nyuuzyou/wikis huggingface dataset:
Significant Figures/Examples/149.80 mm

== Example of Significant Figures ==
:$149 \cdotp 80 \ \mathrm {mm}$ has $5$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(b)}$
Category:Examples of Significant Figures
Significant Figures/Examples/149.8 mm

== Example of Significant Figures ==
:$149 \cdotp 8 \ \mathrm {mm}$ has $4$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(a)}$
Category:Examples of Significant Figures
Significant Figures/Examples/4.0 x 10^3 g

== Example of Significant Figures ==
:$4 \cdotp 0 \times 10^3 \ \mathrm g$ has $2$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(h)}$
Category:Examples of Significant Figures
Significant Figures/Examples/4.5300

== Example of Significant Figures ==
:$4 \cdotp 5300$ has $5$ significant figures.
Beware: It may appear that there are only $3$ significant figures.
However, although $3$ of the digits given are non-zero, the final $2$ zerodigits are significant, because they are specifically stated.
== Sources ==
* : Chapter $1$: Significant Figures: Example 2.
Category:Examples of Significant Figures
Significant Figures/Examples/7.58400 x 10^-5 N

== Example of Significant Figures ==
:$7 \cdotp 584 \, 00 \times 10^5 \ \mathrm N$ has $6$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(i)}$
Category:Examples of Significant Figures
Significant Figures/Examples/Populations of Cities

== Example of Significant Figures ==
Suppose the populations of cities are being quoted to the nearest thousand.
The populations of $3$ cities are given as:
:$1 \, 702 \, 000$
:$814 \, 000$
:$70 \, 000$
Hence:
:the first is reported to $4$ significant figures
:the second is reported to $3$ significant figures
:the third is reported to $2$ significant figures.
== Sources ==
* 
* 
Category:Examples of Significant Figures
Signum Complement Function is Primitive Recursive

== Theorem ==
Let $\overline {\operatorname{sgn}}: \N \to \N$ by defined as the signum-bar function.
Then $\overline {\operatorname{sgn}}$ is primitive recursive.
== Proof ==
From Signum Complement Function on Natural Numbers as Characteristic Function, $\map {\overline {\operatorname{sgn} } } n = \chi_{\set 0} n$.
From Set Containing Only Zero is Primitive Recursive, $\chi_{\set 0}$ is primitive recursive.
Hence the result.
Category:Primitive Recursive Functions
Signum Complement Function on Natural Numbers as Characteristic Function

== Definition ==
<onlyinclude>
Let $\overline {\operatorname{sgn}}: \N \to \N$ be the signum complement function on the natural numbers.
Let $\chi_{\left\{{0}\right\}}: \N \to \left\{{0, 1}\right\}$ be the characteristic function of $\left\{{0}\right\}$.
Let $n \in \N$.
Then:
:$\overline {\operatorname{sgn}} \left({n}\right) = \chi_{\left\{{0}\right\}} \left({n}\right)$
== Proof ==
The signum complement function $\overline {\operatorname{sgn}}: \N \to \N$ is defined as:
:$\forall n \in \N: \operatorname{sgn} \left({n}\right) = \begin{cases}
1 & : n = 0 \\
0 & : n > 0
\end{cases}$
The result follows by definition of the characteristic function.
Category:Number Theory
Signum Function is Completely Multiplicative

== Theorem ==
The signum function on the set of real numbers is a completely multiplicative function:
:$\forall x, y \in \R: \map \sgn {x y} = \map \sgn x \map \sgn y$
== Proof ==
Let $x = 0$ or $y = 0$.
Then:
and either $\map \sgn x = 0$ or $\map \sgn y = 0$ and so:
Let $x > 0$ and $y > 0$.
Then:
and:
Let $x < 0$ and $y < 0$.
Then:
and:
Let $x < 0$ and $y > 0$.
Then:
and:
The same argument, mutatis mutandis, covers the case where $x > 0$ and $y < 0$.
Category:Signum Function
Category:Completely Multiplicative Functions
Signum Function is Primitive Recursive

== Theorem ==
Let $\operatorname{sgn}: \N \to \N$ be defined as the signum function.
Then $\operatorname{sgn}$ is primitive recursive.
=== Signum Complement ===
== Proof ==
By Signum Function on Natural Numbers as Characteristic Function, $\map {\operatorname{sgn} } n = \chi_{\N^*}$, where $\N^* = \N \setminus \set 0$.
By Set of Non-Zero Natural Numbers is Primitive Recursive, $\N^*$ is primitive recursive.
Thus $\operatorname{sgn}$ is primitive recursive by definition of Primitive Recursive Set.
Category:Primitive Recursive Functions
Signum Function is Quotient of Number with Absolute Value

== Theorem ==
Let $x \in \R_{\ne 0}$ be a non-zero real number.
Then:
:$\map \sgn x = \dfrac x {\size x} = \dfrac {\size x} x$
where:
:$\map \sgn x$ denotes the signum function of $x$
:$\size x$ denotes the absolute value of $x$.
== Proof ==
Let $x \in \R_{\ne 0}$.
Then either $x > 0$ or $x < 0$.
Let $x > 0$.
Then:
Similarly:
Let $x < 0$.
Then:
Similarly:
Category:Signum Function
Category:Absolute Value Function
Signum Function on Integers is Extension of Signum on Natural Numbers

== Theorem ==
Let $\sgn_\Z: \Z \to \set {-1, 0, 1}$ be the signum function on the integers.
Let $\sgn_\N: \N \to \set {0, 1}$ be the signum function on the natural numbers.
Then $\sgn_\Z: \Z \to \Z$ is an extension of $\sgn_\N: \N \to \N$.
== Proof ==
Let $n \in \Z: n \ge 0$.
Then by definition of the signum function:
:$\map {\sgn_\Z} n = \begin {cases} 0 & : n = 0 \\ 1 & : n > 0 \end {cases}$
So by definition of the signum function on the natural numbers:
:$\forall n \in \N: \map {\sgn_\Z} n = \map {\sgn_\N} n$
Hence the result, by definition of extension.
Category:Number Theory
Signum Function on Integers is Primitive Recursive

== Theorem ==
Let $\sgn_\Z : \Z \to \set {-1, 0, 1}$ denote the signum function on the integers.
Let $s : \N \to \N$ be defined as:
:$\map s n = m$
where:
:$n$ codes the integer $k$
:$m$ codes the integer $\map {\sgn_\Z} k$
Then $s$ is a primitive recursive function.
== Proof ==
Let $s : \N \to \N$ be defined as:
:$\map s n = \begin{cases}
1 & : k > 0 \\
2 & : k < 0 \\
0 & : \text{otherwise}
\end{cases}$
By:
* Set of Strictly Positive Integers is Primitive Recursive
* Set of Strictly Negative Integers is Primitive Recursive
we have that:
:$k > 0 \iff n \in P$
:$k < 0 \iff n \in N$
are primitive recursive relations.
Thus, by:
* Constant Function is Primitive Recursive
* Definition by Cases is Primitive Recursive/Corollary
the function $s$ defined above is primitive recursive.
First, suppose $k > 0$.
Then, we have:
:$\map {\sgn_\Z} k = 1$
As $1 > 0$, by definition, $m$ codes $1$ :
:$m = 2 \paren 1 - 1 = 1$
But $\map s n = 1$ in this case.
Next, suppose $k < 0$.
Then, we have:
:$\map {\sgn_\Z} k = -1$
As $-1 \le 0$, by definition, $m$ codes $-1$ :
:$m = -2 \paren {-1} = 2$
But $\map s n = 2$ in this case.
Lastly, suppose $k = 0$.
Then, we have:
:$\map {\sgn_\Z} k = 0$
As $0 \le 0$, by definition, $m$ codes $0$ :
:$m = -2 \paren 0 = 0$
But $\map s n = 0$ in this case.
In every case, $s$ satisfies the requirements.
Category:Primitive Recursive Functions
Signum Function on Natural Numbers as Characteristic Function

== Definition ==
Let $\sgn: \N \to \N$ be the signum function on the natural numbers.
Let $\chi_{\N_{>0} }: \N \to \set {0, 1}$ be the characteristic function of $\N_{>0}$, where $\N_{>0} = \N \setminus \set 0$.
Let $n \in \N$.
Then:
: $\sgn \paren n = \chi_{\N_{>0} } \paren n$
== Proof ==
The signum function $\sgn: \N \to \N$ is defined as:
:$\forall n \in \N: \sgn \paren n = \begin{cases}
0 & : n = 0 \\
1 & : n > 0
\end{cases}$
The result follows by definition of the characteristic function.
Category:Number Theory
Similar Figures on Proportional Straight Lines

== Theorem ==
== Proof ==
Let the four straight lines $AB, CD, EF, GH$ be proportional.
That is:
: $AB : CD = EF : GH$
Let there be described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$.
Let there be described on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$.
We need to show that:
: $KAB : LCD = MF : NH$
By Construction of Third Proportional Straight Line‎, let there be taken a third proportional $O$ to $AB, CD$, and a third proportional $P$ to $EF, GH$.
We have that:
: $AB : CD = EF : GH$, and $CD : O = GH : P$
So from Equality of Ratios Ex Aequali:
: $AB : O = EF : P$
But from :
: $AB : O = KAP : LCD$
and:
: $EF : P = MF : NH$
Therefore from Equality of Ratios is Transitive:
: $KAB : LCD = MF : NH$
:
Now suppose $MF : NH = KAB : LCD$.
We need to show that:
: $AB : CD = EF : GH$
Suppose, with a view to obtaining a contradiction, that:
: $EF : GH \ne AB : CD$
Instead, by Construction of Fourth Proportional Straight Line, let:
: $EF : QR = AB : CD$
On $QR$, by Construction of Similar Polygon, let the rectilineal figures $SR$ be described similar and similarly situated to either $MF$ or $NH$.
We have that:
: $AB : CD = EF : QR$
We also have that there have been described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$
We also have that there have been on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$.
From Equality of Ratios is Transitive:
: $MF : SF = MF : NH$
Therefore $MF$ has the same ratio to each of the figures $NH, SR$.
Therefore, by Magnitudes with Same Ratios are Equal:
: $NH = SR$
But it is also similar and similarly situated to it.
Therefore:
: $GH = QR$
But we have that $AB : CD = EF : QR$ while $QR = GH$.
Therefore:
: $AB : CD = EF : GH$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Polygons
Similar Figures on Sides of Right-Angled Triangle

== Theorem ==
== Proof ==
Let $\triangle ABC$ be a right-angled triangle where $\angle BAC$ is a right angle.
We need to show that the area of the figure on $BC$ is equal to the sum of the areas of the similar figures on $AB$ and $AC$.
:
Let $AD$ be drawn perpendicular to $BC$.
From Perpendicular in Right-Angled Triangle makes two Similar Triangles, $\triangle ABD$ and $\triangle ADC$ are similar both to each other and to $\triangle ABC$.
So from , $CB : BA = AB : BD$.
So from the porism to Ratio of Areas of Similar Triangles, the figures on each of these sides are proportional in the same way.
So $BC : BD$ is the same as the ratio of the areas of the figures on $CB$ and $BA$.
Similarly, $BC : CD$ is the same as the ratio of the areas of the figures on $CB$ and $CA$.
So $BC : BD + DC$ equals the ratio of the areas of $BC$ and the sum of the areas of the figures on $CA$ and $AB$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Triangles
Similar Matrices are Equivalent

== Theorem ==
If two square matrices over a ring with unity $R$ are similar, then they are equivalent.
That is:
:every equivalence class for the similarity relation on $\map {\MM_R} n$ is contained in an equivalence class for the relation of matrix equivalence.
where $\map {\MM_R} n$ denotes the $n \times n$ matrix space over $R$.
== Proof ==
If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
Let $\mathbf Q = \mathbf P$.
Then $\mathbf A$ are equivalent to $\mathbf B$, as:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Algebra
Similar Matrices have same Traces

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be square matrices of order $n$.
Let $\mathbf A$ and $\mathbf B$ be similar.
Then:
:$\map \tr {\mathbf A} = \map \tr {\mathbf B}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.
== Proof ==
By definition of similar matrices:
:$\exists \mathbf P: \mathbf P^{-1} \mathbf A \mathbf P = \mathbf B$
where $\mathbf P$ is an nonsingular matrix of order $n$.
Therefore:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
Category:Matrix Similarity
Category:Traces of Matrices
Similar Parallelogram on Half a Straight Line

== Theorem ==
== Proof ==
Let $AB$ be a straight line and let it be bisected at $C$.
Let the parallelogram $\Box AD$ be applied on $AB$ deficient by the parallelogramic figure $\Box DB$ described on $AC$.
We need to show that of all the parallelograms applied to $AB$ and deficient by parallelogramic figures similar and similarly situated to $\Box DB$, $\Box AD$ is the greatest.
:
Let $\Box AF$ be applied to $AB$ deficient by $\Box FB$ similar and similarly situated to $\Box DB$.
We have that $\Box DB$ is similar to $\Box FB$.
So from Parallelogram Similar and in Same Angle has Same Diameter they are about the same diameter.
Let that diameter $DB$ be drawn, and let that figure be described.
From Complements of Parallelograms are Equal:
: $\Box CF = \Box FE$
As $\Box FB$ is common:
: $\Box CH = \Box KE$
But from Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box CH = \Box CG$ since $AC = CB$
Therefore:
: $\Box GC = \Box EK$
Add $\Box CF$ to each.
Therefore $\Box AF$ is equal in area to the gnomon $LMN$.
So $\Box DB$, that is $\Box AD$, is greater than $\Box AF$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Similar Plane Numbers have Same Ratio as between Two Squares

== Theorem ==
== Proof ==
Let $a$ and $b$ be similar plane numbers.
From Between two Similar Plane Numbers exists one Mean Proportional, there exists a mean proportional $m$ between them.
By definition of mean proportional:
:$\left({a, m, b}\right)$
is a geometric sequence.
From Form of Geometric Sequence of Integers:
:$\exists k, p, q \in Z: a = k p^2, b = k q^2, m = k p q$
Thus:
:$\dfrac a b = \dfrac {k p^2} {k q^2} = \dfrac {p^2} {q^2}$
Hence the result.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Similar Polygons are composed of Similar Triangles

== Theorem ==
== Proof ==
Let $ABCDE$ and $FGHKL$ be similar polygons such that $AB$ corresponds to $FG$.
We need to show that $ABCDE$ and $FGHKL$ are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes.
Also that the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$.
:
Join up $BE, EC, GL, LH$.
Since $ABCDE$ and $FGHKL$ are similar:
:$\angle BAE = \angle GFL$
From :
:$BA : AE = GF : FL$
Thus from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle ABE$ is similar to $\triangle FGL$.
So $\angle ABE = \angle FGL$.
But $\angle ABC = \angle FGH$ because $ABCDE$ and $FGHKL$ are similar.
So:
:$\angle EBC = \angle LGH$
Because $\triangle ABE$ is similar to $\triangle FGL$:
:$EB : BA = LG : GF$
Also, because $ABCDE$ and $FGHKL$ are similar:
:$AB : BC = FG : GH$
So from Equality of Ratios Ex Aequali:
:$EB : BC = LG : GH$
So from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle EBC$ is similar to $\triangle LGH$.
For the same reason, $\triangle ECD$ is similar to $\triangle LHK$.
So $ABCDE$ and $FGHKL$ have been divided into similar triangles, and into triangles equal in multitude.
Now let $AC, FH$ be joined.
Because $ABCDE$ and $FGHKL$ are similar:
:$\angle ABC = \angle FGH$
From Triangles with One Equal Angle and Two Sides Proportional are Similar $\triangle ABC$ is similar to $\triangle FGH$.
Therefore $\angle BAC = \angle GFH$ and $\angle BCA = \angle GHF$.
Also, we have that $\angle BAM = \angle GFN$, and $\angle ABM = \angle FGN$.
So from Sum of Angles of Triangle Equals Two Right Angles $\angle AMB = \angle FGN$ and so $\triangle ABM$ is similar to $\triangle FGN$.
Similarly we can show that $\triangle BMC$ is similar to $\triangle GNH$.
Therefore $AM : MB = FN : NG$ and $BM : MC = FN : NH$.
But from Areas of Triangles and Parallelograms Proportional to Base:
:$AM : MC = \triangle ABM : \triangle MBC$.
So from Sum of Components of Equal Ratios:
:$\triangle ABM : \triangle MBC = \triangle ABE : \triangle CBE$
But:
:$\triangle ABM : \triangle MBC = AM : MC$
So:
:$\triangle ABE : \triangle CBE = AM : MC$
For the same reason:
:$FN : NH = \triangle FGL : \triangle GLH$
As $AM : MC = FN : NH$, it follows that:
:$\triangle ABE : \triangle BEC = \triangle FGL : \triangle GLH$
Similarly:
:$\triangle ABE : \triangle FGL = \triangle BEC : \triangle GLH$
We now join $BD$ and $GK$, and by a similar construction show that:
:$\triangle BEC : \triangle LGH = \triangle ECD : \triangle LHK$
From Sum of Components of Equal Ratios:
:$ABE : FGL = ABCDE : FGHKL$
But from Ratio of Areas of Similar Triangles $\triangle ABE$ has to $\triangle FGL$ a ratio duplicate of $AB : FG$.
Therefore the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$.
=== Porism ===
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Similar Triangles
Category:Polygons
Similar Segments on Equal Bases are Equal

== Theorem ==
== Proof ==
:
Let $AEB$ and $CFD$ be similar segments of circles on equal bases $AB$ and $CD$.
Let the segment $AEB$ be applied to $CFD$ so that $A$ be placed on $C$ and $AB$ on $CD$.
Then $B$ will coincide with $D$ as $AB = CD$.
Suppose that segment $AEB$ does not coincide with segment $CFD$.
It will fall in one of three ways:
:$(1) \quad$ Inside it
:$(2) \quad$ Outside it
:$(3) \quad$ Awry, as $CGD$.
If $CFD$ falls inside or outside $AEB$, then by definition $AEB$ and $CFD$ are not similar.
But from Two Circles have at most Two Points of Intersection option $(3)$ is impossible.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Similar Solid Numbers have Same Ratio as between Two Cubes

== Theorem ==
== Proof ==
Let $a$ and $b$ be similar solid numbers.
From Between two Similar Solid Numbers exist two Mean Proportionals, there exists two mean proportionals $m_1$ and $m_2$ between them.
By definition of mean proportional:
:$\left({a, m_1, m_2, b}\right)$
is a geometric sequence.
From Form of Geometric Sequence of Integers:
:$\exists k, p, q \in Z: a = k p^3, b = k q^3, m_1 = k p^2 q, m_2 = k p q^2$
Thus:
:$\dfrac a b = \dfrac {k p^3} {k q^3} = \dfrac {p^3} {q^3}$
Hence the result.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Similarity Dimension/Examples/Koch Snowflake

== Example of Similarity Dimension ==
The similarity dimension $D$ of the Koch snowflake is given by:
:$D = \dfrac {\ln 4} {\ln 3}$
== Proof ==
See Fractal Dimension of Koch Snowflake.
== Sources ==
* 
* 
* 
* 
Category:Examples of Similarity Dimensions
Similarity Dimension and Hausdorff-Besicovitch Dimension Coincide for Self-Similar Set

== Theorem ==
Let $S$ be a set which has a self-similarity.
Then the similarity dimension and the Hausdorff-Besicovitch dimension of $S$ are the same.
== Proof ==
== Sources ==
* 
* 
Category:Hausdorff-Besicovitch Dimension
Similarity Mapping is Automorphism

== Theorem ==
Let $G$ be a vector space over a field $\struct {K, +, \times}$.
Let $\beta \in K$.
Let $s_\beta: G \to G$ be the similarity on $G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
If $\beta \ne 0$ then $s_\beta$ is an automorphism of $G$.
== Proof ==
By definition, a vector space automorphism on $G$ is a vector space isomorphism from $G$ to $G$ itself.
To prove that $s_\beta$ is a automorphism it is sufficient to demonstrate that:
By definition, a vector space isomorphism is a mapping $s_\beta: G \to G$ such that:
:$(1): \quad s_\beta$ is a bijection
:$(2): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
:$(3): \quad \forall \mathbf x \in G: \forall \lambda \in K: \map {s_\beta} {\lambda \mathbf x} = \lambda \map {s_\beta} {\mathbf x}$
It has been established in Similarity Mapping is Linear Operator that $s_\beta$ is a linear operator on $G$.
Hence $(2)$ and $(3)$ follow by definition of linear operator.
It remains to prove bijectivity.
That is, that $s_\beta$ is both injective and surjective.
Let $1_K$ denote the multiplicative identity of $K$.
We have:
Hence it has been demonstrated that $s_\beta$ is injective.
Let $\mathbf y \in G$.
Consider $\beta^{-1} \in K$ defined such that $\beta \beta^{-1} = 1_K$.
By , $\beta^{-1}$ always exists.
Then:
Hence it has been demonstrated that $s_\beta$ is surjective.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Category:Automorphisms (Abstract Algebra)
Similarity Mapping is Linear Operator

== Theorem ==
Let $G$ be a vector space over a field $\struct {K, + \times}$.
Let $\beta \in K$.
Then the similarity $s_\beta: G \to G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
is a linear operator on $G$.
== Proof ==
To prove that $s_\beta$ is a linear operator it is sufficient to demonstrate that:
:$(1): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
:$(2): \quad \forall \mathbf x \in G: \forall \lambda \in K: \map {s_\beta} {\lambda \mathbf x} = \lambda \map {s_\beta} {\mathbf x}$
Indeed:
and:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Category:Linear Operators
Similarity Mapping on Plane Commutes with Half Turn about Origin

== Theorem ==
Let $\beta \in \R_{>0}$ be a (strictly) positive real number.
Let $s_{-\beta}: \R^2 \to \R^2$ be the similarity mapping on $\R^2$ whose scale factor is $-\beta$.
Then $s_{-\beta}$ is the same as:
:a stretching or contraction of scale factor $\beta$ followed by a rotation one half turn
and:
:a rotation one half turn followed by a stretching or contraction of scale factor $\beta$.
== Proof ==
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
From Similarity Mapping on Plane with Negative Parameter, $s_{-\beta}$ is a stretching or contraction of scale factor $\beta$ followed by a rotation one half turn.
Thus:
That is:
:$s_\beta$ is a rotation one half turn followed by a stretching or contraction of scale factor $\beta$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Category:Geometric Rotations
Similarity Mapping on Plane with Negative Parameter

== Theorem ==
Let $\beta \in \R_{<0}$ be a (strictly) negative real number.
Let $s_\beta: \R^2 \to \R^2$ be the similarity mapping on $\R^2$ whose scale factor is $\beta$.
Then $s_\beta$ is a stretching or contraction followed by a rotation one half turn.
== Proof ==
Let $\beta = -\gamma$ where $\gamma \in \R_{>0}$.
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
Then:
Because $\gamma > 0$ we have by definition that $s_\gamma$ is a stretching or contraction.
From Similarity Mapping on Plane with Scale Factor Minus 1, $s_{-1}$ is the plane rotation of the plane about the angle $\pi$.
Hence, by definition of half turn:
:$s_\beta$ is a stretching or contraction followed by a rotation one half turn.
== Also see ==
* Similarity Mapping on Plane Commutes with Half Turn about Origin, where it is seen that $s_\beta$ is also the same as a rotation one half turn followed by a stretching or contraction.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Similarity Mapping on Plane with Scale Factor Minus 1

== Theorem ==
Let $s_{-1}: \R^2 \to \R^2$ be a similarity mapping on $\R^2$ whose scale factor is $-1$.
Then $s_{-1}$ is the same as the rotation $r_\pi$ of the plane about the origin one half turn.
== Proof ==
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
Then:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Category:Geometric Rotations