### 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 $q_0<\cdots 1$ 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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