nyuuzyou/wikis huggingface dataset:
Negative Part of Composition of Functions

== Theorem ==
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $T : X \to X'$ be a $\Sigma/\Sigma'$-measurable mapping.
Let $f : X' \to \overline \R$ be a function.
Then: 
:$\paren {f \circ T}^- = f^- \circ T$
where $\paren {f \circ T}^-$ denotes the positive part of $f \circ T$.
== Proof ==
Let $x \in X$ be such that $\map {\paren {f \circ T} } x = \map f {\map T x} \le 0$.
Then $\map {f^-} {\map T x} = -\map f {\map T x}$ by the definition of the negative part. 
So:
Now let $x \in X$ be such that $\map {\paren {f \circ T} } x = \map f {\map T x} \ge 0$.
Then $\map {f^-} {\map T x} = 0$ and $\map {\paren {f \circ T}^-} x = 0$ by the definition of the positive part.
So: 
:$\map {\paren {f \circ T}^-} x = \map {\paren {f^- \circ T} } x$
for all $x \in X$.
Category:Negative Parts
Negative Part of Continuous Function is Continuous

== Theorem ==
Let $X$ be a topological space.
Let $f : X \to \R$ be continuous.
Let $f^-$ be the negative part of $f$. 
Then $f^-$ is continuous.
== Proof ==
From Constant Function is Continuous, we have that $\mathbf 0 : X \to \R$ defined by $\map {\mathbf 0} x = 0$ for each $x \in X$ is continuous.
From Minimum Rule for Continuous Real-Valued Functions, we have that the function $f \vee \mathbf 0 : X \to \R$ defined by:
:$\map {\paren {f \wedge \mathbf 0} } x = \min \set {0, \map f x}$ for each $x \in X$
is continuous.
From Multiple Rule for Continuous Real-Valued Function, $-\paren {f \wedge \mathbf 0}$ is continuous.
From the definition of the negative part, we have $f^- = -\paren {f \wedge \mathbf 0}$.
Hence $f^-$ is continuous.
== Also see ==
* Positive Part of Continuous Function is Continuous‎
Category:Negative Parts
Category:Continuous Real Functions
Negative Part of Function Bounded above by Absolute Value

== Theorem ==
Let $X$ be a set.
Let $f : X \to \C$ be a function.
Let $f^-$ be the negative part of $f$.
Then $f^- \le \cmod f$.
== Proof ==
By definition, we have $f^- \ge 0$. 
Let $f^+$ be the positive part of $f$. 
By definition, we also have $f^+ \ge 0$.
Hence $f^- \le f^+ + f^-$.
By Sum of Positive and Negative Parts, we have $f^+ + f^- = \cmod f$. 
Hence $0 \le f^- \le \cmod f$.
Category:Negative Parts
Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $y \in Y$. 
Then: 
:$\paren {f^y}^- = \paren {f^-}^y$
where:
:$f^y$ denotes the $y$-horizontal function of $f$
:$f^-$ denotes the negative part of $f$.
== Proof ==
Fix $y \in Y$.
Then, we have, for each $x \in X$:
Category:Horizontal Section of Functions
Category:Negative Parts
Negative Part of Multiple of Function

== Theorem ==
Let $X$ be a set.
Let $f : X \to \overline \R$ be an extended real-valued function.
Let $\alpha$ be a real number. 
Then:
:$\ds \paren {\alpha f}^- = \begin{cases}\alpha f^- & \alpha \ge 0 \\ -\alpha f^+ & \alpha < 0\end{cases}$
where:
:$\paren {\alpha f}^-$ and $f^-$ are the negative parts of $\alpha f$ and $f$ respectively
:$f^+$ is the positive part of $f$.
== Proof ==
Let $x \in X$. 
First take $\alpha \ge 0$.
Suppose that $\map f x \ge 0$.
Then $\alpha \map f x \ge 0$.
So:
:$-\min \set {\alpha \map f x, 0} = 0$
and:
:$-\min \set {\map f x, 0} = 0$
So by the definition of the positive part and negative part, we have:
:$\map {\paren {\alpha f}^-} x = 0$
and:
:$\map {f^-} x = 0$
So:
:$\map {\paren {\alpha f}^-} x = \alpha \map {f^-} x$
if $\map f x \ge 0$.
Now suppose that $\map f x < 0$.
Then $\alpha \map f x < 0$.
Then: 
:$-\min \set {\map f x, 0} = -\map f x$
and:
:$-\min \set {\alpha \map f x, 0} = -\alpha \map f x$ 
So by the definition of the positive part and negative part, we have:
:$\map {f^-} x = -\map f x$
and:
:$\map {\paren {\alpha f}^-} x = -\alpha \map f x$
So:
:$\map {\paren {\alpha f}^-} x = \alpha \map {f^-} x$
if $\map f x < 0$.
So: 
:$\paren {\alpha f}^- = \alpha f^-$
if $\alpha \ge 0$. 
Now take $\alpha < 0$. 
Suppose that $\map f x \ge 0$.
Then $\alpha \map f x \le 0$.
Then: 
:$-\min \set {\alpha \map f x, 0} = -\alpha \map f x$
and:
:$\max \set {\map f x, 0} = \map f x$
So by the definition of the positive part and negative part, we have:
:$\map {\paren {\alpha f}^-} x = -\alpha \map f x$
and:
:$\map {f^+} x = \map f x$
So:
:$\map {\paren {\alpha f}^-} x = -\alpha \map {f^+} x$
if $\map f x \ge 0$.
Now suppose that $\map f x < 0$.
Then $\alpha \map f x > 0$.
We then have: 
:$-\min \set {\alpha \map f x, 0} = 0$
and: 
:$\max \set {\map f x, 0} = 0$
So by the definition of the positive part and negative part, we have:
:$\map {\paren {\alpha f}^-} x = 0$
and:
:$\map {f^-} x = 0$
So:
:$\map {\paren {\alpha f}^-} x = -\alpha \map {f^+} x$
if $\map f x < 0$.
So: 
:$\paren {\alpha f}^- = -\alpha f^+$
if $\alpha \ge 0$. 
Category:Negative Parts
Negative Part of Pointwise Product of Functions

== Theorem ==
Let $X$ be a set. 
Let $f, g : X \to \overline \R$ be extended real-valued functions.
Then: 
:$\paren {f \cdot g}^- = f^- g^+ + f^+ g^-$
where: 
:$f \cdot g$ is the pointwise product of $f$ and $g$
:$\paren {f \cdot g}^-$ denotes the negative part.
== Proof ==
We have: 
Note that we have $\map f x \map g x \le 0$ $\map f x$ and $\map g x$ have the opposite sign.
That is, $\map f x \ge 0$ and $\map g x \le 0$ or $\map f x \le 0$ and $\map g x \ge 0$. 
In the first case, we have $\map {g^+} x = 0$ and $\map {f^-} x = 0$, so:
:$\map {f^+} x \map {g^+} x = \map {f^-} x \map {g^-} x = 0$
In the second case, we have $\map {f^+} x = 0$ and $\map {g^-} x = 0$, so: 
:$\map {f^+} x \map {g^+} x = \map {f^-} x \map {g^-} x = 0$
So, in either case we have: 
:$\map f x \map g x = -\paren {\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x}$
so since: 
:$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x \ge 0$
we obtain: 
:$\map {\paren {f \cdot g}^-} x = \map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x$
if $\map f x \map g x \le 0$. 
If $\map f x \map g x \ge 0$, $\map f x$ and $\map g x$ have the same sign. 
That is, either $\map f x \ge 0$ and $\map g x \ge 0$ or $\map f x \le 0$ and $\map g x \le 0$. 
In the first case, we have $\map {f^-} x = 0$ and $\map {g^-} x = 0$, so: 
:$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x = \map {\paren {f \cdot g}^-} x = 0$ 
In the second case, we have $\map {f^+} x = 0$ and $\map {g^+} x = 0$, so: 
:$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x = \map {\paren {f \cdot g}^-} x = 0$
and hence we are done.
Category:Negative Parts
Negative Part of Real-Valued Function Vanishing at Infinity Vanishes at Infinity

== Theorem ==
Let $X$ be a topological space. 
Let $f : X \to \R$ be a real-valued function vanishing at infinity.
Let $f^-$ be the negative part of $f$. 
Then $f^-$ vanishes at infinity.
== Proof ==
From Negative Part of Function Bounded above by Absolute Value, we have:
:$0 \le f^- \le \cmod f$
From Complex-Valued Function Dominated by Function Vanishing at Infinity also Vanishes at Infinity:
:$f^-$ vanishes at infinity.
Category:Negative Parts
Category:Complex-Valued Functions Vanishing at Infinity
Negative Part of Real-Valued Random Variable is Real-Valued Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a real-valued random variable. 
Then the negative part $X^-$ of $X$ is a real-valued random variable.
== Proof ==
Since $X$ is a real-valued random variable, $X$ is $\Sigma$-measurable.
From Function Measurable iff Positive and Negative Parts Measurable, $X^-$ is $\Sigma$-measurable.
So $X^-$ is a real-valued random variable.
Category:Random Variables
Category:Negative Parts
Negative Part of Simple Function is Simple Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $f^-: X \to \R$, the negative part of $f$ is also a simple function.
== Proof ==
Let $f$ have the following standard representation:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then we see that $f^-$ must satisfy:
:$f^- = \ds \sum_{i \mathop = 0}^n \min \set {a_i, 0} \chi_{E_i}$
as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise.
Since all of the $E_i$ are measurable, it follows that $f^+$ is a simple function.
== Also see ==
* Positive Part of Simple Function is Simple Function, a natural counterpart
== Sources ==
* : $8.7 \ \text{(v)}$
Category:Simple Functions
Category:Negative Parts
Negative Part of Vertical Section of Function is Vertical Section of Negative Part

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $x \in X$. 
Then: 
:$\paren {f_x}^- = \paren {f^-}_x$
where:
:$f_x$ denotes the $x$-vertical function of $f$
:$f^-$ denotes the negative part of $f$.
== Proof ==
Fix $x \in X$.
Then, we have:
Category:Vertical Section of Functions
Category:Negative Parts
Negative Slope indicates Line slopes Downward from Left to Right

== Theorem ==
Let $\LL$ be a straight line with a slope which is negative.
Then $\LL$ slopes downward from left to right.
== Proof ==
Let $\LL$ have a slope which is negative.
Expressed in slope-intercept form, $\LL$ can be written:
:$y = x \tan \psi + c$
where:
:$\psi$ is the angle between $\LL$ and the $x$-axis
:$c$ is the $y$-intercept.
:
By construction:
:$90 \degrees < \psi < 180 \degrees$
Hence by Shape of Tangent Function:
:$\tan \psi < 0$
Hence the result.
== Sources ==
* 
* 
Category:Equations of Straight Lines in Plane
Negative in Integral Domain is Unique

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain.
Let $a \in R$.
Then the negative $-a$ of $a$ is unique.
== Proof ==
From the definition of an integral domain, $\struct {D, +, \times}$ is a ring.
The result follows from Ring Negative is Unique.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $1 \ \text{(iii)}$
Category:Integral Domains
Negative of Absolute Value

== Theorem ==
Let $x \in \R$ be a real number.
Let $\size x$ denote the absolute value of $x$.
Then:
:$-\size x \le x \le \size x$
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Either $x \ge 0$ or $x < 0$.
:If $x \ge 0$, then:
::$-\size x \le 0 \le x = \size x$
:If $x < 0$, then:
::$-\size x = x < 0 < \size x$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.15$
Category:Negative of Absolute Value
Category:Absolute Value Function
Category:Inequalities
Negative of Absolute Value/Corollary 1

== Corollary to Negative of Absolute Value ==
Let $x, y \in \R$ be a real numbers.
Let $\size x$ be the absolute value of $x$.
Then:
:$\size x < y \iff -y < x < y$
== Proof ==
=== Necessary Condition ===
Let $\size x < y$.
Then from Negative of Absolute Value:
:$x \le \size x$
and:
:$\size x \ge -x$
So $x < y$ and $-x < y$, and so $x > -y$ from Ordering of Inverses in Ordered Monoid.
It follows that $-y < x < y$.
=== Sufficient Condition ===
Let $-y < x < y$.
Then $x < y$ and $-x < y$.
For all $x$:
:$\size x = x$
or:
:$\size x = -x$
Thus it follows that $\size x < y$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (1)$
Category:Negative of Absolute Value
Negative of Absolute Value/Corollary 2

== Corollary to Negative of Absolute Value ==
Let $x, y \in \R$ be a real numbers such that $y > 0$.
Let $\size x$ be the absolute value of $x$.
Then:
:$\size x \le y \iff -y \le x \le y$
that is:
:$\size x \le y \iff \begin {cases} x & \le y \\ -x & \le y \end {cases}$
== Proof ==
=== Necessary Condition ===
Let $\size x \le y$.
If $\size x < y$ then from Corollary 1:
:$-y < x < y$
Thus:
:$-y \le x \le y$
Otherwise, if $\size x = y$ then either $x = y$ or $-x = y$.
Hence the result.
=== Sufficient Condition ===
Let $-y \le x \le y$.
If $-y < x < y$ then from Corollary 1:
:$\size x < y$
Hence:
:$\size x \le y$
Otherwise, if either $-y = x$ or $x = y$ then:
:$\size x = y$
Hence the result.
== Sources ==
* : $\text {(iv)}$
* : $\text {(iv)}$
Category:Negative of Absolute Value
Negative of Complex Modulus

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$-\cmod z \le \cmod z$
where $\cmod z$ denotes the complex modulus of $z$.
The equality holds $z = 0$.
== Proof ==
Let $z = x + i y$.
By definition of complex modulus:
:$\cmod z = \sqrt {x^2 + y^2}$
and so:
:$\cmod z \ge 0$
Thus:
:$-\cmod z\le 0$
Hence the result.
== Also see ==
* Negative of Absolute Value for the same result on the real number line.
Category:Complex Modulus
Negative of Division Product

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Then:
:$\forall x \in R: -\dfrac x z = \dfrac {-x} z = \dfrac x {-z}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is the division product of $x$ by $z$.
== Proof ==
Follows directly from Product of Negative with Product Inverse and the definition of division product.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (1)$
Category:Commutative Algebra
Negative of Element in Field is Unique/Proof 2

== Theorem ==
== Proof ==
Let $b, c \in F$ such that both $a + b = 0$ and $a + c = 0$.
Thus:
But also:
So $b = c$ and the result follows.
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $1 \ \text{(iii)}$
Category:Negative of Element in Field is Unique
Negative of Field Negative

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ and let $-a$ be the field negative of $a$.
Then:
:$-\paren {-a} = a$
== Proof ==
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(i)}$
Category:Field Theory
Negative of Infimum is Supremum of Negatives

== Theorem ==
Let $T$ be a non-empty subset of the real numbers $\R$.
Let $T$ be bounded below.
Then:
:$(1): \quad \set {x \in \R: -x \in T}$ is bounded above
:$(2): \quad \ds -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x}$
where $\sup$ and $\inf$ denote the supremum and infimum respectively.
== Proof ==
As $T$ is non-empty and bounded below, it follows by the Continuum Property that $T$ admits an infimum.
Let $B = \inf T$.
Let $S = \set {x \in \R: -x \in T}$.
Since $\forall x \in T: x \ge B$ it follows that:
:$\forall x \in T: -x \le -B$
Hence $-B$ is an upper bound for $S$, and so $\set {x \in \R: -x \in T}$ is bounded above.
Let $C$ be the supremum of $S$.
Then by definition of supremum:
:$C \le -B$
On the other hand:
:$\forall y \in S: y \le C$
Therefore:
:$\forall y \in S: -y \ge -C$
Since $T = \set {x \in \R: -x \in S}$ it follows that $-C$ is a lower bound for $T$.
Therefore $-C \le B$ and so $C \ge -B$.
So $C \ge -B$ and $C \le -B$ and the result follows.
== Also see ==
* Negative of Supremum is Infimum of Negatives
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (3)$
Category:Real Analysis
Negative of Integer

== Theorem ==
Let $x \in \Z$ be an integer.
Let $x = \eqclass {a, b} {}$ be defined from the formal definition of integers, where $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Then:
:$-x = \eqclass {b, a} {}$
== Proof ==
Follows from Inverse for Integer Addition.
Category:Integers
Category:Examples of Inverse Elements
Negative of Logarithm of x plus Root x squared minus a squared

== Theorem ==
Let $x \in \R: \size x > 1$.
Let $x > 1$.
Then:
:$-\map \ln {x + \sqrt {x^2 - a^2} } = \map \ln {x - \sqrt {x^2 - a^2} } - \map \ln {a^2}$
== Proof ==
First we note that if $x > 1$ then $x + \sqrt {x^2 - a^2} > 0$.
Hence $\map \ln {x + \sqrt {x^2 - a^2} }$ is defined.
Then we have:
== Also see ==
* Negative of Logarithm of x plus Root x squared plus a squared
Category:Logarithms
Negative of Logarithm of x plus Root x squared plus a squared

== Theorem ==
Let $x \in \R$ be a real number.
Then:
:$-\map \ln {x + \sqrt {x^2 + a^2} } = \map \ln {-x + \sqrt {x^2 + a^2} } - \map \ln {a^2}$
=== Corollary ===
== Proof ==
We have that $\sqrt {x^2 + a^2} > x$ for all $x$.
Thus:
:$x + \sqrt {x^2 + a^2} > 0$
and so $\map \ln {x + \sqrt {x^2 + a^2} }$ is defined for all $x$.
Then we have:
== Also see ==
* Negative of Logarithm of x plus Root x squared minus a squared
Category:Logarithms
Negative of Logarithm of x plus Root x squared plus a squared/Corollary

== Theorem ==
Let $x \in \R: \size x > 1$.
Let $x > 1$.
Then:
:$-\map \ln {x + \sqrt {x^2 + a^2} } = \ln \size {x - \sqrt {x^2 + a^2} } - \map \ln {a^2}$
== Proof ==
We have that $\sqrt {x^2 + a^2} > x$ for all $x$.
Hence for all $x$:
:$-x + \sqrt {x^2 + a^2} > 0$
and so:
:$x - \sqrt {x^2 + a^2} < 0$
Hence:
:$-x + \sqrt {x^2 + a^2} = \size {x - \sqrt {x^2 + a^2} }$
Then we have:
== Also see ==
* Negative of Logarithm of x plus Root x squared plus a squared
Category:Logarithms
Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives

== Theorem ==
Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$B$ is a lower bound of $S$
:
:$-B$ is an upper bound of $T$.
== Proof ==
Let $B$ be a lower bound of $S$.
That is:
:$\forall x \in S: x \ge B$
Let $x \in T$ be arbitrary.
As $x$ is arbitrary it follows that:
:$\forall x \in T: x \le -B$
That is, $-B$ is an upper bound for $T$.
=== Necessary Condition ===
Let $U$ be an upper bound for $T$.
:$\forall x \in T: x \le U$
Let $x \in S$ be arbitrary.
As $x$ is arbitrary it follows that:
:$\forall x \in S: x \ge -U$
That is, $-U$ is a lower bound for $S$.
== Also see ==
* Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives
Category:Real Analysis
Negative of Natural Logarithm of Euler-Mascheroni Constant

== Example of Natural Logarithm ==
The negative of the natural logarithm of the Euler-Mascheroni constant $\gamma$ is approximately:
:$-\ln \gamma \approx 0 \cdotp 54953 \, 93129 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $0,54953 93129 \ldots$
Category:Examples of Natural Logarithms
Category:Euler-Mascheroni Constant
Negative of Product Inverse

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
where $z^{-1}$ is the ring product inverse of $z$.
== Proof ==
Let the unity of $\struct {R, +, \circ}$ be $1_R$.
Thus:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (1)$
Category:Rings with Unity
Negative of Quotient of Real Numbers

== Theorem ==
:$\forall x \in \R, y \in \R_{\ne 0}: \dfrac {-x} y = -\dfrac x y = \dfrac x {-y}$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(t)}$
Category:Real Division
Negative of Real Function that Decreases Without Bound

== Theorem ==
Let $f: \R \to \R$ be a real function.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} -\map f x = +\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} -\map f x = +\infty$
== Proof ==
Suppose $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty$.
Then by the definition of negative infinite limit at infinity:
:$\forall M < 0: \exists N > 0: x > N \implies \map f x < M$
But:
:$M < 0 \iff -M > 0$
Likewise:
:$\map f x < M \iff -\map f x > -M$
Putting $M' = -M$:
:$\forall M' > 0: \exists N > 0: x > N \implies -\map f x > M'$
The result then follows from the definition of infinite limit at infinity.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous.
== Also see ==
* Negative of Real Function that Increases Without Bound
Category:Unbounded Mappings
Negative of Real Function that Increases Without Bound

== Theorem ==
Let $f: \R \to \R$ be a real function.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} -\map f x = -\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} -\map f x = -\infty$
== Proof ==
Suppose $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
Then by the definition of infinite limits at infinity:
:$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$
But $M > 0 \iff -M < 0$.
Likewise $\map f x > M \iff -\map f x < -M$.
Putting $M' = -M$:
:$\forall M' < 0: \exists N > 0: x > N \implies -\map f x < M'$
The result then follows from the definition of negative infinite limit at infinity.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analagous.
== Also see ==
* Negative of Real Function that Decreases Without Bound
Category:Unbounded Mappings
Negative of Ring Negative

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$ and let $-a$ be the ring negative of $a$.
Then:
:$-\paren {-a} = a$
== Proof ==
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse of Group Inverse.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(i)}$
Category:Ring Theory
Negative of Subring is Negative of Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
For each $x \in R$ let $-x$ denote the ring negative of $x$ in $R$.
Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.
For each $x \in S$ let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.
Then:
:$\forall x \in S: \mathbin \sim x = -x$
== Proof ==
Let $i_S: S \to R$ be the inclusion mapping from $S$ to $R$.
By Inclusion Mapping on Subring is Homomorphism, $i_S$ is a ring homomorphism.
By Ring Homomorphism of Addition is Group Homomorphism, $i_S$ is a group homomorphism on ring addition $+$.
Let $x \in S$.
Then:
Category:Ring Theory
Negative of Sum of Real Numbers

== Theorem ==
:$\forall x, y \in \R: -\paren {x + y} = -x - y$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(h)}$
Category:Real Addition
Category:Real Subtraction
Negative of Sum of Real Numbers/Corollary

== Theorem ==
:$\forall x, y \in \R: -\paren {x - y} = -x + y$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(h)}$
Category:Real Addition
Category:Real Subtraction
Negative of Supremum is Infimum of Negatives

== Theorem ==
Let $S$ be a non-empty subset of the real numbers $\R$.
Let $S$ be bounded above.
Then:
:$(1): \quad \set {x \in \R: -x \in S}$ is bounded below
:$(2): \quad \ds -\sup_{x \mathop \in S} x = \map {\inf_{x \mathop \in S} } {-x}$
where $\sup$ and $\inf$ denote the supremum and infimum respectively.
== Proof ==
As $S$ is non-empty and bounded above, it follows by the Continuum Property that $S$ admits a supremum.
Let $B = \sup S$.
Let $T = \set {x \in \R: -x \in S}$.
By definition, $B$ is an upper bound for $S$.
From Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives:
:$-B$ is a lower bound for $T$.
Thus $\set {x \in \R: -x \in S}$ is bounded below.
Let $C$ be the infimum of $T$.
Then by definition of infimum:
:$C \ge -B$
On the other hand:
:$\forall y \in T: y \ge C$
Therefore:
:$\forall y \in T: -y \le -C$
Since $S = \set {x \in \R: -x \in T}$ it follows that $-C$ is an upper bound for $S$.
Therefore $-C \ge B$ and so $C \le -B$.
So $C \le -B$ and $C \ge -B$ and the result follows.
== Also see ==
* Negative of Infimum is Supremum of Negatives
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (3)$
Category:Real Analysis