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     \mathrm{ Let\ V\ be\ a\ vector\ space\ over\ a\ field\ \mathbb{F}.}
     \mathrm{ Directly\ from\ the\ axioms\ of\ a\ vector\ space\ (carefully\ explaining\ which\ axioms\ you\ are\ using\ at\ each\ step)}
     \mathrm{ prove\ that\ the\ following\ statement\ is\ always\ true}:

     \mathrm{ If\ v\ is\ an\ element\ of\ V\ and\ 2v = v\ , then\ v = 0}

    Now, here's what I did:

    If we look at the RHS:

     2*v = (1 + 1) * v

     (1 + 1) * v = v + v - \mathrm{ By\ the\ left\ distributive\ law}

    So if we have  2*v = v

    Then we have  v + v = v

     v + v + (-v) = v + (-v) - \mathrm{ Because\ addition\ admits\ inverses}

     v + (v - v) = (v -v) - \mathrm{ Because\ addition\ is\ associative}

     v + 0 = 0

     v = 0 - \mathrm{ Because\ addition\ has\ a\ zero\ element}

    Now, I think all of this is right. I followed the axioms, and explained which one I was using at each stage, but this person on my course is saying I'm wrong.

    The thing they aren't accepting is that we can rewrite  2*v as  (1 + 1) * v

    They won't accept that we can just assume that  2 = 1 + 1
    But I don't see how else it can be done?

    Can anyone help?

    Thanks
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    (Original post by RG.)
     \mathrm{ Let\ V\ be\ a\ vector\ space\ over\ a\ field\ \mathbb{F}.}
     \mathrm{ Directly\ from\ the\ axioms\ of\ a\ vector\ space\ (carefully\ explaining\ which\ axioms\ you\ are\ using\ at\ each\ step)}
     \mathrm{ prove\ that\ the\ following\ statement\ is\ always\ true}:

     \mathrm{ If\ v\ is\ an\ element\ of\ V\ and\ 2v = v\ , then\ v = 0}

    Now, here's what I did:

    If we look at the RHS:

     2*v = (1 + 1) * v

     (1 + 1) * v = v + v - \mathrm{ By\ the\ left\ distributive\ law}

    So if we have  2*v = v

    Then we have  v + v = v

     v + v + (-v) = v + (-v) - \mathrm{ Because\ addition\ admits\ inverses}

     v + (v - v) = (v -v) - \mathrm{ Because\ addition\ is\ associative}

     v + 0 = 0

     v = 0 - \mathrm{ Because\ addition\ has\ a\ zero\ element}

    Now, I think all of this is right. I followed the axioms, and explained which one I was using at each stage, but this person on my course is saying I'm wrong.

    The thing they aren't accepting is that we can rewrite  2*v as  (1 + 1) * v

    They won't accept that we can just assume that  2 = 1 + 1
    But I don't see how else it can be done?

    Can anyone help?

    Thanks
    Looks fine to me.
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    (Original post by Mark13)
    Looks fine to me.
    Thank you
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    I suspect it is possible to have a field where 1+1\not= 2 though I can't find an example to support it. We'd need a non-standard definition of "+". But since your question is for a general field, relying on 1+1=2 seems flawed to me.

    Edit: See following.
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    (Original post by ghostwalker)
    I suspect it is possible to have a field where 1+1\not= 2 though I can't find an example to support it. We'd need a non-standard definition of "+". But since your question is for a general field, relying on 1+1=2 seems flawed to me.
    I'm just not sure how else it can be done
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    (Original post by ghostwalker)
    I suspect it is possible to have a field where 1+1\not= 2 though I can't find an example to support it. We'd need a non-standard definition of "+". But since your question is for a general field, relying on 1+1=2 seems flawed to me.
    What would you define '2' to be if not 1+1?
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    (Original post by Mark13)
    What would you define '2' to be if not 1+1?
    A good point.

    We could have the integers in terms of successors 1,2,3,4,...
    But define the "+" operation such that 1=1<>2

    I'm not saying I know how to do it. But, to say 1+1=2 is saying that it can't be done.
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    (Original post by ghostwalker)
    I suspect it is possible to have a field where 1+1\not= 2 though I can't find an example to support it. We'd need a non-standard definition of "+". But since your question is for a general field, relying on 1+1=2 seems flawed to me.
    The point is that for any unital ring 2 is by definition 1+1.

    In general, if R is a unital ring then there is a homomorphism

    \mathbb{Z} \rightarrow R

    given by

    n \mapsto \begin{cases}\sum_{i=1}^n 1_R, & n&gt;0 \\ 0, & n=0 \\ \sum_{i=1}^{-n}-1_R, & n&lt;0\end{cases}

    and when one talks about an integer in an arbitrary ring, the convention is that we mean the image under that map.

    I think what is confusing you is that you could, for example, take the real numbers and switch the symbols for 2 and 3 and then you would have a field where 1+1=3. The point is though, that the 2 in the question is referring to the identity element added to itself once, so in that case '2' would actually be 3.

    This fact does hold in the generality it was stated, even for fields of characteristic 2.
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    (Original post by Mark85)
    The point is that for any unital ring 2 is by definition 1+1.

    In general, if R is a unital ring then there is a homomorphism

    \mathbb{Z} \rightarrow R

    given by

    n \mapsto \begin{cases}\sum_{i=1}^n 1_R, & n&gt;0 \\ 0, & n=0 \\ \sum_{i=1}^{-n}-1_R, & n&lt;0\end{cases}

    and when one talks about an integer in an arbitrary ring, the convention is that we mean the image under that map.

    I think what is confusing you is that you could, for example, take the real numbers and switch the symbols for 2 and 3 and then you would have a field where 1+1=3. The point is though, that the 2 in the question is referring to the identity element added to itself once, so in that case '2' would actually be 3.

    This fact does hold in the generality it was stated, even for fields of characteristic 2.
    I'm now more confused. I don't think I need to go into quite as much detail as that, but I typed the entire question, word for word, and so it didn't mention anything about redefining + etc. Hence I assumed it was ok to simply say that   1 + 1 = 2

    Is my answer wrong?
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    (Original post by RG.)
    I'm now more confused. I don't think I need to go into quite as much detail as that, but I typed the entire question, word for word, and so it didn't mention anything about redefining + etc. Hence I assumed it was ok to simply say that   1 + 1 = 2

    Is my answer wrong?
    Not at all. I was just addressing the other posters concern.

    I was just explaining that if you take a general field and talk about '2' or '3', what you mean is the identity of that field added together that many times.

    So 1+1=2 because this is the definition of what '2' means in a general field.

    Your answer is right.
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    (Original post by Mark85)
    ...
    Thanks for clearing that up. G.

    (Original post by RG.)
    ...
    Sorry for confusing you, but it really did seem flawed to me.
 
 
 
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