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Math Help - resscueeee meee

  1. #1
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    resscueeee meee

    will you pleaseeee show me how to use the axioms given to prove this :

    1) For every natural number i, i1 = i and 1i = i
    2) For every natural number i, i0 = 0 and 0i = 0

    I just think that the number multiply 1 is always the number and anything multiple by 0 is 0 but my professor wants me to prove them...i really dont' know how..
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  2. #2
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    I always thought that the first one was a field axiom already (i.e. considered true implicitly, but not provable).

    The second one can be shown this way:

     i0 = i(0 + 0) = i0 + i0 .

    Since we now have i0 = i0 + i0, there is an axiom (if x+y=x then y=0). So i0 = 0

    Use that knowledge and the commutative axiom to show 0i.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Soltras
    I always thought that the first one was a field axiom already (i.e. considered true implicitly, but not provable).

    The second one can be shown this way:

     i0 = i(0 + 0) = i0 + i0 .

    Since we now have i0 = i0 + i0, there is an axiom (if x+y=x then y=0). So i0 = 0

    Use that knowledge and the commutative axiom to show 0i.
    I think these are to be proven from what are essentially the axioms of Peano
    Arithmetic given in the other thread, namely:

    O is a natural number
    For every natural number n, its sucessor Sn is a natural number.
    For every natural number n, Sn is not equal to 0;
    For every natural number n, for every natural number m, if Sn = Sm, then n = m;
    The principle of mathematical induction;
    For every natural number n, for every natural number m, their sum n + m is a natural number;
    For every natural number n, n + 0 = n;
    For every natural number n, for every natural number m, n + (Sm) = S(n + m);
    For every natural number n, for every natural number m, their product nm is a natural number;
    For every natural number n, n0 = 0; and
    For every natural number n, for every natural number m, n(Sm) = (nm) + n.
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