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Math Help - Using Field Axioms

  1. #1
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    Using Field Axioms

    I don't know if this is where this type of question should be, so sorry if this isn't it.

    I need to show the following using the Field Axioms:

    1. That the identity numbers 0 and 1 are unique.
    2. That 0*a=0.
    3. That the inverses -a and 1/a are unique.
    4. That -1*a = -a
    5. That (-1)(-1)=1
    6. That if a_3\neq 0, then
    \frac{a_1+a_2}{a_3}=\frac{a_1}{a_3}+\frac{a_2}{a_3  }

    -----------

    Could the first be demonstrated like this?
    1. Show that the identity numbers 0 and 1 are unique.

    Suppose there is a number 0', an additive identity, different from zero such that 0'+a=a. Then 0'+a-a=a-a \Longleftrightarrow 0'+0=0 by the definition of additive inverse and 0'+0=0 \Longleftrightarrow 0'=0 by the definition of additive identity which is clearly a contradiction. Therefore, the additive identity 0 is unique.

    Suppose there is a number 1', a multiplicative identity, different from one such that 1'\cdot a = a. Then 1'\cdot a = a \Longleftrightarrow 1'\cdot a \cdot \frac{1}{a} by the definition of multiplicative inverse and 1'\cdot 1 = 1 \Longleftrightarrow 1'=1 by the definition of multiplicative identity which is a contradiction. Therefore, the multiplicative identity 1 is unique.
    Last edited by rualin; July 6th 2007 at 06:35 AM.
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  2. #2
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    Basically a Field F has <F,+> an abelian group and <F,\cdot > and abelian group. So it is necessary and sufficient to prove the following.

    Theorem: In a Group G the identity elemenet is unique.

    Proof: Let e be this identity element. And e' another identity element. Then e=ee'=e'
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  3. #3
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    Perfect! Thanks
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  4. #4
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    Note, this does not show that 1\not = 0.

    For example, consider R=\{a\} defined as a+a=a and aa=a. Then we see that a=0 and a=1. But this example is reffered to as trivial ring. And we do not consider it to be a field by convention, just to avoid this boring counterexample.
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  5. #5
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    I simply want to get a feel for how to demonstrate something step by step. Here's what I've done so far but I'm having trouble justifying the step with the ***** in problem 2. I have no idea how to start with problem 6. Please look over them if possible and correct any mistakes.

    2) Show that 0\cdot a=0.

    (0+\frac{b}{a})\cdot a = 0\cdot a + b by distributive law of multiplication over addition and multiplicative inverse
    0\cdot a + b = b ****
    0\cdot a + b - b = b - b \Longleftrightarrow 0\cdot a + 0 = 0 by definition of additive inverse
    0\cdot a = 0 by definition of additive identity

    ---

    3) Show that in a group G, the inverse of each element a is unique.

    Let G be a group with binary operation * and identity e.

    Suppose there is some element a with two inverses denoted by a^{-1} and a'. Then

    (a^{-1} * a) * a' = a^{-1} * (a * a') by the associative law
    e*a'=a^{-1}*e by definition of inverse
    a'=a^{-1} by definition of identity

    Therefore, the inverse of each element in a group is unique.

    ---

    4) Show that -1\cdot a = -a.

    0=(1-1)\cdot a \Longleftrightarrow 0 = (1\cdot a) + (-1\cdot a) by distributive law
    0=a +(-1\cdot a) by multiplicative identity
    0-a=a-a+(-1\cdot a) \Longleftrightarrow -a=0+(-1\cdot a) by additive inverse
    -a=-1\cdot a by additive identity

    ---

    5) Show that (-1)(-1)=1.

    0=(1-1)(-1) \Longleftrightarrow 0=-1+(-1)(-1) by distributive law and multiplicative identity
    0+1=-1+1+(-1)(-1) \Longleftrightarrow 1=(-1)(-1) by additive inverse

    ---

    Does anyone know the latex for an actual negative that's shorter and higher than the minus sign and for the Real number sign?
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  6. #6
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    Quote Originally Posted by rualin View Post
    2) Show that 0\cdot a=0.
    0a=(1-1)a=a-a=0

    3) Show that in a group G, the inverse of each element a is unique.
    Say a has b,c as inverses.
    So ab=e=ac. Thus, ab=ac, use the left-cancellation law to get b=c.

    4) Show that -1\cdot a = -a.
    -a means the additive inverse of a. So show (-1)a+a=0. By distributive law.

    5) Show that (-1)(-1)=1.
    Try this yourself.
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    Quote Originally Posted by ThePerfectHacker View Post

    4) Show that -1\cdot a=-a.

    -a means the additive inverse of a. So show (-1)a+a=0. By distributive law.
    How did you arrive at (-1)a+a=0?

    Quote Originally Posted by ThePerfectHacker View Post
    5) Show that (-1)(-1)=1.

    Try this yourself.
    (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1
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  8. #8
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    Quote Originally Posted by rualin View Post
    How did you arrive at (-1)a+a=0?
    Because,
    0a=(-1+1)a=(-1)a+a

    (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1
    Correct! You just need to justify,
    (-1)(-1)+(-1)=0
    Which is the distributive law,
    (-1)[(-1)+1]=(-1)(0)=0
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  9. #9
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    Here is a PDF file that I use with students on your question.
    I hope it helps.
    Attached Files Attached Files
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  10. #10
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    Nice! Thanks, Plato.

    I still am not able to follow the Double inverse theorem... maybe you could help make it clearer.
    Let * be an associative binary operation on a set A with identity e, and let x\in A. If x is invertible for *, let [tex]x^{-1}[/math denote the (unique) inverse for x. Then x^{-1} is invertible and (x^{-1})^{-1} = x.

    Proof: If y is the inverse for x, then
    y*x=e=x*y.

    But this is exactly the condition for x to be the inverse for y.
    What confuses me about that proof is the last statement. I don't understand how that is sufficient.
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