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Math Help - Field: If A=B; Prove A+C = B+C

  1. #1
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    Field: If A=B; Prove A+C = B+C

    This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
    Seems like we have to rely of definition rather than a formal prove.

    One approach:
    Let us for the moment assume
    A=B => B-A = 0 Let is call it (1)

    Then
    A+C = A + C + B - A = B + C

    But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

    Any help please?
    Thanks
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  2. #2
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    Hi

    Isn't this just a property of equality?
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  3. #3
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    Quote Originally Posted by aman_cc View Post
    This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
    Seems like we have to rely of definition rather than a formal prove.

    One approach:
    Let us for the moment assume
    A=B => B-A = 0 Let is call it (1)

    Then
    A+C = A + C + B - A = B + C

    But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

    Any help please?
    Thanks
    There is nothing to prove here.
    Addition, is a binary operation +:F\times F\to F.
    Therefore, +(A,C) = +(B,C) since A=B.
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  4. #4
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    Quote Originally Posted by clic-clac View Post
    Hi

    Isn't this just a property of equality?
    Thanks. I am new to Algebra. If you could please elaborate. Sorry if this is too trivial.
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  5. #5
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    Smile My answer

    Quote Originally Posted by aman_cc View Post
    This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
    Seems like we have to rely of definition rather than a formal prove.

    One approach:
    Let us for the moment assume
    A=B => B-A = 0 Let is call it (1)

    Then
    A+C = A + C + B - A = B + C

    But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

    Any help please?
    Thanks

    I think this way will help you .
    Lets assume ,
    A=B => A+C != B+C (!= stands for not equal)
    => A+C > B+C or A+C < B+C
    => A > B or A < B (by definitions of inequalities)
    => contradiction

    What do you think ?
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  6. #6
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    Quote Originally Posted by K A D C Dilshan View Post
    I think this way will help you .
    Lets assume ,
    A=B => A+C != B+C (!= stands for not equal)
    => A+C > B+C or A+C < B+C
    => A > B or A < B (by definitions of inequalities)
    => contradiction

    What do you think ?

    Yes - Makes sense. Though I had to prove to myself the following result
    "=> A+C > B+C or A+C < B+C
    => A > B or A < B (by definitions of inequalities)"

    You can also prove the original from the definition of mapping.
    If x=y then f(x) = f(y), as others suggested.

    Thanks
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  7. #7
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    Lets assume ,
    A=B => A+C != B+C (!= stands for not equal)
    => A+C > B+C or A+C < B+C
    => A > B or A < B (by definitions of inequalities)
    => contradiction
    Doing that, you assume your field is totally ordered, which is often wrong.

    The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").
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  8. #8
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    Quote Originally Posted by clic-clac View Post
    Doing that, you assume your field is totally ordered, which is often wrong.

    The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").
    Yes clic-clac, that makes lot more sense as the result will hold even in fields which are not ordered. Like you said it can be proved/argued based on the logic of equality and definition of mapping ("+"). Atleast I;m convinced with the argument now ! Thanks very much
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  9. #9
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    Unhappy

    Quote Originally Posted by clic-clac View Post
    Doing that, you assume your field is totally ordered, which is often wrong.

    The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").
    But I think I have used only definitions ...
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