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Math Help - Proving -1(a) = -a

  1. #1
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    Proving -1(a) = -a

    Sorry if this is in the wrong section. My multivariable calculus teacher just gave us this question on the first day of class.

    We are asked to prove that for any real number a, -1(a) = -a

    We can use the 11 real number axioms (ie closure, commutative, associative, indentity, inverse), 1 order axiom, and one completeness axiom, along with the theorem we proved in class which states a(0) = 0

    I am completely clueless on this. I worked on it for over an hour and didn't get anywhere.

    Any help or direction appreciated, sorry again if there is a better place for this question.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Shananay View Post
    Sorry if this is in the wrong section. My multivariable calculus teacher just gave us this question on the first day of class.

    We are asked to prove that for any real number a, -1(a) = -a

    We can use the 11 real number axioms (ie closure, commutative, associative, indentity, inverse), 1 order axiom, and one completeness axiom, along with the theorem we proved in class which states a(0) = 0

    I am completely clueless on this. I worked on it for over an hour and didn't get anywhere.

    Any help or direction appreciated, sorry again if there is a better place for this question.
    well, you never stated all the axioms you can use, but i'm pretty sure the distributive law is included. i will give you the procedure, i leave it to you to justify each step by stating what axiom i used.

    -a + a = 0

    \Rightarrow -a + a + (-1 \cdot a) = -1 \cdot a

    \Rightarrow -a + 1 \cdot a + (-1 \cdot a) = -1 \cdot a

    \Rightarrow -a + a(1 - 1) = - 1\cdot a

    \Rightarrow -a + a \cdot 0 = -1 \cdot a

    \Rightarrow -a + 0 = -1 \cdot a

    \Rightarrow -a = -1 \cdot a
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    -a + a = 0

    \Rightarrow -a + a + (-1 \cdot a) = -1 \cdot a

    \Rightarrow -a + 1 \cdot a + (-1 \cdot a) = -1 \cdot a

    \Rightarrow -a + a(1 - 1) = - 1\cdot a

    \Rightarrow -a + a \cdot 0 = -1 \cdot a

    \Rightarrow -a + 0 = -1 \cdot a

    \Rightarrow -a = -1 \cdot a
    I went through this and it looks great. Thank you. Distributive law is good, I forgot that. Thanks again.
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