nyuuzyou/wikis huggingface dataset:
Roots of Complex Number/Examples/z^6 + 1 = root 3 i/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(b)}$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:Solve the equations ... $\text{(b)}$ $z^6 + 1 = \sqrt 3 i$
:Ans. ... $\set {\sqrt [6] 2 \cis 40 \degrees, \sqrt [6] 2 \cis 1000 \degrees, \sqrt [6] 2 \cis 160 \degrees, \sqrt [6] 2 \cis 220 \degrees, \sqrt [6] 2 \cis 280 \degrees, \sqrt [6] 2 \cis 340 \degrees}$
''
== Correction ==
The correct solution is:
:$\set {\sqrt [6] 2 \cis 20 \degrees, \sqrt [6] 2 \cis 80 \degrees, \sqrt [6] 2 \cis 140 \degrees, \sqrt [6] 2 \cis 200 \degrees, \sqrt [6] 2 \cis 260 \degrees, \sqrt [6] 2 \cis 320 \degrees}$
as demonstrated in Roots of $z^6 + 1 = \sqrt 3 i$.
The mistake probably arose from mistaking $\dfrac {2 \pi} 3 = 240 \degrees$ when calculating the arguments in degrees.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(b)}$
Category:Mistakes/Theory and Problems of Complex Variables
Roots of Complex Number/Examples/z^8 + 1 = 0

== Theorem ==
The roots of the polynomial:
:$z^8 + 1 = 0$
are:
:$\set {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8: k \in \set {0, 1, \ldots, 7} }$
== Proof ==
From Euler's Identity:
:$-1 = e^{i \pi}$
Let $b$ be defined as:
From Roots of Complex Number: Corollary:
:$\paren {-1}^{1/8} = \cos \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8} + i \sin \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8}$
for $k = 0$ to $7$.
=== Illustration ===
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $8$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration

== Illustration of Roots of $z^8 + 1 = 0$ ==
The roots of the polynomial:
:$z^8 + 1 = 0$
are illustrated below:
:
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $8$
Category:Examples of Complex Roots
Roots of Complex Number/Exponential Form

== Theorem ==
Let $z := r e^{i \theta}$ be a complex number expressed in exponential form, such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then the $n$th roots of $z$ are given by:
:$z^{1 / n} = \set {r^{1 / n} e^{i \paren {\theta + 2 \pi k} / n}: k \in \set {0, 1, 2, \ldots, n - 1} }$
=== Principal Root ===
== Proof ==
== Sources ==
* : $\S 3$. Roots of Unity
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Roots: $3.7.28$
* : $\S 7$: Operations with Complex Numbers in Polar Form: $7.28$
Category:Complex Roots
Roots of Complex Number/Exponential Form/Principal Root

== Definition ==
Let $z := r e^{i \theta}$ be a complex number expressed in exponential form, such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
The principal $n$th root of $z$ is the value of $r^{1/n} e^{i \theta / n}$ such that:
:$-\pi < \theta \le \pi$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Roots: $3.7.28$
Category:Complex Roots
Roots of Quadratic with Rational Coefficients of form r plus s Root 2

== Theorem ==
Consider the quadratic equation:
:$(1): \quad a^2 x + b x + c = 0$
where $a, b, c$ are rational.
Let $\alpha = r + s \sqrt 2$ be one of the roots of $(1)$.
Then $\beta = r - s \sqrt 2$ is the other root of $(1)$.
== Proof ==
We have that:
Because $a$, $b$, $c$, $r$ and $s$ are rational, it must be that $\paren {2 a + b} s = 0$.
Hence:
and so $\beta$ is also a root of $(1)$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (5)$
Category:Quadratic Equations
Roots of Resolvent of Cubic

== Theorem ==
Let $P$ be the cubic equation:
:$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
Let $\map r P$ be the resolvent equation of $P$, given by:
:$u^6 - 2 R u^3 - Q^3$
Let the roots of $P$ be $\alpha_1, \alpha_2, \alpha_3$.
Then the roots of $\map r P$ can be expressed as:
where $\omega = -\dfrac {-1 + \sqrt {-3} } 2$ is one of the primitive (complex) cube roots of $1$.
== Proof ==
== Sources ==
* : Introduction
Category:Cubic Equations
Roots of Unity under Multiplication form Cyclic Group

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 0$.
The $n$th complex roots of unity under the operation of multiplication form the cyclic group which is isomorphic to $C_n$.
== Proof ==
From Complex Roots of Unity in Exponential Form:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of complex $n$th roots of unity.
Let $\omega = e^{2 i \pi / n}$.
Then we have:
:$U_n = \set {\omega^k: k \in \N_n}$
that is:
:$U_n = \set {\omega^0, \omega^1, \omega^2, \ldots, \omega^{n - 1} }$
Let $\omega^a, \omega^b \in U_n$.
Then $\omega^a \omega^b = \omega^{a + b} \in U_n$.
Either $a + b < n$, in which case $\omega^{a + b} \in U_n$, or $a + b \ge n$, in which case:
So $U_n$ is closed under multiplication.
We have that $\omega_0 = 1$ is the identity and that $\omega^{n - t}$ is the inverse of $\omega^t$.
Finally we note that $U_n$ is generated by $\omega$.
Hence the result, by definition of cyclic group, and from Cyclic Groups of Same Order are Isomorphic:
:$U_n = \gen \omega \cong C_n$.
== Also see ==
* Definition:Multiplicative Group of Complex Roots of Unity
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups: Example $1$
* : $\S 1$: Some examples of groups: Example $1.09$
* : Chapter $2$: Subgroups and Cosets: $\S 44$
* : $\S 3.2$: Groups; the axioms: Exercise $(1)$
* : $\S 1.1$: Example $4$
* : Chapter $1$: Definitions and Examples: Example $1.6$
* : Chapter $4$: Subgroups: Example $4.8$
* 
Category:Multiplicative Groups of Complex Roots of Unity
Category:Complex Roots of Unity
Category:Roots of Unity
Category:Examples of Cyclic Groups
Rotating Frame of Reference is Non-Inertial

== Theorem ==
Let $F$ be a frame of reference.
Let $F$ be rotating.
Then $F$ is a non-inertial frame of reference.
A particle fixed in such a frame has a centripetal component of acceleration relative to an inertial frame.
== Proof ==
== Sources ==
* 
* 
Category:Non-Inertial Frames of Reference
Rotating Indices Property of Vector Triple Product

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Euclidean $3$-space.
Then:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b} = 0$
where $\mathbf a \times \paren {\mathbf b \times \mathbf c}$ denotes the vector triple product operator.
== Proof ==
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 5$: $(12)$
Category:Vector Triple Product
Rotation is Isometry

== Theorem ==
Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions, wher $n = 2$ or $n = 3$.
Let $r_\theta$ be a rotation in $\Gamma$.
Then $r_\theta$ is an isometry.
== Proof ==
== Sources ==
* 
* 
Category:Geometric Rotations
Category:Isometries (Euclidean Geometry)
Rotation of Cartesian Axes around Vector

== Theorem ==
Let $\mathbf r$ be a vector in space.
Let a Cartesian plane $\CC$ be established such that:
:the initial point of $\mathbf r$ is at the origin $O$
:the terminal point of $\mathbf r$ is the point $P$.
Let $\tuple {X, Y}$ be the coordinates of $P$ under $\CC$.
Let $\CC$ be rotated about $O$ to $\CC'$, through an angle $\varphi$ in the anticlockwise direction, while keeping $\mathbf r$ fixed.
Let $\tuple {X', Y'}$ be the coordinates of $P$ under $\CC'$.
Then:
== Proof ==
:
With reference to the above diagram:
:$X P X' = \varphi$
and so:
:$OX' = OX \cos \varphi + PX \sin \varphi$
and:
:$OY' = OY \cos \varphi - PY \cos \varphi$
Hence the result.
== Sources ==
* : Chapter $1$ Vector Analysis $1.2$ Rotation of Coordinates
Category:Cartesian Coordinate Systems
Category:Vectors
Rotation of Plane about Origin is Linear Operator

== Theorem ==
Let $r_\alpha$ be the rotation of the plane about the origin through an angle of $\alpha$.
That is, let $r_\alpha: \R^2 \to \R^2$ be the mapping defined as:
:$\forall x \in \R^2: \map {r_\alpha} x = \text { the point into which a rotation of $\alpha$ carries $x$}$
Then $r_\alpha$ is a linear operator.
== Proof ==
Let $P = \tuple {\lambda_1, \lambda_2}$ be an arbitrary point in $\R^2$.
From Equations defining Plane Rotation:
:$\map {r_\alpha} P = \tuple {\lambda_1 \cos \alpha - \lambda_2 \sin \alpha, \lambda_1 \sin \alpha + \lambda_2 \cos \alpha}$
This demonstrates that $r_\alpha$ can be expressed as an ordered tuple of $4$ real numbers.
The result follows from Linear Operator on the Plane.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.2$
Category:Geometric Rotations
Category:Linear Operators
Rotation of Rectangular Hyperbola from Reduced to Standard Form

== Theorem ==
Let $\KK$ be a rectangular hyperbola embedded in a Cartesian plane in reduced form:
:$x^2 - y^2 = a^2$
Let $\KK$ be rotated $45 \degrees$ clockwise about the origin.
Then $\KK$ is in standard form:
:$x y = c^2$
where $c = \dfrac a {\sqrt 2}$.
== Proof ==
Let $\tuple {x, y}$ denote an arbitrary point on $\KK$ before rotation.
Let $\tuple {x', y'}$ denote the image of $\tuple {x, y}$ after rotation.
Then:
Substituting for $x$ and $y$ in $x^2 - y^2 = a^2$:
Hence the result.
== Sources ==
* 
* 
Category:Rectangular Hyperbolas
Rotation of Unit Cube about Vertical Axis

== Example of Symmetry in context of Geometry ==
Let $C$ be the unit cube embedded in real Cartesian space of $3$ dimensions as follows.
Let the vertices of $C$ be defined as:
Thus:
:$u$ and $d$ stand for up and down respectively
:$f$ and $b$ stand for front and back respectively
:$l$ and $r$ stand for left and right respectively.
Let:
Let $r$ be the rotation by $180 \degrees$ about the axis defined by the line through the points $\tuple {\dfrac 1 2, \dfrac 1 2, 0}$ and $\tuple {\dfrac 1 2, \dfrac 1 2, 1}$.
This is a symmetry of $C$ which induces $3$ permutations:
:$f_0: C_0 \to C_0$
:$f_1: C_1 \to C_1$
:$f_2: C_2 \to C_2$
$f_0$, $f_1$ and $f_2$ can be specified explicitly as follows:
:$\begin{array} {|r|r|}
\hline
v & \map {f_0} v \\
\hline
O = dbl & P_4 = dfr \\
P_1 = dfl & P_2 = dbr \\
P_2 = dbr & P_1 = dfl \\
P_3 = ubl & P_7 = ufr \\
P_4 = dfr & O = dbl \\
P_5 = ufl & P_6 = ubr \\
P_6 = ubr & P_5 = dfl \\
P_7 = ufr & P_3 = ubl \\
\hline
\end{array} \qquad \begin{array} {|r|r|}
\hline
e & \map {f_1} e \\
\hline
uf & ub \\
ur & ul \\
ub & uf \\
ul & ur \\
fr & bl \\
br & fl \\
bl & fr \\
fl & br \\
df & db \\
dr & dl \\
db & df \\
dl & dr \\
\hline
\end{array} \qquad \begin{array} {|r|r|}
\hline
f & \map {f_2} f \\
\hline
F & B \\
B & F \\
U & U \\
D & D \\
L & R \\
R & L \\
\hline
\end{array}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.7$, Ponderable $2.1.7$
Category:Examples of Symmetries (Geometry)
Roth's Theorem on Diophantine Approximation to Algebraic Numbers

== Theorem ==
Let $\alpha$ be a real algebraic number.
Let $\nu$ be a real number greater than $2$.
Then there exist finitely many pairs of integers $\tuple {p, q}$ such that:
:$\size {\alpha - \dfrac p q} < \dfrac 1 {q^\nu}$
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Roth's Theorem
Category:Algebraic Numbers
Category:Diophantine Equations
Rouché's Theorem

== Theorem == 
Let $\gamma$ be a closed contour.
Let $D$ be the region enclosed by $\gamma$. 
Let $f$ and $g$ be complex-valued functions which are holomorphic in $D$. 
Let $\cmod {\map g z} < \cmod {\map f z}$ on $\gamma$.
Then $f$ and $f + g$ have the same number of zeroes in $D$ counted up to multiplicity. 
== Proof == 
Let $N_f$ and $N_{f + g}$ be the number of zeroes of $f$ and $f + g$ in $D$ respectively. 
By the Argument Principle: 
:$\ds N_f = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$
Similarly: 
:$\ds N_{f + g} = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {\paren {f + g}'} z} {\map {\paren {f + g} } z} \rd z$
We aim to show that $N_f = N_{f + g}$. 
From $\cmod {\map g z} < \cmod {\map f z}$ we have that $f$ is non-zero on $\gamma$, otherwise we would have $\cmod {\map g z} < 0$. 
From the fact that $\cmod {\map g z} \ne \cmod {\map f z}$ we also have that $\map g z \ne - \map f z$, so $f + g$ is also non-zero on $\gamma$. 
We have: 
So:
For brevity, write: 
:$F = 1 + \dfrac g f$
As $\cmod {\dfrac g f} < 1$ on $\gamma$, we must have: 
:$\cmod {\map \Re {\dfrac g f} } < 1$
That is:
:$0 < \map \Re F < 2$
on $\gamma$. 
That is, the image of $\gamma$ under $F$ does not encircle $0$. 
So, by the definition of winding number, we have:
:$\map {\mathrm {Ind}_{\map F \gamma} } 0 = 0$
So: 
 
Hence:
:$N_{f + g} = N_f$
Category:Complex Analysis
Round Peg fits in Square Hole better than Square Peg fits in Round Hole

== Theorem ==
A round peg fits better in a square hole than a square peg fits in a round hole.
:
== Proof ==
The situation is modelled by considering the ratios of the areas of:
:a square to the circle in which it is inscribed
:a square to the circle around which it has been circumscribed.
Let a square $S$ be inscribed in a circle $C$ of radius $r$.
Let $A_c$ and $A_s$ be the areas of $C$ and $S$ respectively.
From Area of Circle:
:$A_c = \pi r^2$
The diameter of $S$ is $2 r$.
Thus from Pythagoras's Theorem its side is of length $r \sqrt 2$.
From Area of Square:
:$A_s = 2 r^2$
Thus:
:$\dfrac {A_s} {A_c} = \dfrac {2 r^2} {\pi r^2} = \dfrac 2 \pi \approx 0.6366 \ldots$
Let a square $S$ be circumscribed around a circle $C$ of radius $r$.
Let $A_c$ and $A_s$ be the areas of $C$ and $S$ respectively.
From Area of Circle:
:$A_c = \pi r^2$
The side of $S$ is of length $2 r$.
From Area of Square:
:$A_s = 4 r^2$
Thus:
:$\dfrac {A_c} {A_s} = \dfrac {\pi r^2} {4 r^2} = \dfrac \pi 4 \approx 0.7853 \ldots$
Thus a round peg takes up more space ($0.7853 \ldots$) of a square hole than a square peg takes up ($0.6366 \ldots$) of a round hole.
== Sources ==
* 
* : $9$
* : $9$
Category:Squares
Category:Circles
Rounding/Examples

== Examples of Rounding ==
=== $72 \cdotp 8146$ to $2$ Decimal Places ===
=== $48 \cdotp 6$ to Nearest Unit ===
=== $136 \cdotp 5$ to Nearest Unit ===
=== $2 \cdotp 484$ to Nearest Hundredth ===
=== $0 \cdotp 0435$ to Nearest Thousandth ===
=== $4 \cdotp 500 \, 01$ to Nearest Unit ===
=== $143 \cdotp 95$ to Nearest Tenth ===
=== $368$ to Nearest Hundred ===
=== $24 \, 448$ to Nearest Thousand ===
=== $5 \cdotp 565 \, 00$ to Nearest Hundredth ===
=== $5 \cdotp 565 \, 01$ to Nearest Hundredth ===
=== $12 \cdotp 5664$ to One Decimal Place ===
=== $12 \cdotp 5664$ to Two Decimal Places ===
=== $1 \cdotp 576$ and $1 \cdotp 5751$ to Two Decimal Places ===
=== $1 \cdotp 572$ and $1 \cdotp 5749$ to Two Decimal Places ===
=== $12 \cdotp 5664$ to Two Significant Figures ===
=== $419 \cdotp 8$ to Three Significant Figures ===
=== $0 \cdotp 02006$ to Three Significant Figures ===
Category:Examples of Rounding
Rounding/Examples/0.02006 to Three Significant Figures

== Example of Rounding ==
$0 \cdotp 02006$ rounded to $3$ significant figures is $0 \cdotp 0201$.
This is because $0 \cdotp 02006$ is closer to $0 \cdotp 0201$ than it is to $0 \cdotp 0200$.
Also note that the $0$ immediately to the right of the decimal point does not count as one of the significant figures.
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals: Decimal Places and Significant Figures
Category:Examples of Rounding
Rounding/Examples/0.0435 to Nearest Thousandth

== Example of Rounding ==
$0 \cdotp 0435$ rounded to the nearest thousandth is $0 \cdotp 044$.
We have that $0 \cdotp 0435$ is half way between $0 \cdotp 043$ and $0 \cdotp 044$, and the convention is to round to the nearest even digit.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(d)}$
Category:Examples of Rounding
Rounding/Examples/1.572 and 1.5749 to 2dp

== Examples of Rounding ==
$1 \cdotp 572$ and $1 \cdotp 5749$ rounded to $2$ decimal places are both $1 \cdotp 57$.
This is because they are both closer to $1 \cdotp 57$ than they are to $1 \cdotp 58$.
== Sources ==
* 
* 
Category:Examples of Rounding
Rounding/Examples/1.576 and 1.5751 to 2dp

== Examples of Rounding ==
$1 \cdotp 576$ and $1 \cdotp 5751$ rounded to $2$ decimal places are both $1 \cdotp 58$.
This is because they are both closer to $1 \cdotp 58$ than they are to $1 \cdotp 57$.
== Sources ==
* 
* 
Category:Examples of Rounding
Rounding/Examples/12.5664 to One Decimal Place

== Example of Rounding ==
$12 \cdotp 5664$ rounded to $1$ decimal place is $12 \cdotp 6$.
This is because $12 \cdotp 5664$ is closer to $12 \cdotp 6$ than it is to $12 \cdotp 5$.
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals: Decimal Places and Significant Figures
Category:Examples of Rounding
Rounding/Examples/12.5664 to Two Decimal Places

== Example of Rounding ==
$12 \cdotp 5664$ rounded to $2$ decimal places is $12 \cdotp 57$.
This is because $12 \cdotp 5664$ is closer to $12 \cdotp 57$ than it is to $12 \cdotp 56$.
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals: Decimal Places and Significant Figures
Category:Examples of Rounding
Rounding/Examples/12.5664 to Two Significant Figures

== Example of Rounding ==
$12 \cdotp 5664$ rounded to $2$ significant figures is $13$.
This is because $12 \cdotp 5664$ is closer to $13$ than it is to $12$.
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals: Decimal Places and Significant Figures
Category:Examples of Rounding
Rounding/Examples/136.5 to Nearest Unit

== Example of Rounding ==
$136 \cdotp 5$ rounded to the nearest unit is $136$.
We have that $136 \cdotp 5$ is half way between $136$ and $137$, and the convention is to round to the nearest even digit.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(b)}$
Category:Examples of Rounding
Rounding/Examples/143.95 to Nearest Tenth

== Example of Rounding ==
$143 \cdotp 95$ rounded to the nearest tenth is $144 \cdotp 0$.
We have that $143 \cdotp 95$ is half way between $143 \cdotp 9$ and $144 \cdotp 0$, and the convention is to round to the nearest even digit, which in this case is $0$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(f)}$
Category:Examples of Rounding
Rounding/Examples/2.484 to Nearest Hundredth

== Example of Rounding ==
$2 \cdotp 484$ rounded to the nearest hundredth is $2 \cdotp 48$.
This is because $2 \cdotp 484$ is closer to $2 \cdotp 48$ than $2 \cdotp 49$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(c)}$
Category:Examples of Rounding
Rounding/Examples/24448 to Nearest Thousand

== Example of Rounding ==
$24 \, 448$ rounded to the nearest thousand is $24 \, 000$.
This is because $24 \, 448$ is closer to $24 \, 000$ than $25 \, 000$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(h)}$
Category:Examples of Rounding
Rounding/Examples/4.50001 to Nearest Unit

== Example of Rounding ==
$4 \cdotp 500 \, 01$ rounded to the nearest unit is $5$.
This is because $4 \cdotp 500 \, 01$ is closer to $5$ than $4$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(e)}$
Category:Examples of Rounding
Rounding/Examples/419.8 to Three Significant Figures

== Example of Rounding ==
$419 \cdotp 8$ rounded to $3$ significant figures is $420$.
This is because $419 \cdotp 8$ is closer to $420$ than it is to $419$.
== Sources ==
* : Arithmetic: Chapter $\text I$: Decimals: Decimal Places and Significant Figures
Category:Examples of Rounding
Rounding/Examples/48.6 to Nearest Unit

== Example of Rounding ==
$48 \cdotp 6$ rounded to the nearest unit is $49$.
This is because $48 \cdotp 6$ is closer to $49$ than $48$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(a)}$
Category:Examples of Rounding
Rounding/Examples/5.56500 to Nearest Hundredth

== Example of Rounding ==
$5 \cdotp 565 \, 00$ rounded to the nearest hundredth is $5 \cdotp 56$.
We have that $5 \cdotp 565 \, 00$ is half way between $5 \cdotp 56$ and $5 \cdotp 57$, and the convention is to round to the nearest even digit.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(i)}$
Category:Examples of Rounding
Rounding/Examples/5.56501 to Nearest Hundredth

== Example of Rounding ==
$5 \cdotp 565 \, 01$ rounded to the nearest hundredth is $5 \cdotp 57$.
This is because $5 \cdotp 565 \, 01$ is closer to $5 \cdotp 57$ than $5 \cdotp 56$.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Rounding of Data: $1.3 \ \text{(j)}$
Category:Examples of Rounding
Rounding/Examples/72.8146 to 2dp

== Example of Rounding ==
$72 \cdotp 8146$ rounded to $2$ decimal places is $72 \cdotp 81$.
This is because $72 \cdotp 8146$ is closer to $72 \cdotp 81$ than $72 \cdotp 82$.
== Sources ==
* : Chapter $1$: Rounding of Data
Category:Examples of Rounding
Rounding Error/Examples

== Examples of Rounding Errors ==
=== Example: $73.854 \, \mathrm {mm}$ ===
=== Example: $0.09800 \ \mathrm {m^3}$ ===
=== Example: $3.687 \times 10^8 \, \mathrm {km}$ ===
Category:Examples of Rounding Errors
Rounding Error/Examples/73.854 mm

== Example of Rounding Error ==
Let $x$ be a length expressed as $73.854 \, \mathrm {mm}$ to $5$ significant figures.
The maximum rounding error in $x$ is $0.0005 \, \mathrm {mm}$.
== Proof ==
By definition of rounding, we have:
:$73.8535 \le x \le 73.8545$
Whether the inequalities are strict or not depends on the treatment of the half.
Rounding up, we have:
:$73.8535 \le x < 73.8545$
because $73.8545$ will round up to $73.855$.
Rounding down, we have:
:$73.8535 < x \le 73.8545$
because $73.8535$ will round down to $73.853$.
Rounding to even, we have:
:$73.8535 \le x \le 73.8545$
because both $73.8535$ and $73.8545$ will round to even to $73.854$.
Hence we have:
:$\size {73.854 - 73.8535} \le 0.0005 \ge \size {73.854 - 73.8545}$
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.7 \ \text {(a)}$
Category:Examples of Rounding Errors
Rounding to Even/Examples

== Examples of Rounding to Even ==
=== $72 \cdotp 465$ to $2$ Decimal Places ===
=== $183 \cdotp 575$ to $2$ Decimal Places ===
=== $116 \, 500 \, 000$ to Nearest Million ===
=== $1 \cdotp 575$ to $2$ Decimal Places ===
=== $1 \cdotp 565$ to $2$ Decimal Places ===
Category:Examples of Rounding
Rounding to Even/Examples/1.565 to 2dp

== Example of Rounding to Even ==
$1 \cdotp 565$ rounded to $2$ decimal places using rounding to even is $1 \cdotp 56$.
== Proof ==
We have that $1 \cdotp 565$ is equally far away from $1 \cdotp 56$ as it is from $1 \cdotp 57$.
But $1 \cdotp 56$ ends in an even integer.
That is, $1356$ is even.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Rounding
Rounding to Even/Examples/1.575 to 2dp

== Example of Rounding to Even ==
$1 \cdotp 575$ rounded to $2$ decimal places using rounding to even is $1 \cdotp 58$.
== Proof ==
We have that $1 \cdotp 575$ is equally far away from $1 \cdotp 57$ as it is from $1 \cdotp 58$.
But $1 \cdotp 58$ ends in an even integer.
That is, $1358$ is even.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Rounding
Rounding to Even/Examples/116,500,000 to Nearest Million

== Example of Rounding to Even ==
$116 \, 500 \, 000$ rounded to the nearest million using rounding to even is $116 \, 000 \, 000$.
== Proof ==
We have that $116 \, 500 \, 000$ is equally far away from $116 \, 000 \, 000$ as it is from $117 \, 000 \, 000$.
But $116 \, 000 \, 000$ ends in an even integer.
That is, $\dfrac {116 \, 000 \, 000} {10^6} = 116$ is even.
Hence the result.
== Sources ==
* : Chapter $1$: Rounding of Data
Category:Examples of Rounding
Rounding to Even/Examples/183.575 to 2dp

== Example of Rounding to Even ==
$183 \cdotp 575$ rounded to $2$ decimal places using rounding to even is $183 \cdotp 58$.
== Proof ==
We have that $183 \cdotp 575$ is equally far away from $183 \cdotp 57$ as it is from $183 \cdotp 58$.
But $183 \cdotp 58$ ends in an even integer.
That is, $18 \, 358$ is even.
Hence the result.
== Sources ==
* : Chapter $1$: Rounding of Data
Category:Examples of Rounding