anonymous
  • anonymous
Determine if V is a vector space when: x+y=xy cx=x^c If not state all the vector axioms it fails
Mathematics
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anonymous
  • anonymous
Determine if V is a vector space when: x+y=xy cx=x^c If not state all the vector axioms it fails
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
If yes verify each vector space axiom
TuringTest
  • TuringTest
here's at least one that fails\[c(\vec u+\vec v)\neq c\vec u+c\vec v\]
anonymous
  • anonymous
ok what abt tat scaler multiplication?

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phi
  • phi
x^c is not a linear operation
anonymous
  • anonymous
okkkkk
TuringTest
  • TuringTest
actually it's consistent though... \[c(\vec u+\vec v)= c\vec u+c\vec v\]because we have\[c(\vec u+\vec v)=(uv)^c=u^cv^c= c\vec u+c\vec v\]so I don't see what's wrong with it
anonymous
  • anonymous
well it is definitely not closed under scaler multilication though
TuringTest
  • TuringTest
how so?
anonymous
  • anonymous
because x^c is exponential
TuringTest
  • TuringTest
but that doesn't necessarily take it out of V
TuringTest
  • TuringTest
*the vector space I mean
anonymous
  • anonymous
oh ya? ok I dont get a thing, I feel so stupid
TuringTest
  • TuringTest
aha! it is a vector space I thought I'd seen it befor check it out, the zero vector turns out to be 1 I think http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx
anonymous
  • anonymous
yay lol :D
phi
  • phi
@Tur how do you get from (uv)^c to cu + cv?
TuringTest
  • TuringTest
check out example 5
anonymous
  • anonymous
yup i see it. Thanks :D
TuringTest
  • TuringTest
@ phi that is what you get when you put the two rules together the exponent is distributed because vector addition is defined as multiplication
phi
  • phi
nvm, i'll look at your link...
anonymous
  • anonymous
LOL i htink i am fundamentally lacking basic knowledge
TuringTest
  • TuringTest
I'm not so good at this stuff, I had it wrong at first as well

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