09-240/Classnotes for Thursday September 24: Difference between revisions

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A V.S. over <math>F: V, 0, +, \times</math> s.t.
A V.S. over <math>F: V, 0, +, \cdot</math> s.t.


VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math><br />
VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math><br />
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# <math>O_{\mathcal F(S, F)}(\sigma) = 0_F \forall \sigma \in S</math>
# <math>O_{\mathcal F(S, F)}(\sigma) = 0_F \forall \sigma \in S</math>
# <math>f, g \in \mathcal F(S, F)</math><br /><math>\,\! (f + g)(\sigma) = f(\sigma) + g(\sigma)</math><br /><math>f \in \mathcal F(S, F)</math><br /><math>a \in F \Rightarrow \forall \sigma \in S, S(af)(\sigma) = a \cdot (f(\sigma))</math>
# <math>f, g \in \mathcal F(S, F)</math>
#: <math>(f + g)(\sigma) = f(\sigma) + g(\sigma)</math>
#: <math>f \in \mathcal F(S, F)</math>
#: <math>a \in F \Rightarrow \forall \sigma \in S, S(af)(\sigma) = a \cdot (f(\sigma))</math>


'''Claim''': + is associative. Given <math>f, g, h \in \mathcal F(S, F), (f + g) + h = f + (g + h) \forall \sigma</math>
'''Claim''': + is associative. Given <math>f, g, h \in \mathcal F(S, F), (f + g) + h = f + (g + h) \forall \sigma</math>
: ((f + g) + h)(\sigma) = (f + g)(\sigma) + h(\sigma)
: <math>((f + g) + h)(\sigma) = (f + g)(\sigma) + h(\sigma)</math>
: = (f(\sigma) + g(\sigma)) + h(\sigma)
: <math>= (f(\sigma) + g(\sigma)) + h(\sigma)</math>
: = f(\sigma) + g(\sigma) + h(\sigma) \mbox{ (by F2)}
: <math>= f(\sigma) + g(\sigma) + h(\sigma) \mbox{ (by F2)}</math>
: (f + (g + h))(\sigma) = f(\sigma) + (g + h)(\sigma)
: <math>(f + (g + h))(\sigma) = f(\sigma) + (g + h)(\sigma)</math>
: = f(\sigma) + (g(\sigma) + h(\sigma))
: <math>= f(\sigma) + (g(\sigma) + h(\sigma))</math>
: = f(\sigma) + g(\sigma) + h(\sigma)
: <math>= f(\sigma) + g(\sigma) + h(\sigma)</math>


<ol start="4">
<ol start="4">
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<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb Q</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb Q</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\{0\} \mbox{ is a V.S. over } F</math></li>
<li><math>\,\! \{0\} \mbox{ is a V.S. over } F</math></li>
</ol>
</ol>


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# Cancellation: <math>x + y = x + z \Rightarrow y = z</math> (add ''w'' to both sides s.t. ''x'' + ''w'' = 0)
# Cancellation: <math>x + y = x + z \Rightarrow y = z</math> (add ''w'' to both sides s.t. ''x'' + ''w'' = 0)
# 0 is unique
# 0 is unique
# Negatives are unique: <math>x + y = 0 = x + z \Rightarrow y = 0</math>
# Negatives are unique: <math>x + y = 0 = x + z \Rightarrow y = z</math>
# <math>0x = 0. a \cdot 0 = 0</math>
# <math>0x = 0. a \cdot 0 = 0</math>
# <math>(-a) \cdot x = a \cdot (-x) = -(ax)</math>
# <math>(-a) \cdot x = a \cdot (-x) = -(ax)</math>

Latest revision as of 21:47, 21 October 2009

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.
Convention for today:

A V.S. over s.t.

VS1
VS2
VS3
VS4
VS5
VS6
VS7
VS8

Examples

  1. Let S be a set (F is some field)
    S = Primary colours = {red, green, blue}
    F = F2 = {0, 1}




Claim: + is associative. Given

Dull theorem

  1. Cancellation: (add w to both sides s.t. x + w = 0)
  2. 0 is unique
  3. Negatives are unique: