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Math Help - divisibility

  1. #1
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    divisibility

    Prove that 1 and -1 are only divisible by 1 and -1.

    I think that I'm making this too complicated. Any help would be greatly appreciated.
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  2. #2
    MHF Contributor arbolis's Avatar
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    I do it with 1, do it for -1.
    A number a \in \mathbb{N} is divisible by a number b if and only if a=bq+r with 0\leq r <b.
    It's obvious that a\geq b.
    So in your case you have that a=1. That means that if a was divisible by a number b, it would necessarily be 1. We can try to check if it works : 1=1q+r=1\cdot1 +0, which works.
    I think what I did is not wrong...
    Good luck. (All you have to do is to generalize what I did for the numbers \in \mathbb{N} for any number \in \mathbb{Z}. I suggest you to check this : Division algorithm - Wikipedia, the free encyclopedia out.)
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  3. #3
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    Quote Originally Posted by arbolis View Post
    I do it with 1, do it for -1.
    A number a \in \mathbb{N} is divisible by a number b if and only if a=bq+r with 0\leq r <b.
    The division algorithm just says that given a and b, there exists a unique q and r, 0 \leq r < b such that a = qb + r. Doesn't mention about a dividing b or vice versa but we can see that b divides a when r = 0.

    For this question, just go by definition:
    a \mid 1 \iff ac_{1} = 1 \iff a = \frac{1}{c_{1}} \ \ c_{1} \in \mathbb{Z}

    If \mid c_{1} \mid > 1, we have that the numerator is less than the denominator which cannot be an integer. Thus \mid c_{1} \mid = 1 \ \Rightarrow \ c_{1} = 1, -1. So a must be ...
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