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Math Help - Proof of Theorems

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    Proof of Theorems

    Prove the following two theorems.

    1) If x is any number, then --x = x.

    2) If x is any number distinct from 0, then 1/x is not equal to 0, and 1/(1/x) = x.
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  2. #2
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    (1) \exists a,b \in \mathbb{R} such that  a + b = 0, then b = -a .

    Let b = -a and the conclusion follows.

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    (2) Assume a \neq 0. If we assume \frac{1}{a} = 0, try to arrive at a contradiction. So we know \frac{1}{a} \neq 0 .

    For the second part, use the fact that a \cdot b = 1 \Rightarrow \ b = \frac{1}{a} where a \neq 0, b \in \mathbb{R} and consider a \cdot \frac{1}{a} = 1.
    Last edited by o_O; January 25th 2009 at 06:40 PM.
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    Quote Originally Posted by noles2188 View Post
    Prove the following two theorems.

    1) If x is any number, then --x = x.

    2) If x is any number distinct from 0, then 1/x is not equal to 0, and 1/(1/x) = x.

    These kind of questions although look simple can become very tricky,depending what axioms we choose within the frame of real Nos.

    So if we choose as an axiom the following:

    ....................for all A , A+(-A)=0...........................................1

    Then the -(-x)=x problem can be solved as follows:


    In (1) put A= x and (1) becomes : x + (-x)=0................................2

    And if we put A= -x then (1) becomes: (-x )+ [-(-x)] =0....................3

    Hence from (3) and (2) and using cancellation law : x=-(-x)


    But however if we choose as an axiom the following:

    ................for all x ,there exists y such that ,x+y=0........................4

    then to prove the above we must do the following:

    a) prove the uniqueness of y........

    b) Then introduce in our system a new operation symbol " - " ,which in our 1st approach (equation (1) ) we consider as one of the primitive symbols of our system.

    Hence the following definition:


    ............for all x,y ( -x = y iff x+y =0 ).......................................... 5

    OR ,using different variables for convenience we have:


    ............for all A,B ( -A = B <====> A+B =0 )....................................6

    And if we put now , A= -x.........B= -xTHEN (6) becomes:


    ................-x =-x <====> x + (-x) = 0............................................7


    While if we put A = -x and B= -(-x) ,then 6 becomes:


    .................... -(-x) = -(-x) <=====> (-x) + [-(-x)] =0.......................8


    But since -x = -x and -(-x) = -(-x) ,then (7) and (8) became:



    ..........................x + (-x) =0................................................ ........9



    .................................-(-x) + [-(-x)] = 0......................................10

    Which are exactly the same with equations (2) and (3).That in a way explains why some systems start with (1) as their axiom

    Now for the 2nd part :

    Firstly we choose as our axiom :


    ...............................for all,A (  A\neq 0\rightarrow(1/A).A=1)........................................... ...................................11

    Now to prove that 1/a \neq 0.

    Assume 1/a =0 and a \neq 0 ,then (1/a).a=0 and also by using (11),where we put A=a we have (1/a).a=1 and hence 1=0,a contradiction. Thus 1/a \neq 0


    Since 1/a \neq 0 ,we put in (11) A = 1/a and we get 1/(1/a)*1/a=1. And multiplying that equation by a and using the fact (1/a)*a=1,we get:

    ..................1/(1/a)= a................................................. .........


    Now if we choose as our axiom the following:

    ...................for all A  A\neq 0 there exists B such that AB=1 ,THE PATH of the proof changes completely .


    However following more or less the path shown above we can get the desired result. You try it
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