Commit d479ad6d authored by Jim Hefferon's avatar Jim Hefferon

Finished first chapter; disparity! :-)

parent 836bb296
......@@ -1886,7 +1886,8 @@ Let $p$ be prime. Prove each.\label{ex:EuclidsOtherLemma}
\begin{problem} \notetext{Fundamental Theorem of Arithmetic}
Any number $n>1$ can be expressed as a product of primes
$n=p_0^{e_0}p_1^{e_1}\cdots p_s^{e_s}$.
What's more, this expression is
What's more, under the condition that the primes are distinct,
this expression is
unique:~any two prime factorizations of~$n$ contain the same
primes, to the same powers.
We do this in two parts.
......@@ -1909,72 +1910,66 @@ Let $p$ be prime. Prove each.\label{ex:EuclidsOtherLemma}
\end{answer}
\begin{exercise}
Prove that the factorization is unique.
\hint~you can show by induction on $i$ that for
two products of primes
$p_0\cdots p_i=q_0\cdots q_j$ with $i\leq j$,
the primes on the right are a rearrangement of those
on the left.
\hint~show that
where $n=p_0\cdots p_i$ is a product of primes, if
$q_0\cdots q_j$ is a product of primes that is equal to~$n$ then
the primes are the same, with the same multiplicities,
possibly rearranged; you can use induction on $i$.
\end{exercise}
\begin{answer}
Assume that $n=p_0\cdots p_i=q_0\cdots q_j$ are two products of primes.
One must be of length less than or equal to the other; without
loss of generality we can take $i\leq j$.
We will show that $i=j$ and that the factors on the right are a rearrangement
of those on the left.
We follow the hint: we show by induction on~$i$ that if
$n=p_0\cdots p_i$ be a product of primes.
then any product of primes that is equal to is
$p_0\cdots p_i=q_0\cdots q_j$ consists of the same primes, with the
same multiplicities.
We use induction on~$i$.
The base step is~$i=0$. so that $n=p_0$ is prime.
The base step is~$i=0$, where $n=p_0$ is a prime number.
In this case $p_0=q_0\cdots q_j$ gives a nontrivial factorization of
the prime~$p_0$ unless $j=0$ and $p_o=q_0$.
$p_0$ unless $j=0$ and $p_o=q_0$.
% !! TODO
For the inductive step assume that the factorization is unique for
$s=1$, \ldots, $s=k$ and consider the case of $s=k+1$ primes,
For the inductive step assume that the statement is true when
$i=0$, \ldots, $i=k$ and consider the $i=k+1$ case,
$n=p_0\cdots p_{k}\cdot p_{k+1}$.
Assume that there is another prime factorization $n=q_0\cdots q_{t}$
with $t\geq 0$ and such that $q_0<\cdots <q_{t}$.
We first show that $p_{k+1}$ equals some $q_i$.
Because $p_{k+1}\divides n$ we have $p_{k+1}\divides q_0\cdots q_{t}$ and so
the above result~\ref{ex:EuclidsOtherLemma} shows that
$p_{k+1}$ divides some $q_i$.
But $q_i$ is prime, with only the positive divisors of $1$ and itself,
and so $p_{k+1}=q_i$.
Cancel them:
$n/p_{k+1}=p_0\cdots p_{k}=q_0\cdots q_{i-1}\cdot q_{i+1}\cdots q_{t}$.
By the assumption that $k\geq 0$, omitting the $k+1$-th prime leaves
at least one prime in $p_0\cdots p_{k}$.
So the inductive hypothesis applies and
the ascending sequence of $p$'s without~$p_{k+1}$
equals the ascending sequence of $q$'s without~$q_i$,
$p_0=q_0$, \ldots\@
Obviously then, the entire sequence of $p$'s equals the entire sequence of
$q$'s.
Fix a product of primes $q_0\cdots q_j$ that also equals~$n$.
Because $p_{k+1}\divides n=q_0\cdots q_j$,
Exercise~\ref{ex:EuclidsOtherLemma} gives that
$p_{k+1}$ divides some $q_m$.
But $q_m$ is prime, so its only the positive divisors are $1$ and itself,
and so $p_{k+1}=q_m$.
Finish by cancelling:
$n/p_{k+1}=p_0\cdots p_{k}=q_0\cdots q_{m-1}\cdot q_{m+1}\cdots q_{j}$.
Since this is inductive step~$k+1$,
this omission leaves at least one prime in the product $p_0\cdots p_{k}$,
that is, $n/p_{k+1}\neq 1$.
So the inductive hypothesis applies:
the $q$'s without $q_m$ are the same primes, with the same multiplicities,
as the $p$'s without $p_{k+1}$.
Clearly then, the entirety of $q$'s equals the entirety of
$p$'s.
\end{answer}
\end{exes}
\end{problem}
\noindent\remark this result is why we do not include~$1$ among the primes.
Including~$1$ would require us to change the clause about uniqueness
\noindent\remark that is why we do not include~$1$ among the primes.
Including~$1$ would require us to change the clause about uniqueness,
since we can always multiply by additional~$1$'s.
\begin{problem}[\midlength]
Decide if each is true.
True or false?
% From: http://gowers.wordpress.com/2011/11/13/why-isnt-the-fundamental-theorem-of-arithmetic-obvious/
\begin{items}
\item $5\cdot 7\cdot 19\neq 3\cdot 11\cdot 17$
\item $5\cdot 7\cdot 19= 3\cdot 11\cdot 17$
% \item $47\cdot 863\neq 73\cdot 557$
\item $1357\cdot 4183= 1081\cdot 5251$ % due to David Speyer
\end{items}
\begin{answer}
\begin{items}
\item True since all the numbers are prime and they are different.
\item False, since these are products of primes that are not the same primes.
% \item True since these numbers are prime and different.
\item True (the numbers are not primes).
\item True, both equal $5\, 676\, 331$ (the numbers are not primes;
for example $1357=23\cdot 59$).
\end{items}
\end{answer}
\end{problem}
......@@ -1985,51 +1980,54 @@ Decide if each is true.
% \end{ex}
\begin{problem}[\maxlength]
Let $a,b>1$ and
suppose that $a=p_0^{e_0}\cdots p_{n-1}^{e_{n-1}}$
and $b=p_0^{f_0}\cdots p_{n-1}^{f_{n-1}}$ are their prime factorizations
(to use the same primes $p_0,\ldots,p_{n-1}$ in both
we allow here that some exponents are zero).
Let $a=p_0^{e_0}\cdots p_{n}^{e_{n}}$
and $b=p_0^{f_0}\cdots p_{n}^{f_{n}}$ express each number as a product of
distinct primes;
to use the same primes $p_0,\ldots,p_{n}$ in both
we allow here that some exponents are zero.
Prove that
in the prime factorization of
$\gcd(a,b)$ the exponent of $p_i$ is $\min(\set{e_i,f_i})$.
(Much the same proof shows that in the prime factorization of $\lcm(a,b)$
the exponent of $p_i$ is $\max(\set{e_i,f_i})$.
Taken together the two show that $\gcd(a,b)\cdot\lcm(a,b)=ab$.)
Together these show that $\gcd(a,b)\cdot\lcm(a,b)=ab$.)
\begin{answer}
Let $d=\gcd(a,b)$.
Let $\gcd(a,b)$ be $d$.
First observe that the only primes in the factorization of~$d$ are
among $p_0$, \ldots, $p_{n-1}$:
among $p_0$, \ldots, $p_{n}$:
if $p$ is a prime that divides~$a$ or~$b$ then
Execise~\ref{ex:EuclidsOtherLemma} shows that $p$ is in that list.
Exercise~\ref{ex:EuclidsOtherLemma} shows that $p$ is in that list.
Accordingly,
write $d=p_0^{g_0}\cdots p_n^{g_n}$.
write $d=p_0^{g_0}\cdots p_n^{g_n}$.
If for each $i$ we have $g_i=\min(e_i,f_i)$ then $d$~divides both $a$ and~$b$,
so each exponent $g_i$ is at least as large as $\min(\set{e_i,f_i})$.
To see that it cannot be larger suppose $g_i>\min(\set{e_i,f_i})$
and also suppose without loss of generality that $\min(\set{e_i,f_i})=e_i$.
Expand
$d\cdot k=a$.
To see that it cannot be larger
suppose, without loss of generality, that $\min(\set{e_i,f_i})=e_i$.
The greatest common divisor $\gcd(a,b)$ is a divisor of~$a$, so there
is some $k\in\Z$ with $d\cdot k=a$.
Expand the two into products of primes.
\begin{equation*}
p_0^{g_0}\cdots p_i^{g_i}\cdots p_{n-1}^{g_{n-1}}\cdot k
=p_0^{e_0}\cdots p_i^{e_i}\cdots p_{n-1}^{e_{n-1}}
p_0^{g_0}\cdots p_i^{g_i}\cdots p_{n}^{g_{n}}\cdot k
=p_0^{e_0}\cdots p_i^{e_i}\cdots p_{n}^{e_{n}}
\end{equation*}
This contradicts the uniqueness of prime factorization
because there are more
$p_i$'s on the left than on the right.
Were $g_i$ to be greater than $e_i$ then this would contradict the
uniqueness of prime factorization,
because there would be more $p_i$'s on the left than on the right.
\end{answer}
\end{problem}
\begin{problem}[\midlength] \notetext{Existence of Irrational Numbers}
\begin{exes}
\begin{exercise}
Prove that if a number is a square then in its prime factorization
Prove that in the prime factorization of a square,
each prime is raised to an even power.
\end{exercise}
\begin{answer}
The square is
$n^2=(p_0^{e_0}\cdots p_{s-1}^{e_{s-1}})^2=p_0^{2e_0}\cdots p_{s-1}^{2e_{s-1}}$.
So each prime occurs an even number of times.
$n^2=(p_0^{e_0}\cdots p_{s}^{e_{s}})^2=p_0^{2e_0}\cdots p_{s}^{2e_{s}}$.
Each power is even.
\end{answer}
\begin{exercise}
Prove that $\sqrt{2}$ is irrational.
......@@ -2037,12 +2035,11 @@ Taken together the two show that $\gcd(a,b)\cdot\lcm(a,b)=ab$.)
\begin{answer}
Suppose that $\sqrt{2}=n/d$ for $n,d\in\Z$.
Clearing the fraction and squaring gives $2\cdot n^2=d^2$.
On the right, the prime factorization of~$d^2$ has $2$ raised to an even
power (possibly zero).
On the left, the prime factorization of~$n^2$ has $2$ raised to an even power
also, and there is an additional~$2$.
Thus the left has an odd number of~$2$'s while the right has
and even number of~$2$'s, which
On the right, the factorization of~$d^2$ has $2$ raised to an even
power (possibly this power is zero).
On the left, besides the solo~$2$ the factorization of~$n^2$ has $2$
raised to an even power, and so the left has an odd number of~$2$'s.
This disparity
contradicts the uniqueness of prime factorizations.
\end{answer}
% \begin{exercise}[\maxlength]
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