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Math Help - proof and rules of inferences

  1. #1
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    proof and rules of inferences

    Hi,
    Please can someone help me with this problem.

    show that a,b,c are real numbers and a#0, then there is a unique solution of the equation ax+b=c.

    the uniqueness of the solution is my problem.
    Thank you
    B
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  2. #2
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    Quote Originally Posted by braddy
    Hi,
    Please can someone help me with this problem.

    show that a,b,c are real numbers and a#0, then there is a unique solution of the equation ax+b=c.

    the uniqueness of the solution is my problem.
    Thank you
    B
    Okay.
    If,
    ax+b=c
    There exists preciesly one such number -b such as b+(-b)=0 because the real numbers form a group under addition.
    Thus,
    (ax+b)+(-b)=c+(-b)
    Since addition is associative we have,
    ax+(b-b)=c-b
    Thus,
    ax+0=c-b
    Since,
    0 is an identity element we have,
    ax=b
    Since, a\not = 0 it is an element of the real numbers under multiplication and form a group. Thus, there exist a unique, 1/a such as, a(1/a)=1.
    Thus,
    (1/a)(ax)=(1/a)(c-b)
    Since, multiplication is associative we have,
    ((1/a)a)x=1/a(c-b)
    Thus,
    1x=1/a(c-b)
    Since 1 is the multiplicative identity element we have,
    x=1/a(c-b)
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  3. #3
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    Here is the way that I would expect this to be done.
    Suppose that each of p and q is a solution to the equation ax + b = c,\quad a \not= 0.
    Then
    \begin{array}{rcl}<br />
 ap + b & = & c \\ <br />
 aq + b & = & c \\ <br />
 ap + b & = & aq + b \\ <br />
 ap & = & aq \\ <br />
 p & = & q \\ <br />
 \end{array}
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  4. #4
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    Quote Originally Posted by Plato
    Here is the way that I would expect this to be done.
    Suppose that each of p and q is a solution to the equation ax + b = c,\quad a \not= 0.
    Then
    \begin{array}{rcl}<br />
 ap + b & = & c \\ <br />
 aq + b & = & c \\ <br />
 ap + b & = & aq + b \\ <br />
 ap & = & aq \\ <br />
 p & = & q \\ <br />
 \end{array}
    Yes but you do not show that a solution exists
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  5. #5
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    No, you are correct!
    But you showed a solution existed.
    I simplified its uniqueness.
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  6. #6
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    Quote Originally Posted by Plato
    No, you are correct!
    But you showed a solution existed.
    I simplified its uniqueness.
    Thank you forr your Help!
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