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brinethery Group Title

Linear algebra: Determine whether this is a vector space: The set of all rational numbers, with the usual definitons of addition and scalar multiplication.

  • 2 years ago
  • 2 years ago

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  1. Pippa Group Title
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    this isnt a vector space

    • 2 years ago
  2. Pippa Group Title
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    since a rational number multiplied by an irrational number = irrational number

    • 2 years ago
  3. brinethery Group Title
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    Oooh I see. Thank you very much!

    • 2 years ago
  4. Pippa Group Title
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    did u get it? or like pretending like u understood cuz that wasnt so clear

    • 2 years ago
  5. Pippa Group Title
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    I am actually in middle of taking linear algebra and had the same question like 2 weeks ago :)

    • 2 years ago
  6. brinethery Group Title
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    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.

    • 2 years ago
  7. Pippa Group Title
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    yes :)

    • 2 years ago
  8. brinethery Group Title
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    I SERIOUSLY don't get vector spaces at all. It is all so freaking weird.

    • 2 years ago
  9. Pippa Group Title
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    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

    • 2 years ago
  10. Pippa Group Title
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    It was too confusing and it was killing my brain

    • 2 years ago
  11. brinethery Group Title
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    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.

    • 2 years ago
  12. Pippa Group Title
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    hehehehehe

    • 2 years ago
  13. Callum29 Group Title
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    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...

    • 2 years ago
  14. Pippa Group Title
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    well i thought that F is all rational numbers

    • 2 years ago
  15. Pippa Group Title
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    and i guess the scaler coud be any real number

    • 2 years ago
  16. brinethery Group Title
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    Yeah that's what I was thinking... and that would include decimal numbers so scalar mult. wouldn't hold up.

    • 2 years ago
  17. brinethery Group Title
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    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.

    • 2 years ago
  18. Pippa Group Title
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    hehe i agree. So like y dont u get some classmates to help?

    • 2 years ago
  19. brinethery Group Title
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    I will try. I am so shy in that class, it's been 7 weeks now and I haven't talked with anybody.

    • 2 years ago
  20. Pippa Group Title
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    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

    • 2 years ago
  21. brinethery Group Title
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    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 :-)

    • 2 years ago
  22. Pippa Group Title
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    hehe bye gtg study linear algebra :)

    • 2 years ago
  23. Callum29 Group Title
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    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...

    • 2 years ago
  24. Callum29 Group Title
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    @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...

    • 2 years ago
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