14-240/Classnotes for Monday September 8

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # Week of... Notes and Links 1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday 2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf 3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf 4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf 5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf 6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial 7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday 8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial 9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial 10 Nov 10 HW8, Monday, Tutorial 11 Nov 17 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

We went over "What is this class about?" (PDF, HTML), then over "About This Class", and then over the first few properties of real numbers that we will care about.

Dror's notes above / Students' notes below

The real numbers a set R with 2 binary operations +, *

+:R*R→R

*:R*R→R

in addition 2 special element 0,1∈ R s.t. 0≠1 & furthermore:

R1: The commutative law (for both addition and multiplication)

For every a,b∈R, we have a+b=b+a & ab=ba

R2: The associative law

For every a,b,c∈R, we have (a+b)+c=a+(b+c) (ab)c=a(bc)

in our lives pretty little girls (PL)G≠P(LG)

R3: a+0=a & a*1=a

Wednesday September 10th 2014 - Fields

The real numbers: A set |R with +,x : |R x |R -> |R & $0=/=1$ are elements of |R such that

R1: For every $a,b$ that are elements of |R , $a+b=b+a$ and $ab=ba$

R2: For every $a,b,c$ that are elements of |R, $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$

R3: For every $a$ that is an element of |R, $a+0=a$ and $a*1=a$

R4: For every a that is an element of |R there exists b that is an element of |R such that $a+b=0$ & for every a that is an element of |R and $a=/=0$ there exists b that is an element |R such that $a*b=1$

R5: For every $a,b,c$ that are elements of |R, $(a+b)c=ac+bc$

$(a+b)*(a-b)=a^{2}-b^{2}$ follows from R1-R5

The following is true for the Real Numbers but does not follow from R1-R5 For every a that is an element of |R there exists an $x$ that is an element of |R such that $a=x^{2}ora+x^{2}=0$

However we can see that it does not follow from R1-R5 because we can find a field that obeys R1-R5 yet does not follow the above rule. An example of this is the Rational Numbers |Q. In |Q take $a=2$ and there does not exist $x$ such that $2=x^{2}$ or $2+x^{2}=0$

The Definition Of A Field: A "Field" is a set F along with a pair of binary operations +,x : FxF -> F and along with a pair $0,1$ that are elements of F such that $0=/=1$ and such that R1-R5 hold.

R1: For every $a,b$ that are elements of F , $a+b=b+a$ and $ab=ba$

R2: For every $a,b,c$ that are elements of F, $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$

R3: For every $a$ that is an element of F, $a+0=a$ and $a*1=a$

R4: For every $a$ that is an element of F there exists $b$ that is an element of F such that $a+b=0$ & for every $a$ that is an element of F and $a=/=0$ there exists $b$ that is an element F such that $a*b=1$

R5: For every $a,b,c$ that are elements of F, $(a+b)c=ac+bc$

Example

1. |R is a field (real numbers) 2. |Q is a field (rational numbers) 3. |C is a field (complex numbers) 4. F = {0, 1}

• insert table of addition and multiplication*

Proposition: F is a Field checking F5

etc...

F = {0 , 1} = F2 = Z/2

Do the same for F7

• insert table of addition and multiplication*

"Like remainders when you divide by 7" "like remainders mod 7'

Theorem (that shall remain unproved) : For every prime number P, FP = {0 , 1 , 2 , ... , p-1 } along with + & x defined as above $(a,b)->a+bmodp$ is a field.

---

Theorem: (basic properties of Fields)

Let F be a Field, and let a , b , c denote elements of F Then: 1. $a+b=c+b->a=c$ "Cancellation" still holds 2. $b=/=0,ab=cb->a=c$ 3. If $0'$ is an element of F and satisfies for every $a,a+0'=a$ , then $0'=0$ 4. If $1'$ is "like 1" then $1'=1$

... to be continued...