09-240/Classnotes for Tuesday September 15

09-240/Navigation Panel   Additions to the MAT 240 web site no longer count towards good deed points # Week of... Notes and Links 1 Sep 7 Tue, About, Thu 2 Sep 14 Tue, HW1, HW1 Solution, Thu 3 Sep 21 Tue, HW2, HW2 Solution, Thu, Photo 4 Sep 28 Tue, HW3, HW3 Solution, Thu 5 Oct 5 Tue, HW4, HW4 Solution, Thu, 6 Oct 12 Tue, Thu 7 Oct 19 Tue, HW5, HW5 Solution, Term Test on Thu 8 Oct 26 Tue, Why LinAlg?, HW6, HW6 Solution, Thu 9 Nov 2 Tue, MIT LinAlg, Thu 10 Nov 9 Tue, HW7, HW7 Solution Thu 11 Nov 16 Tue, HW8, HW8 Solution, Thu 12 Nov 23 Tue, HW9, HW9 Solution, Thu 13 Nov 30 Tue, On the final, Thu S Dec 7 Office Hours F Dec 14 Final on Dec 16 To Do List The Algebra Song! Register of Good Deeds Misplaced Material Add your name / see who's in!

WARNING: The notes below, written for students and by students, are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them. It is a bad idea to stop taking your own notes thinking that these notes can be a total replacement - there's nothing like one's own handwriting! Visit this pages' history tab to see who added what and when.

• 

  Yangjiay - Page 1

• 

  Page 2

• 

  Page 3

• 

  Page 4

• 

  Page 5

The real numbers A set $\mathbb {R}$ with two binary operators and two special elements $0,1\in \mathbb {R}$ s.t.

$F1.\quad \forall a,b\in \mathbb {R} ,a+b=b+a{\mbox{ and }}a\cdot b=b\cdot a$

$F2.\quad \forall a,b,c,(a+b)+c=a+(b+c){\mbox{ and }}(a\cdot b)\cdot c=a\cdot (b\cdot c)$

${\mbox{(So for any real numbers }}a_{1},a_{2},...,a_{n},{\mbox{ one can sum them in any order and achieve the same result.}}$

$F3.\quad \forall a,a+0=a{\mbox{ and }}a\cdot 1=a$

$F4.\quad \forall a,\exists b,a+b=0{\mbox{ and }}\forall a\neq 0,\exists b,a\cdot b=1$

${\mbox{So }}a+(-a)=0{\mbox{ and }}a\cdot a^{-1}=1$

${\mbox{(So }}(a+b)\cdot (a-b)=a^{2}-b^{2})$

$\forall a,\exists x,x\cdot x=a{\mbox{ or }}a+x\cdot x=0$

Note: or means inclusive or in math.

$F5.\quad (a+b)\cdot c=a\cdot c+b\cdot c$

Definition: A field is a set F with two binary operators $\,\!+$: F×F → F, $\times \,\!$: F×F → F and two elements $0,1\in \mathbb {R}$ s.t.

$F1\quad {\mbox{Commutativity }}a+b=b+a{\mbox{ and }}a\cdot b=b\cdot a\forall a,b\in F$

$F2\quad {\mbox{Associativity }}(a+b)+c=a+(b+c){\mbox{ and }}(a\cdot b)\cdot c=a\cdot (b\cdot c)$

$F3\quad a+0=a,a\cdot 1=a$

$F4\quad \forall a,\exists b,a+b=0{\mbox{ and }}\forall a\neq 0,\exists b,a\cdot b=1$

$F5\quad {\mbox{Distributivity }}(a+b)\cdot c=a\cdot c+b\cdot c$

Examples

1. $F=\mathbb {R}$
2. $F=\mathbb {Q}$
3. $\mathbb {C} =\{a+bi:a,b\in \mathbb {R} \}$

   $i={\sqrt {-1}}$

   $\,\!(a+bi)+(c+di)=(a+c)+(b+d)i$

   $\,\!0=0+0i,1=1+0i$

4. $\,\!F_{2}=\{0,1\}$
5. $\,\!F_{7}=\{0,1,2,3,4,5,6\}$
6. $\,\!F_{6}=\{0,1,2,3,4,5\}$ is not a field because not every element has a multiplicative inverse.

   Let $a=2.$

   Then $a\cdot 0=0,a\cdot 1=2,a\cdot 3=0,a\cdot 4=2,a\cdot 5=4$

   Therefore F4 fails; there is no number b in F₆ s.t. a · b = 1

Ex. 4 + 0 1 0 0 1 1 1 0	Ex. 4 × 0 1 0 0 0 1 0 1
Ex. 5 + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5	Ex. 5 × 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1

Theorem: F₂ is a field.

In order to prove that the associative property holds, make a table (similar to a truth table) for a, b and c.

a	b	c
0	0	0
0	0	1
0	1	0
0	1	1	(0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
1	0	0
1	0	1
1	1	0
1	1	1

Theorem: $\,\!F_{p}$ for $p>1$ is a field iff (if and only if) $p$ is a prime number

Proof:

$a\in \mathbb {Z}$ has a multiplicative inverse modulo $m$ if and only if a and m are relatively prime.

This can be shown using Bézout's identity:

$\exists x,y{\mbox{ s.t. }}ax+my=1$

$\left(ax+my\right){\pmod {m}}=1{\pmod {m}}$

$ax=1$

$x=a^{-1}$

We have shown that $a$ has a multiplicative inverse modulo m if $a$ and $m$ are relatively prime. It is therefore a natural conclusion that if $m$ is a prime number all elements in the set will be relatively prime to m.

---

Multiplication is repeated addition.

$23\times 27={\begin{matrix}27\\\overbrace {23+23+23+\cdots +23} \end{matrix}}=621$

$27\times 23={\begin{matrix}23\\\overbrace {27+27+27+\cdots +27} \end{matrix}}=621$

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 2³ = 8, but 3² = 9.

Tedious Theorem

1. $a+b=c+b\Rightarrow a=c$ "cancellation property"

   Proof:

   By F4, $\exists d{\mbox{ s.t. }}b+d=0$

   $\,\!(a+b)+d=(c+b)+d$

   $\Rightarrow a+(b+d)=c+(b+d)$ by F2

   $\Rightarrow a+0=c+0$ by choice of d

   $\Rightarrow a=c$ by F3

2. $a\cdot b=c\cdot b,(b\neq 0)\Rightarrow a=c$
3. $a+O'=a\Rightarrow O'=0$

   Proof:

   $\,\!a+O'=a$

   $\Rightarrow a+O'=a+0$ by F3

   $\Rightarrow O'=0$ by adding the additive inverse of a to both sides

4. $a\cdot l'=a,a\neq 0\Rightarrow l'=1$
5. $a+b=0=a+b'\Rightarrow b=b'$
6. $a\cdot b=1=a\cdot b'\Rightarrow b=b'=a^{-1}$

   $\,\!{\mbox{Aside: }}a-b=a+(-b)$

   ${\frac {a}{b}}=a\cdot b^{-1}$

7. $\,\!-(-a)=a,(a^{-1})^{-1}$
8. $a\cdot 0=0$

   Proof:

   $a\cdot 0=a(0+0)$ by F3

   $=a\cdot 0+a\cdot 0$ by F5

   $=0=a\cdot 0$

9. $\forall b,0\cdot b\neq 1$

   So there is no 0⁻¹

10. $(-a)\cdot b=a\cdot (-b)=-(a\cdot b)$
11. $(-a)\cdot (-b)=a\cdot b$
12. (Bonus) $\,\!(a+b)(a-b)=a^{2}-b^{2}$

Quotation of the Day

......