nyuuzyou/wikis huggingface dataset:
Absolute Error of Sum is not Greater than Sum of Absolute Errors

== Theorem ==
Let $X_1, X_2, \ldots, X_n$ be approximations to a collection of (true) values $x_1, x_2, \ldots, x_n$ respectively.
Let $\Delta X_i$ be the absolute error of $X_i$ in $x_i$ for $i \in \set {1, 2, \ldots, n}$.
Then:
:$\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$
That is, the absolute error of a sum of approximations is not greater than the sum of the absolute errors of those approximations.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$
for appropriately defined $X_1, X_2, \ldots, X_n$ and $x_1, x_2, \ldots, x_n$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\map \Delta {X_1 + X_2} \le \Delta X_1 + \Delta X_2$
We have:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \map \Delta {\sum_{i \mathop = 1}^k X_i} \le \sum_{i \mathop = 1}^k \Delta X_i$
from which it is to be shown that:
:$\ds \map \Delta {\sum_{i \mathop = 1}^{k + 1} X_i} \le \sum_{i \mathop = 1}^{k + 1} \Delta X_i$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: n \ge 2: \ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.5$ Absolute and Relative Errors: $3.5.4$
Category:Absolute Error
Absolute Net Convergence Equivalent to Absolute Convergence

== Theorem ==
Let $V$ be a Banach space.
Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.
Let $r \in \R_{\mathop \ge 0}$
Then the following two statements are equivalent:
:$(1): \quad$ the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$
:$(2): \quad$ the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$
== Proof ==
=== Statement $(1)$ implies Statement $(2)$ ===
Let the generalized sum $\ds \sum \set {v_n: n \in \N}$ be absolutely net convergent to $r$.
=== Statement $(2)$ implies Statement $(1)$ ===
Let the series $\ds \sum_{n \mathop = 1}^\infty v_n$ be absolutely convergent to $r$.
Category:Banach Spaces
Category:Nets (Set Theory)
Category:Absolute Net Convergence Equivalent to Absolute Convergence
Absolute Net Convergence Equivalent to Absolute Convergence/Absolute Convergence implies Absolute Net Convergence

== Theorem ==
Let $V$ be a Banach space.
Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.
Let $r \in \R_{\mathop \ge 0}$
Let the series $\ds \sum_{n \mathop = 1}^\infty v_n$ be absolutely convergent to $r$.
Then:
:the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$.
== Proof ==
Let $\epsilon \in \R_{\mathop \ge 0}$.
By definition of absolutely convergent:
:$(2) \quad \exists N \in \N : \forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_m} - r} < \dfrac \epsilon 3$
Let:
:$F = \closedint 0 N$
Let:
:$E \subseteq \N : E \supseteq F : E$ is finite.
Let:
:$m = \max \set{n : n \in E}$
Let:
:$G = \closedint 0 m$
We have:
:$F = \closedint 0 N \subseteq E \subseteq \closedint 0 m = G$
From Set Difference and Intersection form Partition:
:$E = F \cup E \setminus F$
and
:$G = F \cup G \setminus F$
From Set Difference Intersection with Second Set is Empty Set:
:$F \cap E \setminus F = \O$
and
:$F \cap G \setminus F = \O$
From Set Difference over Subset:
:$E \setminus F \subseteq G \setminus F$
We have:
Since $E$ was arbitrary, it follows:
:$\exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\leadsto \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$
Sine $\epsilon$ was arbitrary, it follows:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\leadsto \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$
From Characterization of Convergent Net in Metric Space:
:the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$.
Category:Absolute Net Convergence Equivalent to Absolute Convergence
Absolute Net Convergence Equivalent to Absolute Convergence/Absolute Net Convergence implies Absolute Convergence

== Theorem ==
Let $V$ be a Banach space.
Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.
Let $r \in \R_{\mathop \ge 0}$
Let the generalized sum $\ds \sum \set {v_n: n \in \N}$ be absolutely net convergent to $r$.
Then:
:the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$.
== Proof ==
Let $\epsilon \in \R_{\mathop > 0}$.
From Characterization of Convergent Net in Metric Space:
:$(1) \quad \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\implies \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$
Let $N = \max \set{n : v_n \in F}$.
We have:
:$F \subseteq \closedint 0 N$
Let $m \ge N$.
We have:
:$\closedint 0 m \supseteq \closedint 0 N \supseteq F$
From $(1)$:
:$\size{\ds \sum_{n \mathop \in \closedint 0 m} \norm{v_n} - r} < \epsilon$
By definition of summation over finite index:
:$\size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$
Since $m \ge N$ was arbitrary, it follows that:
:$\forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$
Since $\epsilon$ was arbitrary, it follows that the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$ by definition.
Category:Absolute Net Convergence Equivalent to Absolute Convergence
Absolute Value/Examples

== Examples of Absolute Values ==
=== Absolute Value of $3$ and $-3$ ===
=== Absolute Value of $-2$ ===
=== Absolute Value of $-6$ ===
=== Absolute Value of $\dfrac 3 4$ ===
=== Absolute Value of $3 - 5$ ===
=== Absolute Value of $x - a$ ===
=== Absolute Value $\size x \le 2$ ===
=== Absolute Value of $0$ ===
Category:Examples of Absolute Values
Absolute Value/Examples/-6

== Example of Absolute Value ==
:$\size {-6} = 6 = \size 6$
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers
* : $\S 1$: Real Numbers: $\S 1.14$: Modulus
* 
* 
Category:Examples of Absolute Values
Absolute Value/Examples/3

== Example of Absolute Value ==
:$\size 3 = 3 = \size {-3}$
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 4$: Limits of Functions (footnote $\dagger$)
* : Chapter $1$: Variables and Functions
* : $\S 1$: Real Numbers: $\S 1.14$: Modulus
Category:Examples of Absolute Values
Absolute Value Function is Completely Multiplicative

== Theorem ==
Let $x, y \in \R$ be real numbers.
Then:
:$\size {x y} = \size x \size y$
where $\size x$ denotes the absolute value of $x$.
Thus the absolute value function is completely multiplicative.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 1$: The Complex Numbers: Introduction: $1.1$: The Algebra $\R$ of Real Numbers
* : $\text {(i)}$
* : $\text {(i)}$
Category:Absolute Value Function is Completely Multiplicative
Category:Absolute Value Function
Category:Real Analysis
Category:Completely Multiplicative Functions
Absolute Value Function is Completely Multiplicative/Proof 1

== Theorem ==
== Proof ==
Let either $x = 0$ or $y = 0$, or both.
We have that $\size 0 = 0$ by definition of absolute value.
Hence:
:$\size x \size y = 0 = x y = \size {x y}$
Let $x > 0$ and $y > 0$.
Then:
and:
Let $x < 0$ and $y < 0$.
Then:
and:
The final case is where one of $x$ and $y$ is positive, and one is negative.
, let $x < 0$ and $y > 0$.
Then:
and:
The case where $x > 0$ and $y < 0$ is the same.
Category:Absolute Value Function is Completely Multiplicative
Absolute Value Function is Completely Multiplicative/Proof 2

== Theorem ==
== Proof ==
Let $x$ and $y$ be considered as complex numbers which are wholly real.
That is:
:$x = x + 0 i, y = y + 0 i$
From Complex Modulus of Real Number equals Absolute Value, the absolute value of $x$ and $y$ equal the complex moduli of $x + 0 i$ and $y + 0 i$.
Thus $\cmod x \cmod y$ can be interpreted as the complex modulus of $x$ multiplied by the complex modulus of $y$.
By Complex Modulus of Product of Complex Numbers:
:$\cmod x \cmod y = \cmod {x y}$
As $x$ and $y$ are both real, so is $x y$.
Thus by Complex Modulus of Real Number equals Absolute Value, $\cmod {x y}$ can be interpreted as the absolute value of $x y$ as well as its complex modulus.
Category:Absolute Value Function is Completely Multiplicative
Absolute Value Function is Continuous

== Theorem ==
Let $f$ be the real function defined as:
:$\forall x \in \R: \map f x = \size x$
where $\size x$ denotes the absolute value of $x$.
Then $f$ is a continuous real function.
== Proof ==
Let $a \in \R$.
Let $\epsilon \in \R_{\mathop > 0}$.
Let $\delta \le \epsilon$.
We have:
Since $\epsilon$ was arbitrary, it follows that $f$ is continuous at $a$ by definition.
Since $a$ was arbitrary, it follows that $f$ is continuous everywhere by definition.
Category:Absolute Value Function
Category:Continuous Real Functions
Absolute Value Function is Convex

== Theorem ==
Let $f: \R \to \R$ be the absolute value function on the real numbers.
Then $f$ is convex.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 12.15$ Example
Category:Absolute Value Function
Category:Convex Real Functions
Category:Absolute Value Function is Convex
Absolute Value Function is Convex/Proof 1

== Theorem ==
== Proof ==
Let $x_1, x_2, x_3 \in \R$ such that $x_1 < x_2 < x_3$.
Consider the expressions:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$
:$\dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
The following cases are investigated:
:$(1): \quad x_1, x_2, x_3 < 0$:
Then:
:$(2): \quad x_1, x_2, x_3 > 0$:
Then:
:$(3): \quad x_1 < 0, x_2, x_3 > 0$:
:$(4): \quad x_1, x_2 < 0, x_3 > 0$:
Thus for all cases, the condition for $f$ to be convex is fulfilled.
Hence the result.
Category:Absolute Value Function is Convex
Absolute Value Function is Even Function

== Theorem ==
Let $\size {\, \cdot \,} : \R \to \R$ denote the absolute value function on $\R$:
Then $\size {\, \cdot \,}$ is an even function.
== Proof ==
Recall the definition of the absolute value function:
:$\size x = \begin{cases}
x & : x > 0 \\
0 & : x = 0 \\
-x & : x < 0
\end{cases}$
Testing the $3$ cases in turn:
Hence the result by definition of even function.
Category:Absolute Value Function
Category:Examples of Even Functions
Absolute Value Function on Integers induces Equivalence Relation

== Theorem ==
Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is an equivalence relation.
== Proof ==
$\RR$ is shown to be an equivalence relation thus:
=== Reflexivity ===
:$\forall x \in \Z: \size x = \size x$
Thus $\RR$ is reflexive.
=== Symmetry ===
:$\forall x, y \in \Z: \size x = \size y \implies \size y = \size x$
Thus $\RR$ is symmetric.
=== Transitive ===
:$\forall x, y, z \in \Z: \size x = \size y \land \size y = \size z \implies \size x = \size z$
Thus $\RR$ is transitive.
Thus, by definition, $\RR$ is an equivalence relation.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.1$
Category:Examples of Equivalence Relations
Category:Integers
Category:Absolute Value Function
Absolute Value induces Equivalence Compatible with Integer Multiplication

== Theorem ==
Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \struct {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is a congruence relation for integer multiplication.
== Proof ==
From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation.
Let:
:$\size {x_1} = \size {x_2}$
:$\size {y_1} = \size {y_2}$
Then by Absolute Value Function is Completely Multiplicative:
That is:
:$\paren {x_1 y_1, x_2 y_2} \in \RR$
That is, $\RR$ is a congruence relation for integer multiplication.
== Also see ==
* Absolute Value induces Equivalence not Compatible with Integer Addition
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.1$
Category:Examples of Congruence Relations
Category:Integer Multiplication
Category:Absolute Value Function
Absolute Value induces Equivalence not Compatible with Integer Addition

== Theorem ==
Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is not a congruence relation for integer addition.
== Proof ==
From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation.
However, consider that:
By conventional integer addition:
:$-1 + 2 = 1$
while:
:$1 + 2 = 3$
But it does not hold that:
:$\size 1 = \size 3$
Therefore $\RR$ is not a congruence relation for integer addition.
Hence the result, by Proof by Counterexample.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.1$
Category:Examples of Congruence Relations
Category:Integer Addition
Category:Absolute Value Function
Absolute Value is Bounded Below by Zero

== Theorem ==
Let $x \in \R$ be a real number.
Then the absolute value $\size x$ of $x$ is bounded below by $0$.
== Proof ==
Let $x \ge 0$.
Then $\size x = x \ge 0$.
Let $x < 0$.
Then $\size x = -x > 0$.
The result follows.
== Notes ==
This result applies also to the set of integers $\Z$ and rational numbers $\Q$.
In that context the result is usually included as part of the field of number theory as well as that of analysis.
== Also see ==
* Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.14$: Modulus
Category:Absolute Value Function
Absolute Value is Many-to-One

== Theorem ==
Let $f: \R \to \R$ be the absolute value function:
:$\forall x \in \R: \map f x = \begin{cases}
x & : x \ge 0 \\
-x & : x < 0
\end{cases}$
Then $f$ is a many-to-one relation.
== Proof ==
 $f$ is not a many-to-one relation.
Then there exists $y_1 \in \R$ and $y_2 \in \R$ where $y_1 \ne y_2$ such that:
:$\exists x \in \R: \map f x = y_1, \map f x = y_2$
There are two possibilities:
Suppose $x \ge 0$.
Then:
That is:
:$y_2 = y_1 = x$
Suppose $x < 0$.
Then:
That is:
:$y_2 = y_1 = -x$
So by Proof by Cases we have that $y_1 = y_2$ whatever $x$ may be.
This contradicts our assertion that $y_1 \ne y_2$.
Hence it follows by Proof by Contradiction that $f$ is many-to-one.
== Sources ==
* : $\S 4$. Relations; functional relations; mappings: Exercise $1$
Category:Absolute Value Function
Absolute Value is Norm

== Theorem ==
The absolute value is a norm on the set of real numbers $\R$.
== Proof ==
By Complex Modulus is Norm then the complex modulus satisfies the norm axioms on the set of complex numbers $\C$.
Since the real numbers $\R$ is a subset of the complex numbers $\C$ then the complex modulus satisfies the norm axioms on the real numbers $\R$.
By Complex Modulus of Real Number equals Absolute Value then the absolute value satisfies the norm axioms on set of real numbers $\R$.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:Examples of Norms
Category:Absolute Value Function
Absolute Value of Absolutely Continuous Function is Absolutely Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an absolutely continuous function.
Then $\size f$ is absolutely continuous.
== Proof ==
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
By the Reverse Triangle Inequality, we have: 
:$\size {\map f {b_i} - \map f {a_i} } \ge \size {\size {\map f {b_i} } - \size {\map f {a_i} } }$
Therefore: 
:$\ds \sum_{i \mathop = 1}^n \size {\size {\map f {b_i} } - \size {\map f {a_i} } } \le \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
:$\size f$ is absolutely continuous.
Category:Absolute Value Function
Category:Absolutely Continuous Real Functions
Absolute Value of Absolutely Convergent Product is Absolutely Convergent

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {a_n}$ be a sequence in $\mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converge absolutely to $a \in \mathbb K$.
Then $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges absolutely to $\norm a$.
== Proof ==
By definition of absolute convergence of $\ds \prod_{n \mathop = 1}^\infty a_n$, $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - 1}$ converges absolutely.
By the Triangle Inequality:
:$\size {\norm {a_n} - 1} \le \norm {a_n - 1}$
By the Comparison Test, $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges absolutely.
By Absolute Value is Continuous, its limit is $\norm a$.
== Also see ==
* Absolute Value of Infinite Product, for related results
Category:Infinite Products
Absolute Value of Complex Cross Product is Commutative

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Let $z_1 \times z_2$ denote the (complex) cross product of $z_1$ and $z_2$.
Then:
:$\size {z_1 \times z_2} = \size {z_2 \times z_1}$
where $\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
Hence the result.
== Examples ==
Category:Absolute Value Function
Category:Complex Cross Product
Absolute Value of Complex Cross Product is Commutative/Examples

== Examples of Use of Absolute Value of Complex Cross Product is Commutative ==
=== Example: $\size {\paren {2 + 5 i} \times \paren {3 - i} } = \size {\paren {3 - i} \times \paren {2 + 5 i} }$ ===
Category:Examples of Use of Absolute Value of Complex Cross Product is Commutative
Absolute Value of Complex Cross Product is Commutative/Examples/2+5i cross 3-i

== Example of Use of Absolute Value of Complex Cross Product is Commutative ==
==== Example: $\size {\paren {2 + 5 i} \times \paren {3 - i} }$ ====
==== Example: $\size {\paren {3 - i} \times \paren {2 + 5 i} }$ ====
As can be seen:
:$\size {\paren {2 + 5 i} \times \paren {3 - i} } = \size {\paren {3 - i} \times \paren {2 + 5 i} }$
Category:Examples of Use of Absolute Value of Complex Cross Product is Commutative
Absolute Value of Complex Dot Product is Commutative

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Let $z_1 \circ z_2$ denote the (complex) dot product of $z_1$ and $z_2$.
Then:
:$\size {z_1 \circ z_2} = \size {z_2 \circ z_1}$
where $\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
From Dot Product Operator is Commutative‎:
:$z_1 \circ z_2 = z_2 \circ z_1$
The result follows trivially.
== Examples ==
Category:Absolute Value Function
Category:Complex Dot Product
Absolute Value of Complex Dot Product is Commutative/Examples

== Examples of Use of Absolute Value of Complex Dot Product is Commutative ==
=== Example: $\size {\paren {2 + 5 i} \circ \paren {3 - i} } = \size {\paren {3 - i} \circ \paren {2 + 5 i} }$ ===
Category:Examples of Use of Absolute Value of Complex Dot Product is Commutative
Absolute Value of Complex Dot Product is Commutative/Examples/2+5i dot 3-i

== Example of Use of Absolute Value of Complex Dot Product is Commutative ==
==== Example: $\size {\paren {2 + 5 i} \circ \paren {3 - i} }$ ====
==== Example: $\size {\paren {3 - i} \circ \paren {2 + 5 i} }$ ====
As can be seen:
:$\size {\paren {2 + 5 i} \circ \paren {3 - i} } = \size {\paren {3 - i} \circ \paren {2 + 5 i} }$
Category:Examples of Use of Absolute Value of Complex Dot Product is Commutative
Absolute Value of Components of Complex Number no greater than Root 2 of Modulus

== Theorem ==
Let $z = x + i y \in \C$ be a complex number.
Then:
:$\size x + \size y \le \sqrt 2 \cmod z$
where:
:$\size x$ and $\size y$ denote the absolute value of $x$ and $y$
:$\cmod z$ denotes the complex modulus of $z$.
== Proof ==
Let $z = x + i y \in \C$ be an arbitrary complex number.
 the contrary:
But as $\size x$ and $\size y$ are both real this cannot happen.
Thus our initial assumption $\size x + \size y > \sqrt 2 \cmod z$ is false.
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $138$
Category:Complex Modulus
Absolute Value of Continuous Real Function is Continuous

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $f$ be continuous at a point $a \in \R$.
Then:
:$\size f$ is continuous at $a$
where:
:$\map {\size f} x$ is defined as $\size {\map f x}$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a positive real number.
Because $f$ is continuous at $a$:
:$\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
From Reverse Triangle Inequality:
:$\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$
and so:
:$\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
$\epsilon$ is arbitrary, so:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
That is, by definition:
:$\size f$ is continuous at $a$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 17$
Category:Continuous Real Functions
Category:Absolute Value Function
Absolute Value of Convergent Infinite Product

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converge to $a \in \mathbb K$.
Then $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges to $\norm{a}$.
== Proof ==
By Absolute Value of Limit of Sequence:
:$\ds \prod_{n \mathop = 1}^\infty \norm {a_n} = \norm a$
It remains to show that the product converges.
By the convergence, there exists $n_0 \in \N$ such that $a_n \ne 0$ for $n \ge n_0$.
Then $\norm {a_n} \ne 0$ for $n \ge n_0$.
Let $P_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty a_n$.
Then $\norm {P_n}$ is the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty \norm {a_n}$.
Let $P_n$ converge to $b \in \mathbb K \setminus \set 0$.
By Convergence of Absolute Value of Sequence:
:$\norm {P_n} \to \norm b \in \mathbb K \setminus \set 0$.
Thus $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges.
== Also see ==
* Absolute Value of Infinite Product, for related results
Category:Infinite Products
Absolute Value of Cut is Greater Than or Equal To Zero Cut

== Definition ==
Let $\alpha$ be a cut.
Let $\size \alpha$ denote the absolute value of $\alpha$.
Then:
:$\size \alpha \ge 0^*$
where:
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
== Proof ==
Let $\alpha \ge 0^*$.
Then by definition $\size \alpha = \alpha \ge 0^*$.
Let $\alpha < 0^*$.
Then:
:$\exists \beta: \beta + \alpha = 0^*$
Thus:
:$\alpha = -\beta$
and it follows that $\beta > 0^*$.
The result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition
Category:Cuts
Category:Absolute Value Function