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Math Help - helpppppp!!!!!! math is difficulttttt

  1. #1
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    helpppppp!!!!!! math is difficulttttt

    0 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.

    Using these axioms (and logic), prove the following law of algebra:

    1) For every natural number i, for every natural number j, for every natural number k, ( i + j ) + k = i + ( j + k )

    2) For every natural number i, i + 0 = i and 0 + i = i
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  2. #2
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    Quote Originally Posted by jenjen
    2) For every natural number i, i + 0 = i and 0 + i = i
    Since i+j=j+i we only have half to prove (commutativity was proven last time). But that is one of the Peano axioms nothing to prove.
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  3. #3
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    Quote Originally Posted by jenjen
    For every natural number n, for every natural number m, n + (Sm) = S(n + m);
    Use this peano axiom.

    1) For every natural number i, for every natural number j, for every natural number k, ( i + j ) + k = i + ( j + k )
    Then (by axiom) it is necessay and sufficient to prove,
    S((i+j)+k)=S(i+(j+K))
    But then by top axiom ya have,
    i+j+S(k)=i+S(j+k)
    But then by top axiom ya have,
    i+j+S(k)=i+j+S(k)
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  4. #4
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    Hi,

    Thanks for your reply, but then my professor told me I can't do it that way. From the axioms, I know that for every natural number n, n + 0 = n but not for every natural number n, 0 + n = n. That is why I am having a hard time.
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    Then you can do it by induction.
    S=\{x\in \mathbb{N}| 0+x=x\}

    For n=0,
    0+0=0
    Thus, n\in S

    If, k\in S then,
    0+k=k
    If and only if,
    S(0+k)=S(k)
    Thus,
    0+S(k)=S(k)
    Thus, S(k)\in S

    Conditions for induction are satisfied, thus
    S=\mathbb{N}
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  6. #6
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    so you mean after that is proven by induction, like how you did it, then I can say that "For every natural number i, i + 0 = i and 0 + i = i?"
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  7. #7
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    Quote Originally Posted by jenjen
    so you mean after that is proven by induction, like how you did it, then I can say that "For every natural number i, i + 0 = i and 0 + i = i?"
    Yes, but i+0 does not need to be proven (one of axioms).
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  8. #8
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    Ohh ok, I got it. Thank you so much. How about this problem:

    For every natural number i, for every natural number j, for every natural number k, ( i j ) k = i ( j k ). Which axioms can I use??
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  9. #9
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    Quote Originally Posted by jenjen
    2) For every natural number i, i + 0 = i and 0 + i = i
    This is Lemma A in the other thread

    RonL
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  10. #10
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    Quote Originally Posted by jenjen
    0 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.

    Using these axioms (and logic), prove the following law of algebra:

    1) For every natural number i, for every natural number j, for every natural number k, ( i + j ) + k = i + ( j + k )
    from the other thread :

    Lemma A: for all natural numbers x,\ x+0=x=0+x
    To be proven later

    Lemma B: for all natural numbers x, y,\ x+S(y)=S(x+y)=S(x)+y
    To be proven later.
    We will prove that for all x,\ y,\ z\ \in \mathbb{N} that

    <br />
(x+y)+z=x+(y+z)<br />

    by induction on x.

    Base case:

    For all y,\ z \in \mathbb{N}

    <br />
(0+y)+z=y+z, by Lemma A
    <br />
=0+(y+z)<br />
, again by Lemma A
    .

    So the base case is true.

    Induction step:

    For any y,\ z \in \mathbb{N}, suppose there exists a x such that:

    <br />
(x+y)+z=x+(y+z)<br />

    Now look at the successor of the LHS:

    <br />
S((x+y)+z)=S(x+y)+z=(S(x)+y)+z<br />
double application of Lemma B

    Now look at the successor of the RHS:

    <br />
S((x+(y+z))=S(x)+(y+z)<br />
again by Lemma B

    So: (S(x)+y)+z=S(x)+(y+z), so we have proven asociativity
    for S(x) from its assumption for x which completes the induction step.

    Hence for all x,\ y,\ z \in \mathbb{N} we have proven:

    <br />
(x+y)+z=x+(y+z)<br />

    by induction on x

    RonL
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