text:
\begin{document}

\title[On the Factorization of Two Adjacent Numbers]{On the Factorization of Two Adjacent Numbers in Multiplicatively Closed Sets Generated by Two
Elements}
\author[C.P. Anil Kumar]{C.P. Anil Kumar}
\address{No. 104, Bldg. 23, Lakshmi Paradise, 5th Main, 11th Cross, LN Puram, Bengaluru-560021}
\email{akcp1728@gmail.com}
\subjclass[2010]{Primary: 11A55, Secondary: 11K60}
\keywords{Multiplicatively Closed Sets, Continued Fractions, Primary and Secondary Convergents}
\date{\sc \today}
\begin{abstract}
For two natural numbers $1<p_1<p_2$, with $\ga=\frac{\log(p_1)}{\log(p_2)}$ irrational, we describe, in main Theorem~\ref{theorem:FactRectangles} and in Note~\ref{note:AdjFact}, the factorization of two adjacent numbers in the multiplicatively closed subset $S=\{p_1^ip_2^j\mid i,j\in \mbb{N}\cup\{0\}\}$ using primary and secondary convergents of $\ga$. This suggests general Question~\ref{ques:AdjFact} for more than two generators which is still open.
\end{abstract}
\maketitle
\section{\bf{Introduction}}
Continued fractions have been studied extensively in the theory of diophantine approximation. More so as a tool to prove results in this theory, for example, Hurwitz's theorem. A basic introduction to the theory of continued fractions is given in~\cite{MR0001185},~\cite{MR1083765}. A proof of Hurwitz's theorem is also mentioned in ~\cite{MR1083765} (Chapter $7$). The following is the question which this article concerns and we answer the question using the theory of continued fractions as a tool.

\begin{ques}
\label{ques:AdjacentNumberFactorization}
Let $S=\{1=s_0<s_1<s_2<\cdots\}\subs \mbb{N}$ be a multiplicatively closed set generated by two natural numbers $p_1<p_2$ such that $\frac{\log(p_1)}{\log(p_2)}$ is irrational.
Let $s_k=p_1^ip_2^j\in S$ for some $k\geq 0$. What are the factorizations of the adjacent numbers $s_{k-1},s_{k+1}$ in terms of $p_1,p_2,i,j$?
\end{ques}
We answer this Question~\ref{ques:AdjacentNumberFactorization} using the simple continued fraction expansion of $\frac{\log(p_1)}{\log(p_2)}$ and its primary and secondary convergents.
In C.~P.~Anil~Kumar~\cite{MR3943663} (Theorem $3.2$), a question on the existence of arbitrary large gaps in $S$ has been answered affirmatively by another constructive technique. Also in Article~\cite{MR3943663} ( Lemma $5.4$ and Note $5.5$) a formula for the next number of $p_2^j$ in the multiplicatively closed set $S$ has been found. Here we prove this result as Corollary~\ref{cor:ArbLargeInt} of main Theorem~\ref{theorem:FactRectangles}.

The following question for more than two generators is still open.
\begin{ques}
\label{ques:AdjFact}
Let $T=\{1=t_0<t_1<t_2<\cdots\}\subs \mbb{N}$ be a finitely generated multiplicatively closed
infinite set generated by positive integers $d_1,d_2,\cdots,d_n$ for $n>2$. Let $t_k=d_1^{i_1}d_2^{i_2}\cdots d_n^{i_n}$. How do we construct an explicit
factorization of the elements $t_{k-1},t_{k+1}\in T$ in terms of the positive integers $d_j,i_j, 1\leq j\leq n$?
\end{ques}
Now we proceed to mention some notation, a required definition and state main Theorem~\ref{theorem:FactRectangles}.
\begin{notation}
Throughout this article, let $0<p_1<p_2$ be two positive integers such that $\frac {\log p_1}{\log p_2}$ is irrational. Let $S=\{p_1^ip_2^j\mid i,j\in \mbb{N}\cup \{0\}\}=\{s_0=1<s_1<s_2<\cdots\}$ be the multiplicatively closed set generated by $p_1,p_2$.
\end{notation}
\begin{defn}[Non-negative Integer Co-ordinates of an Element]
	Any integer $n\in S$ can be uniquely expressible as $n=p_1^ip_2^j$. We associate the non-negative integer pair $(i,j)$ to $n$ which are the integer co-ordinates of the element $n$.
	So in particular there is a bijection (co-ordinatisation map) of $S$ with the grid $(\mbb{N}\cup \{0\})\times (\mbb{N}\cup \{0\})$.
\end{defn}
Main Theorem~\ref{theorem:FactRectangles} gives the decomposition of the factorization grids
\equa{&(\mbb{N}\cup \{0\})\times (\mbb{N}\cup \{0\})\text{ and }\\
	&\big((\mbb{N}\cup \{0\})\times (\mbb{N}\cup \{0\})\big)^{*}=\big(\mbb{N}\cup\{0\}\big)\times \big(\mbb{N}\cup\{0\}\big)\bs\{(0,0)\}}
into rectangles which are related by local translations to describe the factorization of the number and its next number in an elegant manner. Now we state the main theorem.
\begin{thmOmega}
	\namedlabel{theorem:FactRectangles}{$\Gom$}
	Let $\{a_0=0,a_1,a_2,\cdots,\}$ be the continued fraction of $\frac {\log p_1}{\log p_2}$. Let $h_0=0,k_0=1$ and let $\{\frac{h_i}{k_i}\mid i\in \mbb{N},gcd(h_i,k_i)=1\}$
	be the sequence of primary convergents of $\frac {\log p_1}{\log p_2}$.
	Consider the integer grid rectangles $\rectangle A^t_iB^t_iC^t_iD^t_i$ for $i\geq 1$ of dimensions $h_{2i}\times k_{2i}$ with co-ordinates given by
	\equa{	A^t_i&=\big(k_{2i-1}+tk_{2i},0\big),\\
			B^t_i&=\big(k_{2i-1}+(t+1)k_{2i}-1,0\big),\\
			C^t_i&=\big(k_{2i-1}+(t+1)k_{2i}-1,h_{2i}-1\big),\\
			D^t_i&=\big(k_{2i-1}+tk_{2i}, h_{2i}-1\big), 0\leq t< a_{2i+1}.}
	The corresponding translated rectangles (translation applied to each point) denoted by $\rectangle \ti{A^t_i}\ti{B^t_i}\ti{C^t_i}\ti{D^t_i}$ of next numbers in the multiplicatively closed set are given by
	\equa{	\ti{A^t_i}&=\big(0,h_{2i-1}+th_{2i}\big),\\
			\ti{B^t_i}&=\big(k_{2i}-1,h_{2i-1}+th_{2i}\big),\\
			\ti{C^t_i}&=\big(k_{2i}-1,h_{2i-1}+(t+1)h_{2i}-1\big),\\
			\ti{D^t_i}&=\big(0,h_{2i-1}+(t+1)h_{2i}-1\big),0\leq t< a_{2i+1}}
	again of the same dimensions $h_{2i}\times k_{2i}$ with translation given by \equ{\rectangle\ti{A^t_i}\ti{B^t_i}\ti{C^t_i}\ti{D^t_i}=\rectangle A^t_iB^t_iC^t_iD^t_i+\big(-k_{2i-1}-tk_{2i},h_{2i-1}+th_{2i}\big).}

	Now consider the integer grid rectangles $\rectangle P^t_iQ^t_iR^t_iS^t_i$ for $i\geq 0$ of dimensions $h_{2i+1} \times k_{2i+1}$ with co-ordinates given by
	\equa{	P^t_i&=\big(0,h_{2i}+th_{2i+1}\big),\\
			Q^t_i&=\big(k_{2i+1}-1,h_{2i}+th_{2i+1}\big),\\
			R^t_i&=\big(k_{2i+1}-1,h_{2i}+(t+1)h_{2i+1}-1\big),\\
			S^t_i&=\big(0,h_{2i}+(t+1)h_{2i+1}-1\big), 0\leq t< a_{2i+2}.}
	The corresponding translated rectangles (translation applied to each point) denoted by $\rectangle \ti{P^t_i}\ti{Q^t_i}\ti{R^t_i}\ti{S^t_i}$ of next numbers in the multiplicatively closed set are given by
	\equa{	\ti{P^t_i}&=\big(k_{2i}+tk_{2i+1},0\big),\\
			\ti{Q^t_i}&=\big(k_{2i}+(t+1)k_{2i+1}-1,0\big)\\
			\ti{R^t_i}&=\big(k_{2i}+(t+1)k_{2i+1}-1,h_{2i+1}-1\big),\\
			\ti{S^t_i}&=\big(k_{2i}+tk_{2i+1},h_{2i+1}-1\big), 0\leq t< a_{2i+2}}
	again of the same dimensions $h_{2i+1}\times k_{2i+1}$
	with translation given by \equ{\rectangle\ti{P^t_i}\ti{Q^t_i}\ti{R^t_i}\ti{S^t_i}=\rectangle P^t_iQ^t_iR^t_iS^t_i+\big(k_{2i}+tk_{2i+1},-h_{2i}-th_{2i+1}\big).}

	Also we have the grid
	\equa{\big(\mbb{N}\cup\{0\}\big)\times \big(\mbb{N}\cup\{0\}\big) &=\\
&\us{i\geq 1}{\bigsqcup}\bigg( \us{0\leq t< a_{2i+1}}{\bigsqcup} \rectangle A^t_iB^t_iC^t_iD^t_i \bigg)
		\us{i\geq 0}{\bigsqcup}\bigg( \us{0\leq t< a_{2i+2}}{\bigsqcup} \rectangle P^t_iQ^t_iR^t_iS^t_i \bigg)}
	and the grid of next numbers (hence origin deleted as next number cannot be origin)
	\equa{\big(\mbb{N}\cup\{0\}\big)&\times \big(\mbb{N}\cup\{0\}\big)\bs\{(0,0)\} =\\
		&\us{i\geq 1}{\bigsqcup }\bigg( \us{0\leq t< a_{2i+1}}{\bigsqcup} \rectangle \ti{A^t_i}\ti{B^t_i}\ti{C^t_i}\ti{D^t_i} \bigg)
		\us{i\geq 0}{\bigsqcup }\bigg( \us{0\leq t< a_{2i+2}}{\bigsqcup} \rectangle \ti{P^t_i}\ti{Q^t_i}\ti{R^t_i}\ti{S^t_i} \bigg).}
\end{thmOmega}
Here in the following note we mention briefly how to get the factorization of the previous number of $s_k\in S$.
\begin{note}
	\label{note:AdjFact}
	Using Theorem~\ref{theorem:FactRectangles} the factorization of the previous number can be obtained because previous number of the next number of a number is the given number. The answer to Question~\ref{ques:AdjacentNumberFactorization} can be obtained by suitably expressing $(i,j)$.

	For the next number express
	$s_k=(i,j)\in (\mbb{N}\cup \{0\})\times (\mbb{N}\cup \{0\})$ as $(k_{2l-1}+tk_{2l}+r,s)$ with $0\leq t<a_{2l+1},0\leq r<k_{2l},0\leq s<h_{2l}$ or as
	$(r,h_{2l}+th_{2l+1}+s)$ with $0\leq t<a_{2l+2},0\leq r<k_{2l+1},0\leq s<h_{2l+1}$. We get the next number $s_{k+1}$ using Theorem~\ref{theorem:FactRectangles}.

	For the previous number express $s_{k}=(i,j)\in (\mbb{N}\cup \{0\})\times (\mbb{N}\cup \{0\})\bs\{0,0\}$ as $(r,h_{2l-1}+th_{2l}+s)$ with $0\leq t<a_{2l+1},0\leq r<k_{2l},0\leq s<h_{2l}$
	or as $(r+k_{2l}+tk_{2l+1},s)$ with $0\leq t<a_{2l+2},0\leq r<k_{2l+1},0\leq s<h_{2l+1}$. We get the previous number $s_{k-1}$ again using Theorem~\ref{theorem:FactRectangles} in the reverse manner.
\end{note}
\section{\bf{Types of Fractions and Convergents associated to an Irrational in $[0,1]$}}
In this section we define various types of fractions and convergents associated to an irrational $\ga \in [0,1]$.

\subsection{Primary and Secondary Convergents}
\begin{defn}[Primary and Secondary Convergents]
Let $\ga \in [0,1]$ be an irrational. Suppose $\{a_0=0,a_1,a_2,a_3,\cdots,\}$ be the sequence denoting the simple continued fraction expansion of $\ga$, i.e.,
\equ{\ga=a_0+\frac 1{a_1+\frac 1{a_2+\frac 1{a_3+ \frac 1{\cdots}}}}.}
Let $h_0=0,k_0=1$. Define \equ{\frac{h_i}{k_i}=a_0+\frac 1{a_1+\frac 1{\ddots+\frac 1{a_i}}}, gcd(h_i,k_i)=1 \text{ for } i\in \mbb{N}.}
Then an element in the sequence $\{\frac{h_i}{k_i}: i\in \mbb{N}\cup\{0\}\}$ is called a primary convergent.
The first few primary convergents with relatively prime numerators and denominators are given by
\equ{\frac 01,\frac 1{a_1},\frac{a_2}{1+a_1a_2},\frac{1+a_2a_3}{a_3+a_1+a_1a_2a_3},\frac{a_2+a_4+a_2a_3a_4}{1+a_3a_4+a_1a_2+a_1a_4+a_1a_2a_3a_4},\cdots.}
By induction, we can show, with these expressions for $\frac{h_i}{k_i}$, that $h_ik_{i+1}-k_ih_{i+1}=\pm 1,i\in \mbb{N}\cup\{0\}$ as polynomials. Also we have as polynomials,
\equ{h_{i+2}=a_{i+2}h_{i+1}+h_i, k_{i+2}=a_{i+2}k_{i+1}+k_i.}
So we actually have polynomial expressions for $h_i,k_i$ in terms of variables $a_i:i\in \mbb{N}\cup\{0\}$ arising from continued fraction of the irrational $\ga$. For any irrational $\ga$
the convergents satisfy \equ{\frac{h_{2j}}{k_{2j}}<\frac{h_{2j+2}}{k_{2j+2}}<\ga < \frac{h_{2l+1}}{k_{2l+1}}<\frac{h_{2l-1}}{k_{2l-1}}, j\in \mbb{N}\cup \{0\}, l\in \mbb{N}.}

Now we define the finite monotonic sequences of new intermediate fractions with relatively prime numerators and denominators given by
\equ{\frac{h_{2j}}{k_{2j}}<\frac{h_{2j}+th_{2j+1}}{k_{2j}+tk_{2j+1}}<\frac{h_{2j+2}}{k_{2j+2}}, 0<t<a_{2j+2}, t,j\in \mbb{N}\cup\{0\}}
and
\equ{\frac{h_{2l+1}}{k_{2l+1}}<\frac{h_{2l-1}+th_{2l}}{k_{2l-1}+tk_{2l}}<\frac{h_{2l-1}}{k_{2l-1}}, 0<t<a_{2l+1}, t,l\in \mbb{N}.}
These new intermediate fractions are called secondary convergents.
\end{defn}

\subsection{Upper and Lower Fractions}
Here we define two sequences of fractions called upper and lower fractions associated to an irrational $\ga\in [0,1]$.

\begin{defn}[Upper and and Lower Fractions]
Let $0<\ga<1$. For $n\in \mbb{N}$, let \equ{f(n)=\bigg\lceil \frac n{\ga} \bigg\rceil, g(n)=\bigg\lfloor \frac n{\ga} \bigg\rfloor.}
Then the sequences $\{f(n):n\in \mbb{N}\},\{g(n):n\in \mbb{N}\}$ are called upper and lower
sequences of $\ga$ respectively. We have \equ{g(n)\ga < n < f(n)\ga, f(n)-g(n)=1, n\in \mbb{N}.} Since $0<\ga<1$ for any $n\in \mbb{N}$, we also observe that
\equ{\lfloor f(n)\ga \rfloor = n, \lceil g(n)\ga \rceil = n.}
A fraction in the sequence $\{\frac n{f(n)}:n\in \mbb{N}\}$ is called a lower fraction associated to $\ga$ and a fraction in the sequence $\{\frac n{g(n)}:n\in \mbb{N}\}$ is called an upper fraction
associated to $\ga$. We need not in general have $gcd(n,f(n))=1$ or $gcd(n,g(n))=1$. However we definitely have \equ{\frac {n}{f(n)}<\ga < \frac{n}{g(n)}.}
\end{defn}
\begin{note}
An element in the upper sequence gives a lower fraction and an element in the lower sequence gives an upper fraction associated to $\ga$.
\end{note}
\subsection{The Upper and Lower Sequences $f$ and $g$}
Here in this section we prove Theorems~\ref{theorem:UpLowSequence}~\ref{theorem:UpLowSeqDifferences} concerning the values of the upper and lower sequences.
\begin{theorem}
\label{theorem:UpLowSequence}
Let $\ga \in [0,1]$ be an irrational with continued fraction expansion $\{a_0=0,a_1,a_2,\cdots\}$.
Let $\{f(n):n\in \mbb{N}\},\{g(n):n\in \mbb{N}\}$ be the upper and lower sequences of $\ga$.
Let $h_0=0,k_0=1$. For $i\in \mbb{N}$, let $h_i,k_i$ be the numerator and denominator of $i^{th}$ primary convergent which are relatively prime.
Then we have
\equa{f(h_{2j}+th_{2j+1})&=k_{2j}+tk_{2j+1},\\
g(h_{2j}+th_{2j+1})&=k_{2j}+tk_{2j+1}-1, 0<t\leq a_{2j+2},j\in \mbb{N}\cup\{0\},t\in \mbb{N}\\
g(h_{2l-1}+th_{2l})&=k_{2l-1}+tk_{2l},\\
f(h_{2l-1}+th_{2l})&=k_{2l-1}+tk_{2l}+1,0 \leq t \leq a_{2l+1},t\in \mbb{N}\cup\{0\},l\in \mbb{N}.}
\end{theorem}
\begin{proof}
To prove the theorem it suffices to prove the following inequalities.
\equa{(k_{2j}+tk_{2j+1}-1)\ga<h_{2j}+th_{2j+1}&<(k_{2j}+tk_{2j+1})\ga,\\
	&0<t\leq a_{2j+2},j\in \mbb{N}\cup\{0\},t\in \mbb{N}\\
(k_{2l-1}+tk_{2l})\ga<h_{2l-1}+th_{2l}&<(k_{2l-1}+tk_{2l}+1)\ga,\\
&0\leq t \leq a_{2l+1},t,l\in \mbb{N}.}
This we prove by induction on $j,l$ simultaneously as follows.

We observe that
\equ{h_0=k_0-1=0,k_1\ga < h_1 \Ra (k_0+tk_1-1)\ga<h_0+th_1\text{ for all }t > 0.}
We also have for
\equ{0 \leq t \leq a_2, \frac{h_0+th_1}{k_0+tk_1}\leq \frac{h_2}{k_2}<\ga \Ra h_0+th_1 < (k_0+tk_1)\ga.} So
\equ{(k_0+tk_1-1)\ga<h_0+th_1<(k_0+tk_1)\ga, 0<t\leq a_2.}
We have
\equ{h_2<k_2\ga\text{ and }\frac 1{a_1+1}<\ga \Ra h_1 < (k_1+1)\ga.}
Hence \equ{\text{for all }t \geq 0, h_1+th_2<(k_1+tk_2+1)\ga.}
We also have
\equ{\text{for }0\leq t \leq a_3, \ga<\frac{h_3}{k_3}\leq \frac{h_1+th_2}{k_1+tk_2} \Ra (k_1+tk_2)\ga < h_1+th_2.}
So
\equ{(k_1+tk_2)\ga < h_1+th_2 < (k_1+tk_2+1)\ga, 0 \leq t \leq a_3.}
This proves the initial step of the induction for $j=0,l=1$.

Now assume that the inequalities follow for $j=r,l=r+1$ for some $r\in \mbb{N}$. We prove for $j=r+1,l=r+2$.
We have
\equ{k_{2r+3}\ga < h_{2r+3}\text{ and for }j=r,t=a_{2r+2},(k_{2r+2}-1)\ga < h_{2r+2}}
which together imply
\equ{(k_{2r+2}+tk_{2r+3}-1)\ga < h_{2r+2}+th_{2r+3}\text{ for }t\geq 0.}
We also have for
\equa{0\leq t\leq a_{2r+4}, \frac{h_{2r+2}+th_{2r+3}}{k_{2r+2}+tk_{2r+3}} &\leq \frac{h_{2r+4}}{k_{2r+4}}<\ga \Ra\\
h_{2r+2}+th_{2r+3}&<(k_{2r+2}+tk_{2r+3})\ga.}
So
\equ{(k_{2r+2}+tk_{2r+3}-1)\ga < h_{2r+2}+th_{2r+3}<(k_{2r+2}+tk_{2r+3})\ga.}
We have \equ{h_{2r+4}<k_{2r+4}\ga\text{ and for }l=r+1,t=a_{2r+3}, h_{2r+3}<(k_{2r+3}+1)\ga}
which together impy
\equ{h_{2r+3}+th_{2r+4}<(k_{2r+3}+tk_{2r+4}+1)\ga\text{ for all }t\geq 0.}
We also have for
\equa{0\leq t\leq a_{2r+5}, \ga < \frac{h_{2r+5}}{k_{2r+5}} &\leq \frac{h_{2r+3}+th_{2r+4}}{k_{2r+3}+tk_{2r+4}} \Ra \\
	(k_{2r+3}+tk_{2r+4})\ga &< h_{2r+3}+th_{2r+4}.}
So
\equ{(k_{2r+3}+tk_{2r+4})\ga<h_{2r+3}+th_{2r+4}<(k_{2r+3}+tk_{2r+4}+1)\ga, 0\leq t\leq a_{2r+5}.}
This proves the induction step for $j=r+1,l=r+2$.

Hence the theorem follows and the values of the sequences $f,g$ at the values of $n$ being the numerator of any primary or secondary convergent are known.
\end{proof}
\begin{note}
\label{note:FracPart}
Now we make an important observation about the monotonic nature of the ceil or rounding up fractional parts $h_{*}-k_{*}\ga$.
The numerators of the secondary and primary convergents associated to the lower sequence $g$ satisfy the following monotonicity.
\equ{h_1<h_1+h_2<\cdots<h_1+a_3h_2=h_3<h_3+h_4<\cdots<h_3+a_5h_4=h_5<\cdots.}
The sequence of differences is given by \equ{h_2,h_2,\cdots,h_2,h_4,h_4,\cdots,h_4,\cdots}
where $h_{2i}$ appears $a_{2i+1}$ times for $i\geq 1$. This sequence is non-decreasing and diverges to infinity.
Now we apply $g$ to the above sequence to obtain the denominators of the secondary and primary convergents associated to the lower sequence $g$ which also satisfy the following monotonicity.
\equ{k_1<k_1+k_2<\cdots<k_1+a_3k_2=k_3<k_3+k_4<\cdots<k_3+a_5k_4=k_5<\cdots.}
The sequence of differences is given by \equ{k_2,k_2,\cdots,k_2,k_4,k_4,\cdots,k_4,\cdots}
where $k_{2i}$ appears $a_{2i+1}$ times for $i\geq 1$. This sequence is non-decreasing and diverges to infinity.
The ceil fractional parts satisfy
\equa{h_1-k_1\ga>(h_1+h_2)&-(k_1+k_2)\ga>\cdots>h_3-k_3\ga>\\
	&(h_3+h_4)-(k_3+k_4)\ga>\cdots>h_5-k_5\ga>\cdots.}

Similarly we make an observation on the monotonic nature of the floor or usual fractional parts $k_{*}\ga-h_{*}$.
The numerators of the secondary and primary convergents associated to the upper sequence $f$ satisfy the following monotonicity.
\equ{h_0<h_0+h_1<\cdots<h_0+a_2h_1=h_2<h_2+h_3<\cdots<h_2+a_4h_3=h_4<\cdots.}
The sequence of differences  is given by \equ{h_1,h_1,\cdots,h_1,h_3,h_3,\cdots,h_3,\cdots}
where $h_{2i-1}$ appears $a_{2i}$ times for $i\geq 1$. This sequence is non-decreasing and diverges to infinity.
Now we apply $f$ to the above sequence to obtain the denominators of the secondary and primary convergents associated to the upper sequence $f$ which also satisfy the following monotonicity.
\equ{k_0+k_1<\cdots<k_0+a_2k_1=k_2<k_2+k_3<\cdots<k_2+a_4k_3=k_4<\cdots.}
The sequence of differences after including $k_0$ in the beginning is given by
\equ{k_1,k_1,\cdots,k_1,k_3,k_3,\cdots,k_3,\cdots}
where $k_{2i-1}$ appears $a_{2i}$ times for $i\geq 1$. This sequence is non-decreasing and diverges to infinity.
The floor fractional parts satisfy
\equa{k_0\ga>(k_0+k_1)\ga&-(h_0+h_1)>\cdots>k_2\ga-h_2>\\
	&(k_2+k_3)\ga-(h_2+h_3)>\cdots>k_4\ga-h_4>\cdots.}
\end{note}
Now we prove a useful lemma regarding fractions.
\begin{lemma}
\label{lemma:fracDetOne}
Let $\frac ab>\frac pq>\frac cd \geq 0$ be three fractions such that $ad-bc=1$. Then we have \equ{q>max(b,d).}
\end{lemma}
\begin{proof}
We have $\frac 1{bd}>\frac ab-\frac pq=\frac {aq-bp}{bq}$. If $aq-bp=1$ then $q>d$. If $aq-bp>1$ and $q\leq d$ then $\frac {aq-bp}{bq}\geq \frac {aq-bp}{bd}>\frac 1{bd}$ a contradiction.
Hence $q>d$.
We also have $\frac 1{bd}>\frac pq-\frac cd=\frac {pd-cq}{dq}$. If $pd-cq=1$ then $q>b$. If $pd-cq>1$ and $q\leq b$ then $\frac {pd-cq}{dq}\geq \frac {pd-cq}{bd}>\frac 1{bd}$ a contradiction.
Hence $q>b$.
So the lemma follows.
\end{proof}
We prove the second theorem.
\begin{theorem}
\label{theorem:UpLowSeqDifferences}
Let $\ga \in [0,1]$ be an irrational.
Let $\{f(n):n\in \mbb{N}\},\{g(n):n\in \mbb{N}\}$ be the upper and lower sequences of $\ga$.
Consider the two sequences of lower and upper fractions of primary and secondary convergents respectively.
\equa{\bigg\{\frac {p_1}{q_1}&<\cdots<\frac {p_i}{q_i}<\cdots\mid i\in \mbb{N}\bigg\}=\\
&\bigg\{\frac {h_0}{k_0}<\frac {h_0+h_1}{k_0+k_1}<\cdots <\frac{h_2}{k_2}<\frac {h_2+h_3}{k_2+k_3}<\cdots <\frac{h_4}{k_4}<\frac {h_4+h_5}{k_4+k_5}<\cdots\bigg\}.}
\equa{\bigg\{\frac {r_1}{s_1}&>\cdots>\frac {r_i}{s_i}>\cdots\mid i\in \mbb{N}\bigg\}=\\
&\bigg\{\frac {h_1}{k_1}>\frac {h_1+h_2}{k_1+k_2}>\cdots >\frac{h_3}{k_3}>\frac {h_3+h_4}{k_3+k_4}>\cdots >\frac{h_5}{k_5}>\frac {h_5+h_6}{k_5+k_6}>\cdots\bigg\}.}
Given a lower fraction $\frac{n}{f(n)}\nin \{\frac {p_j}{q_j}\mid j\in \mbb{N}\}$ there exists a lower fraction $\frac {p_i}{q_i}$ such that
\equ{n>p_i, f(n)>f(p_i)=q_i,f(n)\ga-n>q_i\ga-p_i.}
Given an upper fraction $\frac{n}{g(n)}\nin \{\frac {r_j}{s_j}\mid j\in \mbb{N}\}$ there exists an upper fraction $\frac {r_i}{s_i}$ such that
\equ{n>r_i,g(n)>g(r_i)=s_i,n-g(n)\ga>r_i-s_i\ga.}
\end{theorem}
\begin{proof}
Both the sequences $f,g$ are monotonically increasing and $\frac{h_i}{k_i}-\frac{h_{i+1}}{k_{i+1}}=\frac{(-1)^{i+1}}{k_ik_{i+1}}\lra 0$ as $i\lra \infty$.
Hence \equ{\us{i\lra \infty}{\lim} \frac{p_i}{q_i}=\us{i\lra \infty}{\lim} \frac{h_i}{k_i}= \us{i\lra \infty}{\lim} \frac{r_i}{s_i}=\ga.}
Now $0<\frac n{f(n)} <\ga$. So there exist two consecutive lower fractions $\frac{p_{i-1}}{q_{i-1}},\frac{p_i}{q_i}$ such that \equ{\frac{p_{i-1}}{q_{i-1}}<\frac n{f(n)} <\frac {p_i}{q_i}<\ga.}
Using Lemma~\ref{lemma:fracDetOne} we conclude that $f(n)>max(q_i,q_{i-1})=q_i>q_{i-1}$ since $p_iq_{i-1}-q_ip_{i-1}$ $=1$.
By monotonicity of $f$ we conclude that $n> max(p_i,p_{i-1})=p_i>p_{i-1}$.
Now we have \equ{f(n)\ga-n=f(n)(\ga-\frac n{f(n)})>f(n)(\ga-\frac {p_i}{q_i})>q_i(\ga-\frac {p_i}{q_i})=q_i\ga-p_i.}
Now we prove the other case.  Suppose  $\frac n{g(n)}>\frac {h_1}{k_1}=\frac 1{k_1}>\ga$. Since $n>1,g(n)>k_1$. Choose $\frac{r_i}{s_i}=\frac {h_1}{k_1}$ and we have
\equ{n-g(n)\ga=g(n)(\frac n{g(n)}-\ga)>g(n)(\frac {h_1}{k_1}-\ga)>k_1(\frac {h_1}{k_1}-\ga)=h_1-k_1\ga.}
Suppose $\frac{h_1}{k_1}> \frac n{g(n)} > \ga$. Then there exist two consecutive upper fractions $\frac {r_{i-1}}{s_{i-1}},\frac {r_i}{s_i}$ such that
\equ{\frac {r_{i-1}}{s_{i-1}}>\frac n{g(n)}>\frac {r_i}{s_i}>\ga.}
Again using Lemma~\ref{lemma:fracDetOne} we conclude that $g(n)>max(s_i,s_{i-1})=s_i>s_{i-1}$ since $s_ir_{i-1}-r_is_{i-1}=1$.
By monotonicity of $g$ we conclude that $n>max(r_i,r_{i-1})=r_i>r_{i-1}$.
Now we have \equ{n-g(n)\ga=g(n)(\frac n{g(n)}-\ga)>g(n)(\frac {r_i}{s_i}-\ga)>s_i(\frac {r_i}{s_i}-\ga)=r_i-s_i\ga.}
This proves the theorem.
\end{proof}
\begin{note}
\label{note:MinFracPart}
In Theorem~\ref{theorem:UpLowSeqDifferences} if $\frac{n}{f(n)}=\frac{p_i}{q_i}$ then there exists a positive integer $k$ such that $n=kp_i,f(n)=kq_i$.
So if $k>1$ the we have $n>p_i,f(n)>q_i,f(n)\ga-n>q_i\ga-p_i$. The other case is similar. So we conclude that minimal fractional parts occur exactly at the numerators
and denominators of primary and secondary convergents using Note~\ref{note:FracPart} in the sequences of lower and upper fractions of $\ga$ which is the content of the statement of Theorem~\ref{theorem:FracPart}.
\end{note}
\section{\bf{The proof of the main theorem}}
\label{sec:FactRectangles}
We begin this section with a theorem which is an observation on the rounding up or ceil fractional parts of the lower sequence and the usual fractional parts of the upper sequence.

\begin{theorem}
\label{theorem:FracPart}
Let $\ga\in [0,1]$ be an irrational. Let $f,g$ be the upper and lower sequences associated to $\ga$. Let $z_0=\ga$ and for $n\in \mbb{N}$ let $z_n=-n+f(n)\ga,y_n=n-g(n)\ga$.
Let $n_0=0,z_{n_0}=z_0=\ga,m_1=1,y_{m_1}=y_1=1-g(1)\ga$. Define two subsequences $z_{n_j},y_{m_j}$
with the property that
\equa{z_{n_j}<z_{n_{j-1}}&=\min \{z_0,\cdots, z_{n_j-1}\} \text{ for }j\in \mbb{N},\\
y_{m_j}<y_{m_{j-1}}&=\min \{y_1,\cdots, y_{m_j-1}\} \text{ for }1<j\in \mbb{N}.}
Then the sequence \equa{\{n_0<n_1&<\cdots<n_j<\cdots\mid j\in \mbb{N}\cup\{0\}\}=\\
&\{h_0<h_0+h_1<\cdots<h_0+a_2h_1=h_2<h_2+h_3<\cdots<\\
&h_2+a_4h_3=h_4<h_4+h_5<\cdots\}} and the sequence
\equa{\{m_1<m_2&<\cdots<m_j<\cdots\mid j\in \mbb{N}\}=\\
&\{h_1<h_1+h_2<\cdots<h_1+a_3h_2=h_3<h_3+h_4<\cdots<\\
&h_3+a_5h_4=h_5<h_5+h_6<\cdots\}.}
\end{theorem}
\begin{proof}
This theorem follows by applying Theorem~\ref{theorem:UpLowSeqDifferences} and Notes~\ref{note:FracPart},~\ref{note:MinFracPart} which together imply
that the lesser fractional parts in the sequence occur exactly at numerators of primary and secondary convergents of the lower and upper fractions for
the upper and lower sequences respectively.
\end{proof}
Now we prove main Theorem~\ref{theorem:FactRectangles}.
\begin{proof}
Let $\ga=\frac {\log p_1}{\log p_2}\in \mbb{R}\bs \mbb{Q}$.
Consider a point \equ{(k_{2i-1}+tk_{2i}+r,s)\in \rectangle A^t_iB^t_iC^t_iD^t_i} and its next number \equ{(r,h_{2i-1}+th_{2i}+s)\in \rectangle \ti{A^t_i}\ti{B^t_i}\ti{C^t_i}\ti{D^t_i}}
for any \equ{0\leq t<a_{2i+1},0\leq r <k_{2i},0\leq s< h_{2i}.} Now suppose there exists an integer $p_1^bp_2^a$ such that
\equ{p_1^{k_{2i-1}+tk_{2i}+r}p_2^{s}<p_1^bp_2^a<p_1^{r}p_2^{h_{2i-1}+th_{2i}+s}.}
Then we arrive at a contradiction as follows.

The following sequence of inequalities hold
\equ{\big(k_{2i-1}+tk_{2i}+r\big)\ga+s<b\ga+a<r\ga+h_{2i-1}+th_{2i}+s<\big(k_{2i-1}+tk_{2i}+r+1\big)\ga+s}
because $\ga=\frac{\log p_1}{\log p_2}$.
If $b=r$ then there exist two integers $a-s,h_{2i-1}+th_{2i}$ in between \equ{\big(k_{2i-1}+tk_{2i}\big)\ga,\big(k_{2i-1}+tk_{2i}+1\big)\ga} which is a contradiction.

If $b<r$ then we must have $a>s$ so that $a-s>0$. Hence
\equ{\big(k_{2i-1}+tk_{2i}+r-b\big)\ga<a-s<h_{2i-1}+th_{2i}+(r-b)\ga<\big(k_{2i-1}+tk_{2i}+r-b+1\big)\ga.}
So let $0<z<\ga$ be the positive rounding up ceil fractional part such that
\equ{\big(k_{2i-1}+tk_{2i}+r-b\big)\ga+z=a-s.}
Now suppose $k_{2i-1}+tk_{2i}+r-b<k_{2i-1}+(t+1)k_{2i}$. Then the fractional part $z$ has to satisfy
\equ{z\geq h_{2i-1}+th_{2i}-\big(k_{2i-1}+tk_{2i}\big)\ga}
because minimal fractional parts occur exactly at the numerators and denominators of primary and secondary convergents.
On the other hand \equa{z=(a-s)&-\big(k_{2i-1}+tk_{2i}+r-b\big)\ga<\\
	&h_{2i-1}+th_{2i}+(r-b)\ga-\big(k_{2i-1}+tk_{2i}+r-b\big)\ga=\\
	&h_{2i-1}+th_{2i}-\big(k_{2i-1}+tk_{2i}\big)\ga}
which is a contradiction.
So we have \equ{k_{2i-1}+tk_{2i}+r-b\geq k_{2i-1}+(t+1)k_{2i}\Ra r-b \geq k_{2i} \Ra r\geq k_{2i}+b\Ra r\geq k_{2i}}
which is again a contradiction to $0\leq r\leq k_{2i}-1$.

Now consider the case $b>r$ so that $b-r>0$. We have
\equ{\big(k_{2i-1}+tk_{2i}\big)\ga+s-a<(b-r)\ga<h_{2i-1}+th_{2i}+s-a<\big(k_{2i-1}+tk_{2i}+1)\ga+s-a.}

Let $0<y<\ga$ be the rounding up ceil fractional part such that
\equ{(b-r)\ga+y=h_{2i-1}+th_{2i}+s-a.}
Now suppose $h_{2i-1}+th_{2i}+s-a<h_{2i-1}+(t+1)h_{2i}$. Then the fractional part $y$ has to satisfy
\equ{y\geq h_{2i-1}+th_{2i}-\big(k_{2i-1}+tk_{2i}\big)\ga}
because minimal fractional parts occur exactly at the numerators and denominators of primary and secondary convergents.
On the other hand \equa{y&=h_{2i-1}+th_{2i}+s-a-(b-r)\ga<\\
	&h_{2i-1}+th_{2i}+s-a-\big((k_{2i-1}+tk_{2i})\ga+s-a\big)=\\
&h_{2i-1}+th_{2i}-\big(k_{2i-1}+tk_{2i}\big)\ga}
which is a contradiction. So we have
\equ{h_{2i-1}+th_{2i}+s-a \geq h_{2i-1}+(t+1)h_{2i} \Ra s-a\geq h_{2i} \Ra s\geq h_{2i}+a \Ra s\geq h_{2i}}
which is again a contradiction to $0\leq s\leq h_{2i}-1$.

This proves that the next number of $p_1^{k_{2i-1}+tk_{2i}+r}p_2^{s}$ is $p_1^{r}p_2^{h_{2i-1}+th_{2i}+s}$ for
\equ{0\leq t<a_{2i+1},0\leq r <k_{2i},0\leq s<h_{2i}.}

Now we consider the second set of rectangles. Consider a point
\equ{(r,h_{2i}+th_{2i+1}+s)\in \rectangle P^t_iQ^t_iR^t_iS^t_i} and its next number \equ{(r+k_{2i}+tk_{2i+1},s)\in \rectangle \ti{P^t_i}\ti{Q^t_i}\ti{R^t_i}\ti{S^t_i}}
for any \equ{0\leq t<a_{2i+2},0\leq r<k_{2i+1},0\leq s< h_{2i+1}.} Now suppose there exists an integer $p_1^bp_2^a$ such that
\equ{p_1^{r}p_2^{h_{2i}+th_{2i+1}+s}<p_1^bp_2^a<p_1^{r+k_{2i}+tk_{2i+1}}p_2^{s}.}
Then we arrive at a contradiction as follows in a similar manner.

The following sequence of inequalities hold
\equ{\big(k_{2i}+tk_{2i+1}+r-1\big)\ga+s<r\ga+h_{2i}+th_{2i+1}+s<b\ga+a<\big(k_{2i}+tk_{2i+1}+r\big)\ga+s}
because $\ga=\frac{\log p_1}{\log p_2}$. This implies
\equ{\big(k_{2i}+tk_{2i+1}+r-b-1\big)\ga<h_{2i}+th_{2i+1}+(r-b)\ga<a-s<\big(k_{2i}+tk_{2i+1}+r-b\big)\ga.}
If $a=s$ then we observe that $\big(k_{2i}+tk_{2i+1}+r-b\big)\ga\geq \ga$ and $\big(k_{2i}+tk_{2i+1}+r-b-1\big)\ga\leq -\ga$ which is a contradiction.
If $a>s$ so that $a-s>0$ then let $0<x<\ga$ be the positive fractional part such that
\equ{(a-s)+x=\big(k_{2i}+tk_{2i+1}+r-b\big)\ga.}
Now suppose $\big(k_{2i}+tk_{2i+1}+r-b\big)<k_{2i}+(t+1)k_{2i+1}$. Then the fractional part $x$ has to satisfy
\equ{x\geq (k_{2i}+tk_{2i+1})\ga-(h_{2i}+th_{2i+1})}
because minimal fractional parts occur exactly at the numerators and denominators of primary and secondary convergents.
On the other hand
\equa{x&=\big(k_{2i}+tk_{2i+1}+r-b\big)\ga-(a-s)<\\
	&\big(k_{2i}+tk_{2i+1}+r-b\big)\ga-\big(h_{2i}+th_{2i+1}+(r-b)\ga\big)=\\
&(k_{2i}+tk_{2i+1})\ga-(h_{2i}+th_{2i+1})}
which is a contradiction. So we have
\equa{\big(k_{2i}+tk_{2i+1}+r-b\big)&\geq k_{2i}+(t+1)k_{2i+1} \Ra\\
	 r-b&\geq k_{2i+1}\Ra r\geq k_{2i+1}+b\Ra r\geq k_{2i+1}}
which is a contradiction to $0\leq r\leq k_{2i+1}-1$.

Now suppose $a<s$ so that $s-a>0$. Then we have
\equ{\big(k_{2i}+tk_{2i+1}-1\big)\ga+s-a<h_{2i}+th_{2i+1}+s-a<(b-r)\ga<\big(k_{2i}+tk_{2i+1}\big)\ga+s-a.}
So $b-r>0$ and let $0<u<\ga$ be the positive fractional part such that
\equ{h_{2i}+th_{2i+1}+s-a+u=(b-r)\ga.}
Now suppose $h_{2i}+th_{2i+1}+s-a<h_{2i}+(t+1)h_{2i+1}$. Then the fractional part $u$ has to satisfy
\equ{u\geq (k_{2i}+tk_{2i+1})\ga-(h_{2i}+th_{2i+1})}
because minimal fractional parts occur exactly at the numerators and denominators of primary and secondary convergents.
On the other hand
\equa{	u&=(b-r)\ga-(h_{2i}+th_{2i+1}+s-a)<\\
		&\big(k_{2i}+tk_{2i+1}\big)\ga+s-a-(h_{2i}+th_{2i+1}+s-a)=\\
		&(k_{2i}+tk_{2i+1})\ga-(h_{2i}+th_{2i+1})}
which is a contradiction. So we have
\equa{h_{2i}+th_{2i+1}+s-a&>h_{2i}+(t+1)h_{2i+1}\Ra\\
	s-a&\geq h_{2i+1}\Ra s\geq h_{2i+1}+a \Ra s\geq h_{2i+1}}
which is again a contradiction to $0\leq s\leq h_{2i+1}-1$.

This proves that the next number of $p_1^{r}p_2^{h_{2i}+th_{2i+1}+s}$ is $p_1^{r+k_{2i}+tk_{2i+1}}p_2^{s}$ for
\equ{0\leq t<a_{2i+2},0\leq r <k_{2i+1},0\leq s<h_{2i+1}.}

Now we prove that the rectangles cover the grid $(\mbb{N}\cup\{0\})\times(\mbb{N}\cup\{0\})$. For this we need to prove that given
$(x,y)\in (\mbb{N}\cup\{0\})\times(\mbb{N}\cup\{0\})$ either there exists $j \geq 0$ such that
\equ{h_{2j}+th_{2j+1} \leq y < h_{2j}+(t+1)h_{2j+1}, 0 \leq x < k_{2j+1}\text{ for some }0\leq t<a_{2j+2}}
or there exists $i\geq 1$ such that
\equ{k_{2i-1}+\ti{t}k_{2i}\leq x <k_{2i-1}+(\ti{t}+1)k_{2i},0\leq y<h_{2i} \text{ for some }0\leq \ti{t}<a_{2i+1}}

Now there always exist $j\geq 0,0\leq t<a_{2j+2}$ such that \equ{h_{2j}+th_{2j+1} \leq y < h_{2j}+(t+1)h_{2j+1}.}
If $0 \leq x < k_{2j+1}$ then we are done. Otherwise $x\geq k_{2j+1}$. Hence there exist $i\geq j+1, 0\leq \ti{t}<a_{2i+1}$ such that
\equ{k_{2i-1}+\ti{t}k_{2i}\leq x <k_{2i-1}+(\ti{t}+1)k_{2i}.} Now we have \equ{0\leq h_{2j}+th_{2j+1} \leq y < h_{2j}+(t+1)h_{2j+1}\leq h_{2j+2}\leq h_{2i} \Ra 0\leq y<h_{2i}.}
So the rectangles cover the grid. The rest of the proof for the grid with origin deleted is similar.

This completes the proof of Theorem~\ref{theorem:FactRectangles}.
\end{proof}
We mention the following corollary of Theorem~\ref{theorem:FactRectangles} with a very brief proof as it is straight forward.
\begin{cor}
\label{cor:ArbLargeInt}
Let $S$ be a multiplicatively closed set with two generators $1<p_1<p_2$ such that $\frac{\log(p_1)}{\log(p_2)}$ is irrational.  Then there exist arbitrarily large
gap integer intervals of $S$ with end points in $S$.
\end{cor}
\begin{proof}
To obtain arbitrarily large gap intervals, we apply Theorem~\ref{theorem:FactRectangles} for the largest values of $r\in \{k_{2i}-1,k_{2i+1}-1\}, s\in\{h_{2i}-1,h_{2i+1}-1\}$
which tend to infinity as $i$ tends to infinity.
\end{proof}


TITLE: Factor trinomials dividing by the common GCF
QUESTION [0 upvotes]: I have a doubt with the following problem I found in a book. You have to simplify a polynomial using the GCF. Now, this is the problem I am not able to grasp:
$$6x^2-19x-7$$
According to the book, the solution is
$$(2x-7)(3x+1)$$
And, in fact, if you multiply those two terms you'll get:
$$6x^2-21x+2x-7=6x^2-19x-7$$
So it is clear that both expressions are equivalent. What I fail to grasp is which method they have followed to come from $6x^2-19x-7$ to $(2x-7)(3x+1)$.
Thanks in advance,

REPLY [0 votes]: Ok, I think they have actually used the AC method, that's the one that makes more sense here.
For a general explanation of such method:
http://www.virtualnerd.com/algebra-1/polynomials-and-factoring/trinomial-factorization/leading-coefficient-not-1/factor-by-a-c-method
So, for this case we have:
$$6x^2-19x-7$$
Hence:
$$a=6, b=-19, c=-7$$
So:
$$ac = -42$$
Then, we calculate the factors of 42 and their sum following the AC method (I tried to use a Latex table but it is a pain in the ass, so sorry about the format.):
$$1\times-42; 1+(-42)=41$$
$$2\times-21; 2+(-21)=-19$$
As $b=-19$ we can stop right there. Our factors are 2 and -21, then $b=2x-21x$
Now, rewriting the trinome with such factors for b, we have then:
$$6x^2+2x-21x-7$$
Now we factor by grouping by pairs:
$$2x(3x+1)-7(3x+1)$$
And now, we have a $(3x+1)$ common factor there, so:
$$(3x+1)(2x-7)$$
Tadaa!
PS. It is worth to add that the book itself ask explicitly to do the factorization dividing by the GCF –hence the question title–, but that's clearly impossible in this case, so it is probably an errata.


TITLE: strong law of large numbers when mean goes to infinity
QUESTION [1 upvotes]: Let $X_1,X_2,...,X_n$ be independent and identically distributed (iid) random variables with mean $\mathbb{E}[X_i] = \sqrt{n}$. I am a bit confused about using the strong law of large numbers (SLLN) in this case, because $\mathbb{E}[X_i] = \sqrt{n}$ goes to infinity as $n\to\infty$. I wonder if the assumptions of SLLN holds here and whether we can conclude that
$$
\frac{\sum_{i=1}^n X_i}{n} \to \infty
$$
almost surely or not?
I appreciate any response/comment.

REPLY [2 votes]: The strong law of large numbers requires that you have an infinite sequence of random variables, from which you can average any number of terms you like. This gives us an infinite sequence
$$
   X_1, \frac{X_1 + X_2}{2}, \frac{X_1 + X_2 + X_3}{3}, \dots
$$
and the strong law of large numbers is about the limit of this sequence.
Here, you have a finite sequence of random variables, and you aren't just extending it as you increase $n$: if you change $n$, the distributions also change, because the expected value changes. So the strong law of large numbers doesn't apply to this situation.
For an example where the conclusion also does not hold, suppose that each of $X_1, \dots, X_n$ is $n^2$ with probability $\frac1{n\sqrt n}$ and $0$ otherwise. Then with probability $(1 - \frac1{n \sqrt n})^n \ge 1 - \frac1{\sqrt n}$, we have $X_1 = X_2 = \dots = X_n = 0$. As a result, $\frac{\sum_{i=1}^n X_i}{n}$ converges in probability to $0$, not $\infty$. (It does not almost surely converge to anything because, in particular, we don't have a common sample space for different values of $n$.)


TITLE: Probability and Statistics Expected Weekly Loss
QUESTION [0 upvotes]: So I have this exercise (picture below) and I have the mean calculated and the variance of the  random variable X, so I'm left with the formula for Loss, should I simplu substitute the X in the E(10X + 200) with the value of E(X) below? (1.904) in which case it would be 219.04.
Same for the variance?

REPLY [1 votes]: Yes, the expectation of $aX+b$ is $aE(X)+b$ and the variance of $aX+b$ is $a^2$ times the variance of $X$.
Remark: I suggest recomputing $E(X)$ and $\text{Var}(X)$.


TITLE: Does global choice allow to extend the natural well-order on On to a well-order of the whole universe
QUESTION [3 upvotes]: Suppose that T is a set theory with global choice (for example ZF+ global choice).
Question: Does T prove the existence of a well-order on the whole universe V whose restriction to the class On of ordinals is the natural well-order on ON ?
Gérard Lang

REPLY [6 votes]: Yes, use global choice to choose, for each ordinal $\alpha$, a well-ordering $<_\alpha$ of the sets of rank $\alpha$ in the cumulative hierarchy.  Then well-order all the sets by putting $x$ before $y$ if either $x$ has lower rank than $y$ or they have the same rank $\alpha$ and $x<_\alpha y$.


TITLE: Prove that the column space of $AC$ is contained in the column space of $A$.
QUESTION [2 upvotes]: Prove that the column space of $AC$ is contained in the column space of $A$.
The only way I can think of is to write out column vectors of $AC$, and notice that the column vectors of $AC$ are linear combinations of column vectors of $A$. (Or, we don't even have to "write out" to notice that, by definition column vectors of $AC$ are indeed linear combinations of column vectors of $A$.) Therefore, the column space of $AC$ is contained in the column space of $A$.
First of all, is this proof correct?
Second, is there a better way to prove it?
Thanks!

REPLY [4 votes]: Your idea for a proof is good. A second approach which may be faster is the following. The column space of matrix $A$ is:
\begin{align}
\text{Col}(A)=\{ b\in V | \exists x\in W: Ax=b\}
\end{align}
Where $V$ and $W$ are some vector spaces, for example $V=\mathbb{R}^n$. Let $b\in \text{Col}(AC)$ then there exists $x$ such that $ACx=b$. Take $y=Cx$ so there exists $y$ such that $Ay=b$ hence $b\in \text{Col}(A)$. And that means $\text{Col}(AC)\subset \text{Col}(A)$.

REPLY [2 votes]: It's a an absolutely correct answer.


TITLE: Asymptotic formula for “Self Numbers” $s_n$
QUESTION [5 upvotes]: While digging through some old notebooks today, I found a problem from a long time ago that I was never able to solve. It involves a sequence of positive integers called the “self numbers” defined such that $s_n$ is the nth positive integer for which the equation
$$k+\text{digit sum of }k = s_n$$
has no solution (this is is base 10, by the way). This sequence is A003052 on OEIS.
It appears that this sequence grows asymptotically linearly, and I tried (but failed) to find the asymptotic rate at which it grows:
$$r=\lim_{n\to\infty}\frac{s_n}{n}$$
Can anyone figure out how to calculate $r$, even in series or integral form?
Note: The best I was able to prove is that $r < 25/2$.

REPLY [3 votes]: U. Zannier in his paper `On the Distribution of Self-Numbers' proves a theorem about the number of self numbers less than $x$, specifically that $A(x)= Lx + O(\log^2(x))$ for some positive $L$. You can invert this function to get a statement about the $n$th self number growing linearly (in the same way that one can invert the prime number theorem $\pi(x) \sim x/\log(x)$ into $p_n\sim n\log(n)$, being careful with analytic asymptotic inverses). I'm not sure if an exact way to compute the constant is mentioned but that would be somewhere to look to get a sense of how to do what you want to do. His $L$ for decimal (or what is referred to in the paper as scale $g=10$) appears to be $\approx 0.0977$, which you invert to get the $r$ in your formula to be $\approx 10.24$. I don't know if there's a nice analytic form.


TITLE: Calculate marginal distribution, $P(X=0|Y > 0)$, expected value and variance.
QUESTION [3 upvotes]: Let $X$ be a random variable with values $0$ and $1$.
Let $Y$ be  a random variable with  values in $\mathbb{N_0}$.
Let  $ p \in (0,1)$ and $ P(X=0, Y=n) = p \cdot \frac{e^{-1}}{n!} $ and $ P(X=1, Y=n) = (1-p) \cdot \frac{2^{n}e^{-2}}{n!} $.
a) Calculate the marginal distribution $p_x$ and $p_y$.
b) Calculate $P$($X=0$ | $Y > 0$)
c) Calculate $E(Y)$ and $Var(Y)$. (expected value and variance)
I think that we have to start with $p_x = \sum_y P(X=x,Y=y)= \sum_y P(X=x|Y=y) \cdot P(Y=y) $, because this is formula for marginal distribution? Edit a) and b) are clear now.
for c): So We want to use the law of total expectation. $$ E(Y) = E(E(Y|X))= \sum_xE(Y|X=x) \cdot P(X=x)= E(Y|X=0) \cdot p + E(Y|X=1) \cdot (1-p)= \sum_{n=1} n\cdot P(Y=n|X=0)  \cdot p+ \sum_{n=1} n\cdot P(Y=n|X=1)\cdot (1-p) = \sum_{n=1} n\cdot \frac{e^{-1}}{n!} \cdot p +\sum_{n=1} n\cdot \frac{e^{-2}\cdot 2^n}{n!} \cdot(1-p)= e^{-1} \cdot e \cdot p + 2\cdot(1-p)\cdot e^{-2} \cdot \sum_{n=1} \frac{2^{n-1}}{(n-1)!} = p + 2\cdot (1-p) e^{-2} \cdot e^2= p+2 \cdot (1-p)= 2-p $$

REPLY [2 votes]: First, you can compute $P(X=0)$ using the law of total probability.
\begin{align*}
 P(X=0) &= \sum_{n}P(X=0,Y=n) \\
        &= p \sum_{n}{\frac{\exp(-1)}{n!}}
        = p
\end{align*}
Next, note that
\begin{align*}
 P(Y=n|X=0) &= \frac{P(X=0,Y=n)}{P(X=0)} \\
            &= \frac{\exp(-1)}{n!}
\end{align*}
That is $Y|X=0 \sim \text{Poisson}(1)$. Also,
\begin{align*}
P(Y=n|X=1) &= \frac{P(X=1,Y=n)}{P(X=1)} \\
           &= \frac{\exp(-2)2^n}{n!}
\end{align*}
That is, $Y|X=1 \sim \text{Poisson}(2)$.
a) $X \sim \text{Bernoulli}(p)$ and $Y \sim p \cdot \text{Poisson}(1) + (1-p) \cdot \text{Poisson}(2)$.
b) Your equality is incorrect. Instead, you can use Bayes' Theorem:
\begin{align*}
 P(X=0|Y>0)
 &=\frac{P(X=0)P(Y>0|X=0)}{P(X=0)P(Y>0|X=0)+P(X=1)P(Y>0|X=1)}
\end{align*}
c) You can use the conditional distribution of $Y$ to compute $E[Y]$ and $Var[Y]$, that is:
\begin{align*}
 E[Y]
 &= E[E[Y|X]]
 & \text{Law of total expectation} \\
 Var[Y]
 &= Var[E[Y|X]] + E[Var[Y|X]]
 & \text{Law of total variation} \\
\end{align*}


TITLE: How the order of $G/2G$ is $2^n$?
QUESTION [1 upvotes]: Let $G $ be an abelian group. In this section we shall use additive notation
for $G$, so the group operation will be denoted by ' $+$ ' the identity by $0$,
the inverse of $g$ by $-g$ and powers of $g$ by $2g, 3g, ....$
Let $2G$ be the subgroup of $G$ consisting of all elements of the form $g + g$ $(g \in  G)$. If $G$ has a basis of $n$ elements, then $G/2G$ is a group of order $2^n$.
How the order of $G/2G$ is  $2^n$? Plz show in an elementary way.
This an excerpt from "Algebraic number theory and Fermats last theorem by  Ian Stewart, David Tall (3rd ed.).

REPLY [2 votes]: "... G has a basis of $n$ elements..." to me implies that $G$ is a free Abelian group of rank $n$, i.e. there is a basis made of $n$ linearly independent generators $g_1, g_2,\ldots,g_n\in G$; every element of $G$ can be uniquely represented as $m_1g_1+m_2g+\ldots+m_ng_n$ where $m_i\in\mathbb Z$ are "co-ordinates" in this basis.
(Recall we use the same terms in linear algebra, but the coefficients there come from a field, while here they come from the ring $\mathbb Z$ which is not a field. This makes calculations slightly more complicated.)
Now, note $2G$ is the subset of those elements of $G$ which have even co-ordinates, i.e. all $m_i$'s are even. That means that $G/2G$ is made up of "cosets" of the form $\lambda_1g_1+\ldots+\lambda_ng_n+2G$ where $\lambda_i\in\{0,1\}$. (Every $m_1g_1+\ldots+m_ng_n$ belongs to the coset $\lambda_1g_1+\ldots+\lambda_ng_n+2G$ where $\lambda_i =\text{ remainder in dividing }m_i\text{ by }2$). Thus, the number of those cosets (which is the order of $G/2G$) is $2^n$.
Hope this helps!


TITLE: Meaning of the phrase "Line $Y$ spanned by $\vec x$" and "Plane $D$ spanned by $\vec x$, $\vec y$, and $\vec z$"
QUESTION [0 upvotes]: If I say the Line $Y$ spanned by $\vec x$ in $\mathbb{R}^2$ = $\begin{bmatrix}3 \\2\end{bmatrix}$, then do I mean that $\vec x$ is parallel or perpendicular to Line $Y$?
If I say the Plane $D$ spanned by $\vec x$  = $\begin{bmatrix}a \\b\\c\end{bmatrix}$, $\vec y$ = $\begin{bmatrix}d \\e\\f\end{bmatrix}$, and $\vec z$ = $\begin{bmatrix}g \\h\\i\end{bmatrix}$  then do I mean that $\vec x$, $\vec y$, and $\vec z$  are parallel or perpendicular to Plane $D$?

REPLY [1 votes]: First question
The vector $\begin{bmatrix} a \\ b \end{bmatrix}$ is the vector that starts at the origin $\big($in $\mathbb{R}^2$, this is the point $(0, 0)$ $\big)$ and ends at the point $(a, b)$. Therefore, we identify a vector with its ending point.
The line spanned by $\vec x = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ is formed by the endpoints of the vectors $\lambda \vec x, \lambda \in \mathbb{R}$ (the real multiples of $\vec x$). For example, the point $(3\sqrt{2}, 2\sqrt{2})$ is in the line because it is the endpoint of $\sqrt{2} \vec x$. Just drawing a few of these endpoints one can see that $\vec x$ is parallel (actually coincides on) the line spanned by $\vec x$ itself.
Second question
Two vectors $v_1, v_2$ span (or generate) a plane consisting of points of the form $\lambda_1 v_1 + \lambda_2 v_2$. (An exception to this is when $v_1$ and $v_2$ lie on the same line, when $v_2$ can be written as $\lambda' v_1$).
In general, $n$ possibly (but not necessarily) different vectors $v_1, \dots, v_n$ generate a set consisting of points of the form $\lambda_1 v_1+ \dots+ \lambda_nv_n$ (the so-called linear combinations) with $\lambda_1, \dots, \lambda_n \in \mathbb{R}$.
Back to the question, a third vector $v_3$ may not lie on the plane spanned by $v_1, v_2$ (see the $x, y, z$-axis example in the comments of the question). If $v_3$ does lie on the plane, that means for some $\mu_1, mu_2$ it holds that $v_3=\mu_1v_1+\mu_2v_2$. Then what is spanned by $v_1, v_2$ and $v_3$ is of the form
$$\lambda_1v_1+\lambda_2v_2+\lambda_3v_3 = \lambda_1v_1+\lambda_2v_2+\lambda_3(\mu_1v_1+\mu_2v_2)$$
which is in what is spanned by only $v_1,v_2$ (now you can read about linear dependency)


TITLE: Choosing axiom schemes for a logical theory
QUESTION [2 upvotes]: In a Hilbert system, there are many ways that we can choose axiom schemes. My question is:
1- How do we know that we have defined enough schemes? What would happen if I remove a scheme from the list?
2- How do we decide on the form of these schemes?
3- How can we determine that a particular choice of axioms results in a consistent theory?
Please also let me know if you know a source that discusses such issues.

REPLY [2 votes]: I think you are talking about adding axiom schemes to the deductive system. The ultimate goal for the deductive system is soundness and completeness: a sentence should be provable if and only if it is true in every model. So there are "enough" schemes when this is provable for the system that has been defined.
If you remove schemes from a typical system, the resulting system is not complete - typical systems in books don't include redundant schemes. But it would be possible to include redundant axiom schemes if you wanted.
There is no algorithm to decide on the form of the schemes. Investigation leads to ideas for possible schemes, and then in the end the key is to prove soundness and completeness. But there are many different Hilbert systems based on different schemes.
There are two general ways that we prove a set of schemes is consistent. One is to show that the set of true formulas in a model is preserved under the schemes: we cannot take a set of true formulas in any model and deduce a formula that is false in that model.   The other way, which is part of the field of proof theory, is to analyze the structure of a derivation using the schemes, and somehow show that no such derivation can end in an inconsistent result. However, is it usually straightforward to show that the logical axiom schemes are consistent - the usual challenge is with the non-logical axioms of a particular theory.


TITLE: Complemented subspaces of ultrapowers
QUESTION [5 upvotes]: It's a famous result of Maurey that a Banach space $E$ is finitely representable in $X$ if and only if it is a subspace of some ultrapower of $X$. Is there an analogous result for complemented subspaces? More specifically,
Let $E$ be a Banach space with an FDD $(E_n)$. Suppose that $E$ is isomorphic to a complemented subspace of some ultrapower $X^U$. Are the subspaces $(E_n)$ uniformly complemented in $X$?

REPLY [5 votes]: It is proved (but not explicitly stated) in
Heinrich, Stefan Ultraproducts in Banach space theory. J. Reine Angew. Math. 313 (1980), 72–104.
The result you want follows from what Heinrich calls "the local reflexivity of ultrapowers", which is Theorem 7.3 in the paper.
EDIT: Here is how to apply Heinrich's Theorem 7.3.  It is an elementary observation that an $n$ dimensional subspace $E$ of a Banach space $Y$ is $C$-complemented iff there is an $n$ dimensional subspace $F$ of $Y^*$ that $C$-norms $E$. Suppose that $E \subset X^U$ is $n$ dimensional and $F \subset (X^U)^*$ is $n$ dimensional and $C$-norms $E$.  If $F\subset (X^*)^U$, there is no problem pulling back to $X$, and Heinrich's local reflexivity principle for ultrapowers says that you can take $F\subset (X^*)^U$ (which is strictly smaller than $(X^U)^*$ when $X$ is not superreflexive).


TITLE: Is frequency of Electromagnetic wave it's intrinsic property?
QUESTION [1 upvotes]: I was solving a simple refraction problem where the wavelength and velocity of the sodium light needed to be calculated after refracting through glass from air. Now after calculation I found the frequency is same for the both while we know velocity = frequency $\times$ wavelength.
So is the frequency being an intrinsic property of EM waves? I know it is pretty simple but could not find a satisfactory answer.

REPLY [5 votes]: Normally, yes. Some generator provides a wave of a given frequency $\omega$, and the other things (wave vectors $\vec{k}$, etc.) are calculated depending on the medium where the wave travels.


TITLE: Cardinality if set $A$ equals $n|A|$ implies $n|A| = \aleph_0|A|$
QUESTION [2 upvotes]: Suppose $|A| = n|A|$, for any $n\in \mathbb{N}$. Then $|A|=\aleph_0 |A|$. How does one prove this? I'm feeling lost. Would appreciate some hints. Is Zorn's lemma useful here?
Intuitively, I would think that since $n$ is arbitrarily large, and $|A|=n|A|$, $n|A|=\aleph_0|A|$. But this doesn't seem quite rigorous.

REPLY [6 votes]: Here's a proof without the axiom of choice.
We assume that $|A|=2|A|=|\{0,1\}\times A|.$
I.e., there is an injective map $e:\{0,1\}\times A\to A.$
For $a\in A$ define $f(a)=e(0,a)$ and $g(a)=e(1,a)$;
then $f:A\to A$ and $g:A\to A$ are injective maps with disjoint ranges.
For $n\in\mathbb N=\{0,1,2,\dots\}$ and $a\in A$ define $h(n,a)=f^ng(a),$
i.e., $h(0,a)=g(a),\ h(1,a)=f(g(a)),\ h(2,a)=f(f(g(a))),$ etc.
It's easy to see that $h:\mathbb N\times A\to A$ is an injective map, showing that $|\mathbb N\times A|\le|A|.$ Combining this with the obvious inequality $|A|\le|\mathbb N\times A|$ and the Cantor–Bernstein theorem, we get
$|A|=|\mathbb N\times A|=\aleph_0\cdot|A|.$


TITLE: Quantum Mechanics state space
QUESTION [11 upvotes]: In Quantum Mechanics one often deals with wavefunctions of particles. In that case, it is natural to consider as the space of states the space $L^2(\mathbb{R}^3)$. On the other hand, on the book I'm reading, there's a construction which it's quite elegant and general, however it is not rigorous. For those interested in seeing the book, it's "Quantum Mechanics" by Cohen-Tannoudji.
The book proceeds as follows: the first postulate of Quantum Mechanics states that for every quantum system there is one Hilbert space $\mathcal{H}$ whose elements describe the possible states of the system. The idea then is that $\mathcal{H}$ doesn't necessarily is a space of functions.
Indeed, Cohen defines (or doesn't define) $\mathcal{H}$ as the space of kets $|\psi\rangle\in \mathcal{H}$, being the kets just vectors encoding the states of the system.
The second postulate states that for each physically observable quantity there is associated one hermitian operator $A$ such that the only possible values to be measured are the eigenvalues of $A$ and such that

If $A$ has discrete spectrum $\{|\psi_n\rangle : n \in \mathbb{N}\}$ then the probability of measuring the eigenvalue $a_n$ on the state $|\psi\rangle$ is $\langle \psi_n | \psi\rangle$ considering that $|\psi\rangle$ is normalized.
If $A$ has continuous spectrum $\{|\psi_{\lambda}\rangle : \lambda \in \Lambda\}$ then the probability density on state $|\psi\rangle$ for the possible eigenvalues is $\lambda \mapsto \langle \psi_\lambda | \psi\rangle$

If, for example, the position operator $X$ for particle in one-dimension, exists, and if its eigenvectors are $|x\rangle$ with eigenvalues $x$, for each $x\in \mathbb{R}$, the probability density of position is $\langle x |\psi\rangle$ which is a function $\mathbb{R}\to \mathbb{C}$ and we recover the wavefunction.
This formulation, though, seems to be more general. In that case, wavefunction is just the information about one possible kind of measurement which we can obtain from the postulates. There is nothing special with it.
Now, although quite elegant and simple, this is not even a little rigorous. For example: the position operator hasn't been defined! It is just "the operator associated to position with continuous spectrum", but this doesn't define the operator. On the book, it is defined on the basis $\{|x\rangle\}$, but this set is defined in terms of it, so we get circular.
Another problem is that usually we are dealing with unbounded operators which are not defined on the whole of $\mathcal{H}$. And an even greater problem is that $\mathcal{H}$ was never defined!
I've been looking forward to find out how to make this rigorous, but couldn't find anything useful. Many people simply say that the right way is to consider always $L^2(\mathbb{R}^3)$, so that all of this talk is nonsense. But I disagree, I find it quite natural to consider this generalized version.
The only thing I've found was the idea of rigged Hilbert spaces, known also as Gel'fand triple. I've found not much material about it, but anyway, I didn't understand how it can be used to make this rigorous.
In that case, how does one make this idea of space of states, or space of kets, fully rigorous, overcoming the problems I found out, and possibly any others that may exist? Is it through the Gel'fand triple? If so, how is it done?

REPLY [6 votes]: There are two mathematically rigorous (and quite general) Hilbert Space formalisms you might be looking for. Both can be seen as attempts to salvage the engine of Dirac's original bra-ket algorithm, while avoiding its mathematical embarrassments.
The first - created by von Neumann - replaces Dirac's kets by vectors in an abstract Hilbert space $\mathscr{H}$, and replaces his transformations with linear operators on this space. To avoid the need of Dirac's (in)famous delta functions, von Neumann rejects the fundamental status of eigenvectors in Quantum Mechanics. Instead, the central player in von Neumann's game is spectral decomposition (specifically, of self-adjoint operators, although this can be generalized). This approach reduces to eigenvectors and eigenvalues when the operators happen to be finite dimensional (or, in the infinite dimensional case, compact), but handles the continuous case without batting an eyelid.
In the end, however, physicists were reluctant to abandon the highly intuitive eigenvector-based formalism of Dirac, and definitely did not want to muck around with all the subtleties of operator-domains inherent in von Neumann's approach.
And that is why the next fellow entered the fray...
The second - created by Gelfand et al - attempts to preserve more of Dirac's legacy, while straying as little as possible from von Neumann's Hilbert Space. Specifically, they proceed by starting with a Hilbert Space $\mathscr{H}$ and then constructing two auxiliary spaces from it. First a dense inner-space $\Phi$, and then an outer-space $\Phi^*$ (being the dual of the inner-space). Taken together, this framework is referred to as a Rigged Hilbert Space or Gelfand Triple:
$$\Phi\subset\mathscr{H}\subset\Phi^*$$
Very briefly, the reasons they do this are:

The inner-space $\Phi$ supports operator algebra without all the hassle of domains inherent with von Neumann's approach.
The continuous spectrum can be handled in a rigorous eigenvector-eigenvalue fashion because the Dirackian delta 'functions', which don't live in the Hilbert Space, can be found in the dual $\Phi^*$ (technically, as distributions rather than functions).

Where does this leave the poor mathematical physicist with a desire for rigor?
Well, the truth is, nobody really uses Gelfand triples. Just to construct one triple you need to construct an infinite sequence of topologies! Not only that, but the whole construction depends on the specific form of the Hamiltonian, so it is nowhere near as general as von Neumann's approach. (Indeed, finding a Gelfand triple for a specific quantum system would easily qualify you for a PhD!)
To me, the Rigged Hilbert Space approach is reminiscent of the non-standard analysis discovered in the 1960s. Nobody actually uses it, but it makes them feel better about cancelling the $du$ in:
$$\frac{dy}{du}\frac{du}{dx}$$
But let's be honest, you were going to do that anyway right?
Bottom line?
Learn von Neumann's rigorous Hilbert Space quantum mechanics.
There are plenty of excellent books out there. Even von Neumann's seminal Mathematical Foundations of Quantum Mechanics (1932) is a fantastic resource (although the old-fashioned pre-LaTeX notation is a bit intimidating). There were lots of excellent modern books published in the 1970s and 80s on the subject (Prugovecki's Quantum Mechanics in Hilbert Space comes to mind). For a superb up-to-the-minute treatment of Quantum Mechanics in the thorough style von Neumann would salute is Valter Moretti's Spectral Theory and Quantum Mechanics (which I happen to be reading at the moment!).
Note on the issue of generality...
Much of your question concerns the issue of generality. Von Neumann's approach has the virtue of not committing to any specific $L^2$ space, and as a result produces theorems that hold for all quantum mechanical systems. But that is not always advantageous. Sometimes you want to find results about specific (types of) quantum mechanical systems. In this case, you should choose a specific Hilbert Space (often, but not always, an $L^2$ space). But even then, you know you can rely on all the tools of the trade you learned/developed in a general Hilbert Space.


TITLE: Why is $\bigwedge^2( \mathbb{Z}[x,y])=0$
QUESTION [0 upvotes]: According to a comment in Dummit and Foote, $\bigwedge^2( \mathbb{Z}[x,y])=0$.  Why is this true?
I can't convince myself that monomials such that $x \otimes y$ or $x \otimes 1$ as elements of $\mathbb{Z}[x,y] \otimes_{\mathbb{Z}} \mathbb{Z}[x,y]$ would go to zero once we quotient out by the ideal generated by $p(x,y) \otimes p(x,y)$.

REPLY [6 votes]: This is only true if you are taking the wedge square of $\mathbb{Z}[x,y]$ as a $\mathbb{Z}[x,y]$-module, not as a $\mathbb{Z}$-module.  The wedge square is then defined as the quotient of $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]$ by the $\mathbb{Z}[x,y]$-submodule $K$ generated by elements of the form $p(x,y)\otimes p(x,y)$.  Taking $p(x,y)=1$, $1\otimes 1$ is an element of the submodule $K$.  But $1\otimes 1$ generates $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]$ as a $\mathbb{Z}[x,y]$-module, since any simple tensor $p(x,y)\otimes q(x,y)$ is equal to the scalar product $(p(x,y)q(x,y))\cdot (1\otimes 1)$.  (Indeed, $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]\cong\mathbb{Z}[x,y]$ by sending $p(x,y)\otimes q(x,y)$ to $p(x,y)q(x,y)$, and $1\otimes 1$ maps to $1\in\mathbb{Z}[x,y]$ under this isomorphism.)
(Note there is nothing special about $\mathbb{Z}[x,y]$: this argument shows that for any commutative ring $R$, $\bigwedge_R^2(R)=0$.)


TITLE: How does the interaction of an photon with a free electron in metal and with a isolated free electron differs?
QUESTION [1 upvotes]: I am curious to know this,its confusing me a lot.How does it differ,What is the interaction cross section for both the cases if an infrared photon incidents on it?.

REPLY [4 votes]: A free electron in a metal is actually a Bloch electron interacting with the periodic potential of the lattice and all the other electrons which are there. It turns out that the electrons in a band near the Fermi surface can be approximated rather well as being isolated (they are “dressed” quasiparticles), but there’s an important difference. A real isolated free electron is subject to no additional interactions. A free electron in a metal is subject to many. These include electron-electron scattering, electron-phonon scattering, defect scattering, surface scattering, collective oscillations (plasmons), interband transitions, etc. In a simple case, say, Au with photon energies in the infrared range, the scattering can be lumped into a phenomenological parameter called the Drude scattering time (or momentum relaxation time).  Condensed matter computations of the light scattering cross-section often focus on calculating these scattering rates, based on various assumptions about the metal, after computing the band structure. This is because the scattering adds the degree of freedom to allow for absorption/phase shift of light in interaction with the metal. Light interacting with a totally isolated electron cannot be absorbed or experience a phase shift because there’s no scattering of electrons to conserve energy and momentum.