Finished work from last semesterHEADmaster

1 files changed, 20 insertions, 8 deletions

diff --git a/gupta/portfolio.tex b/gupta/portfolio.tex
index 25b41b3..44e71a5 100644
--- a/gupta/portfolio.tex
+++ b/gupta/portfolio.tex
@@ -66,19 +66,31 @@ Squaring both sides, we find
 $$6 = {p^2\over q^2} \to p^2 = 6q^2 = 2(3q^2),$$ from which $3q^2\in\bb
 Z$ tells us $2\mid p^2.$
 
-{\bf Lemma.}
+{\bf Lemma 1.}
 We will show that $2\mid p^2$ implies $2\mid p$ by contrapositive.
 Let $2\nmid p,$ so $p = 2k + 1$ for some $k\in\bb Z.$
 We then compute $p^2 = 4k^2 + 4k + 1 = 2(2k^2+2k) + 1,$ and
 $2k^2+2k\in\bb Z,$ so $2\nmid p^2.$
 \endproof
 
-By the lemma, $2\mid p$ and $4\mid p^2,$ so there exists $k\in\bb Z$
-s.t. $p^2 = 4k.$
-We then get $4k = 6q^2$ and $2k = 3q^2.$ % TODO missing proof that 2k=nm.
-So $q^2$ must be even, and as we've shown $2\mid q^2$ gives $2\mid q.$
-Thus, $\gcd(p,q) = 2 \neq 1,$ giving a contradiction, meaning $\sqrt 6$
-is irrational.
+{\bf Lemma 2.}
+We will show that if $2\mid nm,$ then $2\mid n$ or $2\mid m$ by
+contrapositive.
+Let $2\nmid n$ and $2\nmid m,$ or by definition of odd, $n=2r+1$ and
+$m=2s+1.$
+Then,
+$$nm = (2r+1)(2s+1) = 4rs+2r+2s+1 = 2(2rs+r+s)+1,$$
+and $2rs+r+s\in\bb Z$ by integer closure, so $2\nmid nm$ by definition
+of odd.
+\endproof
+
+By the first lemma, $2\mid p$ and $4\mid p^2,$ so there exists $k\in\bb
+Z$ s.t. $p^2 = 4k.$
+We then get $4k = 6q^2$ and $2k = 3q^2.$
+By the second lemma, $2\nmid 3,$ and $k\in\bb Z,$ so $q^2$ must be even,
+and as we've shown $2\mid q^2$ gives $2\mid q.$
+Thus, $\gcd(p,q) \geq 2 \neq 1,$ giving a contradiction, meaning $\sqrt
+6$ is irrational.
 \endproof
 
 \problem{4.} Equivalence proof
@@ -145,7 +157,7 @@ $k>2,$ $F_k = F_{k-1} + F_{k-2}.$
 
 We will show this identity for all $n\in\bb Z,$ where $n\geq 1,$ by
 induction.
-For $n=1,$ $\sum_{k=1}^n F_k^2 = F_1^2 = 1 = 1\cdot 1 = F_1F_2.$
+For $n=1,$ we find $\sum_{k=1}^n F_k^2 = F_1^2 = 1 = 1\cdot 1 = F_1F_2.$
 
 We now assume for some $j\geq 1,$
 $$\sum_{k=1}^j F_k^2 = F_jF_{j+1},$$