nyuuzyou/wikis huggingface dataset:
L-Infinity Norm is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space for $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ space on $\struct {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$. 
Then the $L^\infty$ norm: 
:$\ds \norm {\eqclass f \sim}_\infty = \norm f_\infty$
is well-defined.
== Proof ==
We show that for $E \in \map {L^\infty} {X, \Sigma, \mu}$, $\norm E_\infty$ is independent of the representative chosen for $E$.
Let:
:$E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
We show that: 
:$\norm f_\infty = \norm g_\infty$
From Equivalence Class Equivalent Statements, we have: 
:$f \sim g$
So, from the definition of the almost-everywhere equality relation, we have: 
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
That is, there exists a $\mu$-null set $N$ such that: 
:if $\map f x \ne \map g x$ then $x \in N$.
We show that: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = \map \mu {\set {x \in X : \size {\map g x} \ge c} }$ 
From this it will follow that: 
:$\inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0} = \inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map g x} \ge c} } = 0}$
That is: 
:$\norm f_\infty = \norm g_\infty$
Clearly: 
:$\set {x \in N : \size {\map f x} \ge c} \subseteq N$
So, from Null Sets Closed under Subset:
:$\map \mu {\set {x \in N : \size {\map f x} \ge c} } = 0$
Swapping $f$ for $g$, we also obtain: 
:$\map \mu {\set {x \in N : \size {\map g x} \ge c} } = 0$
Since: 
:$\set {x \in N : \size {\map f x} \ge c} \subseteq N$
and:
:$\set {x \in X \setminus N : \size {\map f x} \ge c} \subseteq X \setminus N$
we have that: 
:$\set {x \in N : \size {\map f x} \ge c}$ and $\set {x \in X \setminus N : \size {\map f x} \ge c}$ are disjoint for all real numbers $c > 0$. 
Then:
as required.
Category:L-Infinity Norm
L1 Mean Ergodic Theorem

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Let $\map {L^1_\C} \mu$ be the complex-valued $L^1$ space of $\mu$.
Then for each $f \in \map {L^1_\C} \mu$ there is a $T$-invariant function $\tilde f \in \map {L^1_\C} \mu$ such that:
:$\ds \lim_{N \mathop \to \infty} \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} f \circ T^n = \tilde f$ converges in the $L^1$-norm
== Proof ==
For $c > 0$ let:
:$f_c = f \cdot \chi_{\set {\cmod f \mathop \le c} }$
where $\chi_B$ is the characteristic function of $B$.
Then $f_c \in \map {L^2_\C} \mu$ and:
Let $\epsilon > 0$ be arbitrary.
Let $c > 0$ be such that:
:$\norm {f - f_c}_1 \le \dfrac \epsilon 3$
By Mean Ergodic Theorem and Convergent Sequence is Cauchy Sequence, there exists an $N_0 \in \N_{>0}$ such that:
:$\ds \forall M, N \ge N_0 : \norm {\dfrac 1 M \sum_{n \mathop = 0}^{M - 1} f_c \circ T^n - \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} f_c \circ T^n}_1 \le \frac \epsilon 3$
since $\norm \cdot_1 \le \norm \cdot_2^2$ in view of Schwarz's Inequality.
Thus by :
:$\ds \forall M, N \ge N_0 : \norm {\dfrac 1 M \sum_{n \mathop = 0}^{M - 1} f \circ T^n - \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} f \circ T^n}_1 \le \epsilon$
That is:
:$\ds \sequence { \frac 1 N \sum_{n \mathop = 0}^{N - 1} f \circ T^n }_N$
is a Cauchy sequence in $\map {L^1_\C} \mu$.
By Riesz-Fischer Theorem there exists an $\tilde f \in \map {L^1_\C} \mu$ such that:
:$\ds \lim_{N \mathop \to \infty} \norm { \dfrac 1 N \sum_{n \mathop = 0}^{N - 1} f \circ T^n - \tilde f}_1 = 0$
In addition:
That is:
:$\tilde f \circ T = \tilde f$
== Sources ==
* $2.5$ The Mean Ergodic Theorem
Category:Mean Ergodic Theorem
L1 Metric is Topologically Equivalent to Supremum Metric on Bounded Continuous Real Functions

== Theorem ==
Let $\FF$ be the set of all real functions which are also bounded on the closed interval $\closedint a b$.
Let $d: \FF \times \FF \to \R$ be the $L^1$ metric on $\closedint a b$:
:$\ds \forall f, g \in \FF: \map d {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Let $d': \FF \times \FF \to \R$ be the supremum metric on $\closedint a b$:
:$\ds \forall f, g \in \FF: \map {d'} {f, g} := \sup_{x \mathop \in S} \size {\map f x - \map g x}$
Then $d$ and $d'$ are topologically equivalent metrics.
== Proof ==
Let $U$ be an upper bound of $\set {\size {\map f x - \map g x} }$.
Then:
:$\ds U \ge \sup_{x \mathop \in S} \size {\map f x - \map g x}$
Hence:
:$\ds \max_{x \mathop \in \closedint a b} \set {\size {\map f x - \map g x} } = \map {d'} {f, g}$
Then:
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $6$
Category:L1 Metric
Category:Supremum Metric
L1 Metric on Closed Real Interval is Metric

== Theorem ==
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $d: S \times S \to \R$ be the $L^1$ metric on $\closedint a b$:
:$\ds \forall f, g \in S: \map d {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Then $d$ is a metric.
== Proof ==
=== ===
So holds for $d$.
=== ===
So holds for $d$.
=== ===
So holds for $d$.
=== ===
From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
:$\map d {f, g} = 0 \implies f = g$
on $\closedint a b$.
So holds for $d$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $4$
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.9$
Category:Definite Integrals
Category:L1 Metric
L2 Metric on Closed Real Interval is Metric

== Theorem ==
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $d_2: S \times S \to \R$ be the $L^2$ metric on $\closedint a b$:
:$\ds \forall f, g \in S: \map {d_2} {f, g} := \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}$
Then $d_2$ is a metric.
== Proof ==
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
It is required to be shown:
:$\map {d_2} {f, g} + \map {d_2} {g, h} \ge d_2 \map {d_2} {f, h}$
for all $f, g, h \in S$.
Let:
:$(1): \quad t \in \closedint a b$
:$(2): \quad \map f t - \map g t = \map r t$
:$(3): \quad \map g t - \map h t = \map s t$
Thus we need to show that:
:$\ds \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2} \ge \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}$
We have:
So holds for $d_2$.
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
:$\map {d_2} {f, g} = 0 \implies f = g$
on $\closedint a b$.
So holds for $d_2$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.11$
* 
* 
Category:L2 Metric
LCM Divides Common Multiple

== Theorem ==
Let $a, b \in \Z$ such that $a b \ne 0$.
Let $n$ be any common multiple of $a$ and $b$.
That is, let $n \in \Z: a \divides n, b \divides n$.
Then:
:$\lcm \set {a, b} \divides n$
where $\lcm \set {a, b}$ is the lowest common multiple of $a$ and $b$.
== Proof ==
Let $m = \lcm \set {a, b}$.
Then $a \divides m$ and $b \divides m$ by definition.
Suppose $n$ is some other common multiple of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$).
Then from the Division Theorem:
:$n = k m + r$
for some integer $k$ and with $0 < r < m$.
Then since $r = n - k m$, using $a \divides n$ and $a \divides m$:
: $a \divides r$
Similarly:
: $b \divides r$
Then $r$ is a common multiple of $a$ and $b$.
But we have that $r < m$.
This contradicts the fact that $m$ is the lowest common multiple of $a$ and $b$.
So, by contradiction, it follows that $m \divides n$.
== Sources ==
* : Book $\text{VII}$. Propositions
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \gamma$
Category:Lowest Common Multiple
LCM equals Product iff Coprime

== Theorem ==
Let $a, b \in \Z_{>0}$ be strictly positive integers.
Then:
:$\lcm \set {a, b} = a b$
:
:$a$ and $b$ are coprime
where $\lcm$ denotes the lowest common multiple.
== Proof ==
=== Necessary Condition ===
Let $a$ and $b$ be coprime.
Then:
=== Sufficient Condition ===
Let $\lcm \set {a, b} = a b$.
Then:
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Lowest Common Multiple
Category:Coprime Integers
LCM from Prime Decomposition

== Theorem ==
Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
That is, the primes given in these prime decompositions may be divisors of either of the numbers $a$ or $b$.
Then:
:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
where $\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$.
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also see ==
* GCD from Prime Decomposition
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $7$
* : Chapter $1$: Properties of the Natural Numbers: $\S 24 \alpha$
* 
Category:Lowest Common Multiple
Category:Prime Numbers
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples

== Examples of Use of LCM from Prime Decomposition ==
=== $125$ and $150$ ===
=== $132$ and $154$ ===
=== $39$ and $143$ ===
=== $253$ and $506$ ===
=== $111$ and $1221$ ===
=== $39$, $102$ and $75$ ===
=== $p^2 q$ and $p q r$ ===
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/111 and 1221

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $111$ and $1221$ is:
:$\lcm \set {111, 1221} = 1221$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(e)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/125 and 150

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $125$ and $150$ is:
:$\lcm \set {125, 150} = 750$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(a)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/132 and 154

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $132$ and $154$ is:
:$\lcm \set {132, 154} = 924$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(b)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/253 and 506

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $253$ and $506$ is:
:$\lcm \set {253, 506} = 506$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(d)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/39, 102 and 75

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $39$, $102$ and $75$ is:
:$\lcm \set {39, 102, 75} = 33 \, 150$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $11$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/39 and 143

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $39$ and $143$ is:
:$\lcm \set {39, 143} = 429$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(c)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/Examples/p^2 q and p q r

== Example of Use of LCM from Prime Decomposition ==
The lowest common multiple of $p^2 q$ and $p q r$, where $p$, $q$ and $r$ are all primes, is:
:$\lcm \set {p^2 q, p q r} = p^2 q r$
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $9 \ \text {(f)}$
Category:LCM from Prime Decomposition
LCM from Prime Decomposition/General Result

== Theorem ==
Let $n \in \N$ be a natural number such that $n \ge 2$.
Let $\N_n$ be defined as:
:$\N_n := \set {1, 2, \dotsc, n}$
Let $A_n = \set {a_1, a_2, \dotsc, a_n} \subseteq \Z$ be a set of $n$ integers.
From Expression for Integers as Powers of Same Primes, let:
:$\ds \forall i \in \N_n: a_i = \prod_{p_j \mathop \in T} {p_j}^{e_{i j} }$
where:
:$T = \set {p_j: j \in \N_r}$
such that:
:$\forall j \in \N_{r - 1}: p_j < p_{j - 1}$
:$\forall j \in \N_r: \exists i \in \N_n: p_j \divides a_i$
where $\divides$ denotes divisibility.
Then:
:$\ds \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$
where $\map \lcm {A_n}$ denotes the greatest common divisor of $a_1, a_2, \dotsc, a_n$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\ds \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{1 j}, e_{2 j} } }$
This is GCD from Prime Decomposition.
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:
:$\ds \map \lcm {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_k} }$
from which it is to be shown that:
:$\ds \map \lcm {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_{k + 1} } }$
=== 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:
:$\forall n \in \Z_{\ge 2}: \ds \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: i \in \N_n} }$
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $10$
* 
* 
Category:Lowest Common Multiple
Category:Prime Decompositions
LCM from Prime Decomposition/Proof 2

== Theorem ==
== Proof ==
Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.
Let $a \divides m$.
Then:
:$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$
:$a \divides l \iff \forall i: 1 \le i \le r, 0 \le l_i \le h_i$
So:
:$a \divides m \land b \divides m \iff \forall i: 1 \le i \le r, 0 \le \max \set {k_i, l_i} \le h_i$
For $m$ to be at its smallest, we want the smallest possible exponent for each of these primes.
So for each $i \in \closedint 1 r$, $h_i$ needs to equal $\max \set {k_i, l_i}$.
Hence the result:
:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
Category:LCM from Prime Decomposition
LCM iff Divides All Common Multiples

== Theorem ==
Let $a, b \in \Z$ such that $a b \ne 0$.
Let $m \in \Z: d > 0$.
Then $m = \lcm \set {a, b}$ :
: $(1): \quad a \divides m \land b \divides m$
: $(2): \quad a \divides n \land b \divides n \implies m \divides n$
That is, in the set of positive integers, $m$ is the LCM of $a$ and $b$ $m$ is a common multiple of $a$ and $b$, and $m$ also divides any other common multiple of $a$ and $b$.
== Proof ==
=== Necessary Condition ===
Suppose $m = \lcm \set {a, b}$.
Then from LCM Divides Common Multiple:
: $a \divides n \land b \divides n \implies m \divides n$
=== Sufficient Condition ===
Now suppose $a \divides m \land b \divides m$, and $m$ also divides any $n$ that $a$ and $b$ also divide.
From $a \divides m \land b \divides m$, we see that $m$ is a common multiple of $a$ and $b$.
From $a \divides n \land b \divides n$, we see that $n$ is also a common multiple of $a$ and $b$.
Also, we have that $m \divides n$.
From Absolute Value of Integer is not less than Divisors, we see that (in the domain of $\Z_{>0}$) $m \divides n \implies m \le n$.
Thus, whatever $m$ may be, it is no larger than $n$.
Therefore, $m$ must be the least of all the common multiples of $a$ and $b$.
Category:Lowest Common Multiple
LCM of 3 Integers in terms of GCDs of Pairs of those Integers

== Theorem ==
Let $a, b, c \in \Z_{>0}$ be strictly positive integers.
Then:
:$\lcm \set {a, b, c} = \dfrac {a b c \gcd \set {a, b, c} } {d_1 d_2 d_3}$
where:
:$\gcd$ denotes greatest common divisor
:$\lcm$ denotes lowest common multiple
:$d_1 = \gcd \set {a, b}$
:$d_2 = \gcd \set {b, c}$
:$d_3 = \gcd \set {a, c}$
=== Lemma ===
== Proof ==
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $12$
Category:Lowest Common Multiple
Category:Greatest Common Divisor
Category:LCM of 3 Integers in terms of GCDs of Pairs of those Integers
LCM of 3 Integers in terms of GCDs of Pairs of those Integers/Lemma

== Theorem ==
Let $a, b, c \in \Z_{>0}$ be strictly positive integers.
Then:
:$\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = \gcd \set {a, b, c}$
== Proof ==
Let $\gcd \set {a, b, c} = d_1$.
From definition:
:$d_1 \divides a$, $d_1 \divides b$ and $d_1 \divides c$.
By Common Divisor Divides GCD:
:$d_1 \divides \gcd \set {a, b}$ and $d_1 \divides \gcd \set {a, c}$.
By Common Divisor Divides GCD again:
:$d_1 \divides \gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$.
On the other hand, let $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = d_2$.
From definition:
:$d_2 \divides \gcd \set {a, b}$ and $d_2 \divides \gcd \set {a, c}$.
From definition again:
:$d_2 \divides a$, $d_2 \divides b$ and $d_2 \divides c$.
Hence $d_2 \divides \gcd \set {a, b, c}$.
Since $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$ and $\gcd \set {a, b, c}$ divide each other, by Absolute Value of Integer is not less than Divisors they must be equal.
Category:Greatest Common Divisor
Category:LCM of 3 Integers in terms of GCDs of Pairs of those Integers
LCM of Three Numbers

== Theorem ==
Let $a, b, c \in \Z: a b c \ne 0$.
The lowest common multiple of $a, b, c$, denoted $\lcm \set {a, b, c}$, can always be found.
== Proof ==
Let $d = \lcm \set {a, b}$.
This exists from .
Either $c \divides d$ or not, where $\divides$ denotes divisibility.
Suppose $c \divides d$.
But by definition of lowest common multiple, $a \divides d$ and $b \divides d$ also.
Suppose $a, b, c$ are divisors of $e$ where $e < d$.
Then $a, b$ are divisors of $e$.
That is, $e$ is a common divisor of $a$ and $b$ which is lower than $d$.
But by :
:$d \divides e$
which is impossible.
It follows that there can be no such $e$.
Therefore $d = \lcm \set {a, b, c}$.
Now suppose $c \nmid d$.
Let $e = \lcm \set {c, d}$.
This exists from .
Since $a$ and $b$ are both divisors of $d$, it follows that:
:$a \divides e$
:$b \divides e$
But we have that $c \divides e$ as well.
Suppose $a, b, c$ are divisors of $f$ where $f < e$.
Then $a, b$ are divisors of $f$.
So by :
:$d = \lcm \set {a, b} \divides f$
But also $c \divides f$.
Therefore $c$ and $d$ are divisors of $f$.
By :
:$e = \lcm \set {c, d} \divides f$
But this is impossible as by hypothesis $f < e$.
Therefore $a, b, c$ are divisors of no number smaller than $e$.
Therefore $e = \lcm \set {a, b, c}$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Lowest Common Multiple
LU Factorization/Examples/Arbitrary Example 1

== Example of LU Factorization ==
The matrix:
:$\begin {pmatrix} 2 & 1 \\ -1 & 2 \end {pmatrix}$
has this LU factorization:
:$\begin {pmatrix} 1 & 0 \\ -1 & 1 \end {pmatrix} \begin {pmatrix} 2 & 1 \\ 0 & 3 \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of LU Factorizations
Ladies' Diary/Age Problem

== Puzzle ==
:If to my age there added be,
:One half, one third, and three times three,
:Six score and ten the sum you'll see,
:Pray find out what my age will be.
== Solution ==
Let $x$ be the age to be found.
Translating from flowery language into an algebraic expression, we get:
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $137$
Category:Ladies' Diary
Ladies' Diary/Dutchmen's Three Wives

== Puzzle ==
:There came $3$ dutchmen of my acquaintance to see me, being lately married;
:they brought their wives with them.
:The men's names were Hendrick, Claas, and Cornelius;
:the women's Geertrick, Catriin, and Anna;
:but I forget the name of each man's wife.
:They told me that they had been at market, to buy hogs;
:each person bought as many hogs as they gave shillings for each hog;
:Hendrick bought $23$ hogs more than Catriin,
:and Claas bought $11$ more than Geertrick;
:likewise, each man laid out $3$ guineas more than his wife.
:I desire to know the name of each man's wife.
== Solution ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Ladies' Diary
Category:Dutchmen's Three Wives
Ladies' Diary/Dutchmen's Three Wives/Proof 1

== Puzzle ==
== Solution ==
== Proof ==
Let the number of hogs bought by any one of the women be $x$.
Let the number of hogs bought by her husband be $x + n$.
Then:
:the money paid by the woman is $x^2$ shillings
:the money paid by her husband is $\paren {x + n}^2$ shillings
There are $21$ shillings to the guinea, so:
:$\paren {x + n}^2 = x^2 + 63$
and so:
:$x = \dfrac {63 - n^2} {2 n}$
We see that in order for $x$ to be an integer it is necessary for $n$ to be odd.
Now we can plug various values of $n$ and see where this gets us.
It turns out by trying various $n$ that $3$ such values work, as follows:
If $n = 1$ then $x = 31$ and $x + n = 32$
If $n = 3$ then $x = 9$ and $x + n = 12$
If $n = 7$ then $x = 1$ and $x + n = 8$.
Now we can see that:
:Hendrick bought $32$ hogs and Catriin bought $9$ hogs
:Claas bought $12$ hogs and Geertrick bought $1$ hog
and the allocation of men with their wives then becomes apparent.
== Historical Note ==
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Dutchmen's Three Wives
Ladies' Diary/Dutchmen's Three Wives/Proof 1/Historical Note

== Historical Note on Dutchmen's Three Wives: Proof 1 ==
This proof of the problem of the Dutchmen and their $3$ wives is reported as being the work of J. Hill, who may well have been the same who published in $1716$.
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Historical Notes
Ladies' Diary/Dutchmen's Three Wives/Proof 2

== Puzzle ==
== Solution ==
== Proof ==
It is immediate from the last condition that each man paid $63$ shillings more than his wife.
It is also immediate from the first condition that the amount paid for the hogs by each person is a square number.
So we are looking for pairs of square numbers which differ by $63$.
This leads us to the pairs:
:$\tuple {1^2, 8^2}$, that is $\tuple {1, 64}$
:$\tuple {9^2, 12^2}$, that is $\tuple {81, 144}$
:$\tuple {31^2, 32^2}$, that is $\tuple {961, 1024}$
The relations between Hendrick and Catriin, and Claas and Geertrick, provide us with the information we need to match husband and wife for all three.
== Historical Note ==
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Dutchmen's Three Wives
Ladies' Diary/Dutchmen's Three Wives/Proof 2/Historical Note

== Historical Note on Dutchmen's Three Wives: Proof 2 ==
This proof of the problem of the Dutchmen and their $3$ wives is reported as being the work of Mr N. Farrer, whose identity is not known.
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Historical Notes
Ladies' Diary/Dutchmen's Three Wives/Solution

== Puzzle ==
== Solution ==
Note that a guinea is $21$ shillings.
:Hendrick bought $32$ hogs and his wife Anna bought $31$
:Claas bought $12$ hogs and his wife Catriin bought $9$
:Cornelius bought $8$ hogs and his wife Geertrick bought $1$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : The Ladies' Diary or Woman's Almanac, : $141$
Category:Dutchmen's Three Wives