anonymous
  • anonymous
Linear algebra: Determine whether this is a vector space: The set of all rational numbers, with the usual definitons of addition and scalar multiplication.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
this isnt a vector space
anonymous
  • anonymous
since a rational number multiplied by an irrational number = irrational number
anonymous
  • anonymous
Oooh I see. Thank you very much!

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anonymous
  • anonymous
did u get it? or like pretending like u understood cuz that wasnt so clear
anonymous
  • anonymous
I am actually in middle of taking linear algebra and had the same question like 2 weeks ago :)
anonymous
  • anonymous
No, I get it. It's not closed under scalar multiplication because you can multiply it by an irrational number and the answer won't be rational.
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
I SERIOUSLY don't get vector spaces at all. It is all so freaking weird.
anonymous
  • anonymous
lol i know so i got a tutor to come over and we sat for 3 hours and he explained everything and now it is so clear
anonymous
  • anonymous
It was too confusing and it was killing my brain
anonymous
  • anonymous
Nice. My instructor has no life (his officer hours are all over the place) so I think I'll just irritate the hell out of him tomorrow :-D.
anonymous
  • anonymous
hehehehehe
anonymous
  • anonymous
hmm, maybe you could include the field from which your scalars are from. Usually we talk of 'vector space over F' where F is some field, could be the reals, could be the rationals or could be the complex numbers. If the scalars are from the rationals, then it is a vector space as it will be closed under multiplication and the axioms would hold, if real/complex then no...
anonymous
  • anonymous
well i thought that F is all rational numbers
anonymous
  • anonymous
and i guess the scaler coud be any real number
anonymous
  • anonymous
Yeah that's what I was thinking... and that would include decimal numbers so scalar mult. wouldn't hold up.
anonymous
  • anonymous
Anyway, thanks Pippa. I really need to get some serious one-on-one time with someone that really knows what they're doing. The book just doesn't help in this case.
anonymous
  • anonymous
hehe i agree. So like y dont u get some classmates to help?
anonymous
  • anonymous
I will try. I am so shy in that class, it's been 7 weeks now and I haven't talked with anybody.
anonymous
  • anonymous
ohhhhhhhhh i see. Ya its easier to get a tutor. they know how to explain better but not many ppl are familiar with linear algebra
anonymous
  • anonymous
It's mostly guys in the class so I just have to get out of my shell and talk to them. You're right, and the books/lecture notes online can be difficult to follow. Anyway, I won't take up any more of your time. I appreciate your help :-)
anonymous
  • anonymous
hehe bye gtg study linear algebra :)
anonymous
  • anonymous
Well in an abstract sense, vectors are simply 'objects' (can be numbers, may not be) which may be added together and multiplied by scalars. In your case, the set \[\mathbb{Q}\] is where your vectors are from, and scalars are from some field F. Vector addition is just field addition, scalar multiplication is just field multiplication, and if that field is the rationals, then you have a vector space! Note that \[\mathbb{Q}\] will have two subspaces which are \[\{0\}\] and itself...
anonymous
  • anonymous
@brinethery do you see why the rationals are a vector space over themselves, but not a vector space over the reals? If I multiply a rational by a rational I get another rational, if I multiply a rational by a real, I do not necessarily get a rational...

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