Two Powers
Difference of Two Odd Powers

== Theorem ==
Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then for all $a, b \in \mathbb F$:
== Proof ==
A direct application of Difference of Two Powers:
and setting $n \to 2 n + 1$.
== Also see ==
* Factors of Difference of Two Odd Powers
* Difference of Two Even Powers
== Sources ==
* : $\S 2$: Special Products and Factors: $2.20$
* : $\S 2$: Special Products and Factors: $2.20.$
Category:Difference of Two Powers
Difference of Two Odd Powers/Mistake

== Source Work ==
:Chapter $2$: Special Products and Factors
== Mistake ==
:$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \paren {x^{n - 1} + x^{n - 2} y + x^{n - 3} y^2 + \dotsb} \paren {x^{n - 1} - x^{n - 2} y + x^{n - 3} y^2 - \dotsb}$
== Correction ==
This result is true only if $n$ is an odd integer.
When $n$ is even, we get:
and there is no obvious way to factorise $\ds \sum_{j \mathop = 0}^{n - 1} x^{2 \paren {n - j - 1} } y^{2 j}$
As an example, we examine $x^8 - y^8$:
Using the stated formula, we obtain:
which is not the same thing at all.
That they are indeed not the same can be calculated directly.
Let $x = 2, y = 1$.
We have that $2^8 - 1 = 255$.
Then we investigate what the formulae give us:
Using the wrong formula:
== Sources ==
* : $\S 2$: Special Products and Factors: $2.22$
* : $\S 2$: Special Products and Factors: $2.22.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Difference of Two Powers

== Theorem ==
Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.
Let $n \in \N$ such that $n \ge 2$.
Then for all $a, b \in \mathbb F$:
For convenience of applicability, these results are sometimes separated into two cases for odd and even indices:
=== Difference of Two Odd Powers ===
=== Difference of Two Even Powers ===
=== General Commutative Ring ===
The result can also be extended into the general abstract algebraic context as follows:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Examples ==
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $3$
Category:Algebra
Category:Polynomial Theory
Category:Difference of Two Powers
Difference of Two Powers/Examples/Difference of Two Cubes

== Theorem ==
:$x^3 - y^3 = \paren {x - y} \paren {x^2 + x y + y^2}$
=== Corollary ===
== Proof ==
From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
The result follows directly by setting $n = 3$.
== Sources ==
* : $\S 1.5$: The Importance of Variables in Mathematics
* : $\S 2$: Special Products and Factors: $2.12$
* : $\S 2$: Special Products and Factors: $2.12.$
Category:Third Powers
Category:Difference of Two Cubes
Category:Examples of Use of Difference of Two Powers
Difference of Two Powers/Examples/Difference of Two Cubes/Corollary

== Theorem ==
:$x^3 - 1 = \paren {x - 1} \paren {x^2 + x + 1}$
== Proof ==
From Difference of Two Cubes:
:$x^3 - y^3 = \paren {x - y} \paren {x^2 + x y + y^2}$
The result follows directly by setting $y = 1$.
== Sources ==
* : $\S 3$. Roots of Unity: $(3.5)$
Category:Third Powers
Category:Difference of Two Cubes
Difference of Two Powers/General Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a, b \in R$.
Let $n \in \N$ such that $n \ge 2$.
Then:
== Proof ==
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j$.
This can also be written:
:$\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j \circ a^{n - j - 1}$
Consider:
:$\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j$
Taking the first term (where $j = 0$) out of the summation, we get:
:$\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
Similarly, consider:
:$\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j}$
Taking the first term (where $j = 0$) out of the summation:
:$\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
This is equal to:
:$\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j \circ b^{n - j}$
by Permutation of Indices of Summation.
So:
Category:Commutative Rings
Category:Polynomial Theory
Category:Difference of Two Powers
Difference of Two Powers/Proof 1

== Theorem ==
== Proof ==
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$.
This can also be written:
:$\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j a^{n - j - 1}$
Consider:
:$\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j$
Taking the first term (where $j = 0$) out of the summation, we get:
:$\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$
Similarly, consider:
:$\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}$
Taking the first term (where $j = 0$) out of the summation:
:$\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$
This is equal to:
:$\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j b^{n - j}$
by Permutation of Indices of Summation.
So:
Category:Difference of Two Powers
Difference of Two Powers/Proof 4

== Theorem ==
== Proof ==
The proof will proceed by the Principle of Complete Finite Induction on $\Z_{>0}$.
Let $S$ be the set defined as:
:$\ds S := \set {n \in \Z_{>0}: a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j}$
That is, $S$ is to be the set of all $n$ such that:
:$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
=== Basis for the Induction ===
We have that:
:$\ds a^1 - b^1 = \paren {a - b} \sum_{j \mathop = 0}^0 a^{1 - 0 - 1} b^j$
So $1 \in S$.
This is the basis for the induction.
=== Induction Hypothesis ===
It is to be shown that if $r \in S$ for all $r$ such that $1 \le r \le k$, then it follows that $k + 1 \in S$.
This is the induction hypothesis:
:$\forall r \in \Z_{>0}: 1 \le r \le k: \ds a^r - b^r = \paren {a - b} \sum_{j \mathop = 0}^{r - 1} a^{r - j - 1} b^j$
It is to be demonstrated that it follows that:
:$\ds a^{k + 1} - b^{k + 1} = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j$
=== Induction Step ===
This is the induction step:
So $\forall r \in S: 0 \le r \le k: r \in S \implies k + 1 \in S$ and the result follows by the Principle of Complete Finite Induction:
:$\forall n \in \Z_{>0}: \ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $3$
Category:Difference of Two Powers
Difference of Two Sixth Powers

== Theorem ==
:$x^6 - y^6 = \paren {x - y} \paren {x + y} \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
== Proof ==
== Sources ==
* : $\S 2$: Special Products and Factors: $2.17$
* : $\S 2$: Special Products and Factors: $2.17.$
Category:Sixth Powers
Category:Examples of Use of Difference of Two Powers
Difference of Two Squares

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $x, y \in R$.
Then:
:$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:
:$x^2 - y^2 = \paren {x + y} \paren {x - y}$
== Algebraic Proof 1 ==
== Algebraic Proof 2 ==
== Geometric Proof 1 ==
== Geometric Proof 2 ==
== Sources ==
* : Chapter $\text I$ Introductory: $2$. The Use of Symbols
* : $\S 2$: Special Products and Factors: $2.11$
* 
* : $\S 2$: Special Products and Factors: $2.11.$
Category:Difference of Two Squares
Category:Commutative Rings
Category:Square Function
Category:Polynomial Theory
Difference of Two Squares/Algebraic Proof 2

== Theorem ==
== Proof ==
This is a special case of Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \cdots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
The result follows by setting $n = 2$.
Category:Difference of Two Squares
Difference of Two Squares/Geometric Proof 1

== Theorem ==
:$\forall x, y \in \R: x^2 - y^2 = \paren {x + y} \paren {x - y}$
== Proof ==
:
Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.
Then the rectangle contained by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.
(That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.)
This is proved as follows.
Construct the square $CBFE$ on $CB$, and join $BE$.
Construct $DG$ parallel to $CE$ through $G$, and let $DG$ cross $BE$ at $H$.
Construct $KM$ parallel to $AB$ through $H$.
Construct $AK$ parallel to $BF$ through $A$.
From Complements of Parallelograms are Equal:
:$\Box CDHL = \Box FGHM$.
Add the square $DBMH$ to each.
Then $\Box CBML = \Box DBFG$.
But as $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that:
:$\Box ACLK = \Box CBML$
Add $\Box CDHL$ to each.
Then $\Box ADHK$ is equal in area to the gnomon $CBFGHL$.
But $\Box ADHK$ is the rectangle contained by $AD$ and $DB$, because $DB = DH$.
So the gnomon $CBFGHL$ is equal in area to the rectangle contained by $AD$ and $DB$.
Now $\Box LHGE$ is equal to the square on $CD$.
Add $\Box LHGE$ to each of the gnomon $CBFGHL$ and $\Box ADHK$.
Then the gnomon $CBFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $DB$ and the square on $CD$.
But the gnomon $CBFGHL$ together with $\Box LHGE$ is the square $CBFE$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Difference of Two Squares
Difference of Two Squares/Geometric Proof 2

== Theorem ==
== Proof ==
:
Let $\Box ABCD$ be a square of side length $x$.
Let $\Box DEFG$ be a square of side length $y$ where $y < x$
Let $EF$ be produced to $H$.
The area of $\Box ABCD$ is seen to be equal to the sum of:
:the area of the rectangle $AEHB$
:the area of the rectangle $FGCH$
:the area of the square $DEFG$
From Area of Square:
:the area of $\Box ABCD$ is equal to $x^2$
:the area of $\Box DEFG$ is equal to $y^2$
From Area of Rectangle:
:the area of $AEHB$ is equal to $x \paren {x - y}$
:the area of $FGCH$ is equal to $y \paren {x - y}$
Hence:
Category:Difference of Two Squares
Difference of Two Squares cannot equal 2 modulo 4

== Theorem ==
Let $n \in \Z_{>0}$ be of the form $4 k + 2$ for some $k \in \Z$.
Then $n$ cannot be expressed in the form:
:$n = a^2 - b^2$
for $a, b \in \Z$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $10$
* : $10$
Category:Square Numbers
Category:Difference of Two Squares cannot equal 2 modulo 4
Difference of Two Squares cannot equal 2 modulo 4/Proof 1

== Theorem ==
== Proof ==
Let $n = a^2 - b^2$ for some $a, b \in \Z$.
By Square Modulo 4, both $a$ and $b$ are of the form $4 k$ or $4 k + 1$ for some integer $k$.
There are $4$ cases:
;$a \equiv b \equiv 0 \pmod 4$:
Then:
:$a^2 - b^2 \equiv 0 \pmod 4$
and so $n$ is in the form $4 k$.
;$a \equiv 0 \pmod 4$, $b \equiv 1 \pmod 4$:
Then:
:$a^2 - b^2 \equiv -1 \pmod 4 \equiv 3 \pmod 4$
and so $n$ is in the form $4 k + 3$.
;$a \equiv 1 \pmod 4$, $b \equiv 0 \pmod 4$:
Then:
:$a^2 - b^2 \equiv 1 \pmod 4$
and so $n$ is in the form $4 k + 1$.
;$a \equiv b \equiv 1 \pmod 4$:
Then:
:$a^2 - b^2 \equiv 0 \pmod 4$
and so $n$ is in the form $4 k$.
Thus it is never the case that $a^2 - b^2 = 4 k + 2$.
Category:Difference of Two Squares cannot equal 2 modulo 4
Difference of Two Squares cannot equal 2 modulo 4/Proof 2

== Theorem ==
== Proof ==
Let $n_0 = c^2 - d^2$ for some $c, d \in \Z$.
Then:
:$n_0 = \paren {c + d} \paren {c - d}$
and so:
:$\paren {c + d} - \paren {c - d} = 2 d$
Therefore $n$ must be expressible as a product of two integers whose difference is even.
Now consider the integer $n \in \Z$ that satisfies $n \equiv 2 \pmod 4$.
$n$ is an even number that is not divisible by 4.
This implies that any even number that divides $n$ will produce an odd number.
Consider the ordered pairs of integers $\tuple {a, b}, a < b$ such that $ab = n$.
If $a$ is odd, then $b = \dfrac n a$ must in turn be even.
Hence the difference will not be even.
If $a$ is even, $b$ must be odd by the implication mentioned earlier.
Hence the difference will once again not be even.
Hence, $n$ cannot be expressed as a product of two integers whose difference is even.
In turn it cannot be expressed as a Difference of Two Squares.
Category:Difference of Two Squares cannot equal 2 modulo 4
Difference of Unbounded Closed Intervals

== Theorem ==
Let $a, b \in \R$ have $a < b$. 
Then: 
:$\hointl {-\infty} b \setminus \hointl {-\infty} a = \hointl a b$
where $\setminus$ denotes set difference. 
== Proof ==
Note that: 
:$x \in \hointl {-\infty} b \setminus \hointl {-\infty} a$
: 
:$x \in \hointl {-\infty} b$ but $x \not \in \hointl {-\infty} a$.
That is: 
:$x \le b$ but it is not the case that $x \le a$. 
So this is equivalent to: 
:$x \le b$ and $x > a$.
That is: 
:$x \in \hointl a b$
So:
:$\hointl {-\infty} b \setminus \hointl {-\infty} a = \hointl a b$
Category:Real Intervals
Difference of Unions is Subset of Union of Differences

== Theorem ==
Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
where $S_\alpha \setminus T_\alpha$ denotes set difference.
== Proof ==
Let $\ds x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$.
Then by definition of set difference:
By definition of set union, it follows that:
and so:
:$\exists \beta \in I: x \in S_\beta \setminus T_\beta$
Hence:
:$\ds x \in \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
by definition of set union.
The result follows by definition of subset.
== Sources ==
* : $\S 1$: Sets and Functions: Problem $5$
Category:Set Union
Category:Set Difference
Different Latitude and Longitude on Same Meridian

== Problem ==
:There are $3$ remarkable places on the globe that differ in latitude,
:as well as in longitude;
:and yet, all of them lie under the same meridian.
== Solution ==
Let one point be one of the poles.
This has a latitude of $90 \degrees$, either north or south.
Let another point be a number of degrees of latitude different from $90 \degrees$.
Let a third point be a different number of degrees of latitude, again different from $90 \degrees$, $180 \degrees$ of longitude different from the second point.
All $3$ points have a different latitudes by construction.
The pole has no defined longitude.
Hence all $3$ points have different longitude.
But all $3$ points are on the same great circle of Earth.
Hence, using the definition of meridian to mean that complete great circle, the puzzle is seen to have this solution.
It is of course to be noted that, apart from the pole, the points can be chosen arbitrarily to suit the conditions.
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $159$
Category:Geography Puzzles