09-240/Classnotes for Tuesday September 15

• 

  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.

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

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

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

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

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

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

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

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

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 0 2 0 2 4 6 1 3 1 3 0 3 6 2 5 1 2 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 4 6 0 6 5 4 3 2 5

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

Tedious Theorem

1. $a+b=c+d\Rightarrow a=c$ "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. $\,\!a+O'=a\Rightarrow O'=0$

   $a+O'=a$

   $\Rightarrow a+O'=a+0$ 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) $\,\!(a+b)(a-b)=a^{2}-b^{2}$