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@ -596,101 +596,10 @@ emptyset$. But $b in emptyset$ is a contradiction by its definition.
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$
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]
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= Solutions to selected exercises and problems
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Solutions to selected problems and exercises.
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#linebreak()
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*@euclid.* We begin by considering primes $p_1, p_2, ..., p_n$. Let $P = p_1 dot p_2 dot ... dot p_n$. Then let $q = P + 1$.
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Then if $q$ is prime, we have an additional prime not in the original list.
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Otherwise, $q$ is not prime and we have a unique prime factorization of $q$.
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Without loss of generality, take one such prime to be $p_k$. $p_k$ cannot be in
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the original list $p_1, p_2, ..., p_n$.
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If $p_k$ were in the original list, then since $P$ is divisible by $p_k$, and $P
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+ 1$ is also divisible by $p_k$, 1 must be divisible by $p_k$ which is
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impossible. So $p_k$ is a new prime.
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For completeness, let's finish the proof explicitly. Start with primes $p_1$,
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$p_2$. The method above implies the existence of another prime, which we denote
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$p_3$. Repeat this to find additional primes $p_(k+1)$.
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*@perfectsquare.* This is a generalization of the proof $sqrt(6)$ is
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irrational. Seeking a contradiction, suppose $sqrt(a)$ is irrational.
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$
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exists p,q in ZZ, sqrt(a) = p / q \
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p^2 = a q^2
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$
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Then by the fundamental theorem of arithmetic,
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$ a = b_1 dot b_2 dot ... dot b_n $
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where $b_i$ is prime.
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$ p^2 = (Pi^n_(i=1) b_1) dot q^2 $
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Notice all $b_i$ are unique (again by the same theorem) and without loss of
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generality, choose a $b_k$, $1 <= k <= n$.
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Then $p$ has $j = 1,2,...$ $hash$ of $b_k$ in its factors. Then $p^2$ has $2j$
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$hash$ of $b_k$. Similarly, $q$ has $L = 1,2,...$ $hash$ of $b_k$, and $q^2$
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has $2L$. Then $(Pi ^n _(i=1)) q^2$ has $2L + 1$ $hash$ of $b_k$. But
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$ p^2 = (Pi^n_(i=1)) q^2 $
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and by unique factorization they must have the same $hash$ of the prime factor $b_k$, so $sqrt(a)$ is irrational.
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Note that if $a$ was a perfect square, TODO
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*@rational-between.* Effectively we are asked to show that given $x,y in QQ$,
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where $x < y$, $exists z in QQ$ such that $x < z < y$.
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First, let us take the difference between $x$ and $y$.
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$
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exists a,b,c,d in ZZ \
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x = a / b \
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y = c / d \
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y - x = (b c - a d) / (b d)
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$
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So we know that
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$
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y = x + (b c - a d) / (b d)
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$
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We see that if we can find any nonzero integer less than $(b c - a d)/(b d)$,
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adding it to $x$ gives us the number between $x$ and $y$ we desire. One option is
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$
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(b c - a d) / (b d + 1) < (b c - a d) / (b d) \
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x < x + (b c - a d) / (b d + 1) < x + (b c - a d) / (b d) = y \
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$
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So one such $z$ is
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$ z = x + (b c - a d) / (b d + 1) $
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#remark[
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There is a minor hole in this proof, if $b d + 1 = 0$, $z$ is undefined. We
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can easily avoid this by replacing all instances of $b d + 1$ with, say, $b d
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+ 2$, when $b d + 1 = 0$.
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]
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#remark[
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A much easier and less roundabout proof is to take
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$ z = (x + y) / 2 $
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Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
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]
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= Induction
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== Weak and strong
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Induction is a way to prove that a statement is true for all $NN$. Formally, it
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is introduced like this: say you have a set $S subset.eq NN$ such that $1 in
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S$. Then, if $n in S => n + 1 in S$, $S = NN$. Usually we plug $n$ into a
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@ -839,7 +748,9 @@ We demonstrate the WOP in an alternative proof of
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]
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]
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== Relations, partitions
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= Relations, partitions
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== Basic notions
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#definition[
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A relation on a set $A$ is a set of ordered pairs $(a,b)$ where $a,b in A$. A
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@ -1007,3 +918,199 @@ nonempty subsets of $A$ whose union is $A$.
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A = union.big_(x in A) overline(x)
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$
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]
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= Lecture #datetime(day: 19, year: 2025, month: 2).display()
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== Functions
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Let $A$ and $B$ be sets. A relation $R$ from $A$ to $B$ is a subset $R subset.eq A times B$.
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#definition[
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A *function* $f$ from $A$ to $B$ relates each element of $"Dom"(R)$ to to exactly
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one element of $"Rng"(R)$.
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]
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#fact[
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A function from $A$ to $B$ is written
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$
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f : A -> B
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$
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]
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#fact[
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$
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(x,y) in f <=> f(x) = y <=> x f y
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$
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]
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We say $f : A -> A$ is a function on $A$.
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#example[
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Let $A = {1,2,3}$, $B = {x,y}$. A relation from $A$ to $B$ is a subset of $A
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times B$, which has 6 elements. $A$ and $B$ admit $2^6 = 64$ distinct
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relations.
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When asking about functions instead, we note that $f$ admits either $(1, x)$
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or $(1, y)$, but not both and so on. So there are $2^3 = 8$ distinct
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functions.
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]
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#definition[
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A function $f : A -> B$ is called a *map/mapping*. $A$ is the *domain*, $B$ is
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the *codomain*. The *range* of $f$ $"Rng"(f) = {y in B : (exists x in A)(f(x) =
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y)}$.
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]
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More terminology. Let $f : A -> B$, $(a,b) in f$, $f(x) = y$. Then
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- $y$ is the *value* of $f$ at $x$
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- $y$ is the *image* of $x$ under $f$
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- $x$ is the *preimage* of $y$ under $f$
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#fact[
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$f : A -> B = g : C -> D$ if and only if $f subset.eq g$ and $g subset.eq f$.
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]
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#theorem[
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$f : A -> B = g : C -> D$ if and only if
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1. $"Dom"(f) = "Dom"(g), ($A = C$)$
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2. $f(x) = g(x)$, $forall x in A = C$
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]
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#definition[
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The *identity function* on $A$, $I_A : A -> A$ is defined by $I_A (x) = x, forall x in A$.
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]
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#definition[
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If $A subset.eq B$, the *inclusion function* $i : A -> B$ is defined by
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$
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i(x) = x, forall x in A
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$
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Sometimes it is written $iota : A arrow.r.hook B$.
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]
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Why consider the inclusion function rather than simply the identity on $A$? Say
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we have $iota : A -> B$ and $g : B -> C$. Then we can compose these $g compose
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iota$ which is an inclusion function $iota' : A -> C$.
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#definition[
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Let $R$ be an equivalence relation on $A$. Recall $A\/R$ is the set of all
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equivalence classes. The *canonical map* is
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$
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f : A -> A\/R, f(x) = overline(x), forall x in A
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$
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]
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#example[
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Consider the canonical map $f : ZZ - >ZZ_3 = {overline(0), overline(1),
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overline(2)}$, $f(x) = overline(x)$. Then $f(2) = overline(2)$, $f(8) =
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overline(2)$, and so on. So $f$ is not _one-to-one_.
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]
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== Logistics: exam 2
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- One non-proof free response on set operations and families (2.2-2.3, $union$,
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$sect$, etc)
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- 4 proof based questions
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- Set operations/families (2.2-2.3)
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- Principle of mathematical induction, but no complete induction or WOP (2.4)
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- Relations. $"Dom"(R)$, $"Rng"(R)$, etc. (3.1-3.2)
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- Equivalence relations (3.3)
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- T/F questions
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- Extra credit question
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= Solutions to selected exercises and problems
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Solutions to selected problems and exercises.
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#linebreak()
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*@euclid.* We begin by considering primes $p_1, p_2, ..., p_n$. Let $P = p_1 dot p_2 dot ... dot p_n$. Then let $q = P + 1$.
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Then if $q$ is prime, we have an additional prime not in the original list.
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Otherwise, $q$ is not prime and we have a unique prime factorization of $q$.
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Without loss of generality, take one such prime to be $p_k$. $p_k$ cannot be in
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the original list $p_1, p_2, ..., p_n$.
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If $p_k$ were in the original list, then since $P$ is divisible by $p_k$, and $P
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+ 1$ is also divisible by $p_k$, 1 must be divisible by $p_k$ which is
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impossible. So $p_k$ is a new prime.
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For completeness, let's finish the proof explicitly. Start with primes $p_1$,
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$p_2$. The method above implies the existence of another prime, which we denote
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$p_3$. Repeat this to find additional primes $p_(k+1)$.
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*@perfectsquare.* This is a generalization of the proof $sqrt(6)$ is
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irrational. Seeking a contradiction, suppose $sqrt(a)$ is irrational.
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$
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exists p,q in ZZ, sqrt(a) = p / q \
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p^2 = a q^2
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$
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Then by the fundamental theorem of arithmetic,
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$ a = b_1 dot b_2 dot ... dot b_n $
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where $b_i$ is prime.
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$ p^2 = (Pi^n_(i=1) b_1) dot q^2 $
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Notice all $b_i$ are unique (again by the same theorem) and without loss of
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generality, choose a $b_k$, $1 <= k <= n$.
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Then $p$ has $j = 1,2,...$ $hash$ of $b_k$ in its factors. Then $p^2$ has $2j$
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$hash$ of $b_k$. Similarly, $q$ has $L = 1,2,...$ $hash$ of $b_k$, and $q^2$
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has $2L$. Then $(Pi ^n _(i=1)) q^2$ has $2L + 1$ $hash$ of $b_k$. But
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$ p^2 = (Pi^n_(i=1)) q^2 $
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and by unique factorization they must have the same $hash$ of the prime factor $b_k$, so $sqrt(a)$ is irrational.
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Note that if $a$ was a perfect square, TODO
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*@rational-between.* Effectively we are asked to show that given $x,y in QQ$,
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where $x < y$, $exists z in QQ$ such that $x < z < y$.
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First, let us take the difference between $x$ and $y$.
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$
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exists a,b,c,d in ZZ \
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x = a / b \
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y = c / d \
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y - x = (b c - a d) / (b d)
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$
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So we know that
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$
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y = x + (b c - a d) / (b d)
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$
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We see that if we can find any nonzero integer less than $(b c - a d)/(b d)$,
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adding it to $x$ gives us the number between $x$ and $y$ we desire. One option is
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$
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(b c - a d) / (b d + 1) < (b c - a d) / (b d) \
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x < x + (b c - a d) / (b d + 1) < x + (b c - a d) / (b d) = y \
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$
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So one such $z$ is
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$ z = x + (b c - a d) / (b d + 1) $
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#remark[
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There is a minor hole in this proof, if $b d + 1 = 0$, $z$ is undefined. We
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can easily avoid this by replacing all instances of $b d + 1$ with, say, $b d
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+ 2$, when $b d + 1 = 0$.
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]
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#remark[
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A much easier and less roundabout proof is to take
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$ z = (x + y) / 2 $
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Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
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]
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|
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