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Math Help - Proof of the Theorem a0=0

  1. #1
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    Proof of the Theorem a0=0

    I'm trying to prove this a different way, but I thought I have it but noticed a problem I think that kills it and I need to make sure it's correct that I can't do this. This is what I have...

    1) 0=0
    2) a0=a0
    3) a0=a(1-1)
    4) a0=a-a
    5) a0=0

    The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?
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  2. #2
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    Re: Proof of the Theorem a0=0

    Quote Originally Posted by JSB1917 View Post
    The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?
    Yes, that looks like a problem.
    I don't have a book with me, but I assume this is the standard way:
    Let x=a0. Let y be the additive inverse of x (so that x+y=y+x=0). Since 0 is the additive identity, 0 + 0 = 0.
    Now x = a0 = a(0+0) = a0 + a0 = x + x. Then adding y to both sides gives:
    x + y = (x + x) + y = x + (x + y) = x + 0 = x. Since x + y = 0, have shown that 0 = x. Thus a0 = 0.
    Another way:
    a = a(1) = a(0+1) = a0 + a(1) = a0 + a. Then add a's additive inverse (-a, but to avoid confusion I'll call it b... thus a+b=b+a=0) to both sides to get:
    a + b = ( a0 + a ) + b = a0 + ( a + b ) = a0 + 0 = a0. But then a0 = a + b = 0. Thus a0 = 0.
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  3. #3
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    Re: Proof of the Theorem a0=0

    Yeah, the other way i did it was 0+1=1, then multiply by a and use the additive inverse of a.
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  4. #4
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    Re: Proof of the Theorem a0=0

    Quote Originally Posted by JSB1917 View Post
    I'm trying to prove this a different way, but I thought I have it but noticed a problem I think that kills it and I need to make sure it's correct that I can't do this. This is what I have...

    1) 0=0
    2) a0=a0
    3) a0=a(1-1)
    4) a0=a-a
    5) a0=0

    The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?
    Here is a standad proof.
     \begin{align*}0&=a\cdot 0+[-(a\cdot 0)] \\&= a\cdot[0+0]+[-( a\cdot 0)]  \\&=[a\cdot 0+ a\cdot 0]+[-( a\cdot 0)] \\&=a\cdot 0+ ([a\cdot 0]+[-( a\cdot 0)]) \\&=[a\cdot 0]+0 \\&=a\cdot 0\end{align*}

    You can supply the reasons(axioms).
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