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Thread: Canonical representation?

  1. #1
    Member aldrincabrera's Avatar
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    Cool Canonical representation?

    ,.good day everyone,.,need ur help with how to prove this,.
    Let n>=2 be an integer with canonical representation n=p₁^{a₁}p₂^{a₂}p₃^{a₃}...p_{k}^{a_{k}}. an integer m>=1 is a positive divisor of n if and only if m=p₁^{b₁}p₂^{b₂}p₃^{b₃}...p_{k}^{b_{k}}, where 0<=b_i<=a_i for all 1<=i<=k.
    thnx in advance...
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Canonical representation?

    If $\displaystyle \{p_i\}$ is the set of all prime numbers, we can express $\displaystyle n,m, s$ positive integers as unique products

    $\displaystyle n=\prod{p_i^{\alpha_i}}$ , $\displaystyle m=\prod {p_i^{\beta_i}}$ , $\displaystyle s=\prod {p_i^{\gamma_i}} $

    where $\displaystyle \alpha_i\geq 0,\beta_i\geq 0,\gamma_i\geq 0$ are finitely many non zero integers.

    If $\displaystyle \beta_i\leq \alpha_i$ for all $\displaystyle i$ evidently $\displaystyle m|n$ . If $\displaystyle m|n$ there exists $\displaystyle s$ positive integer such that $\displaystyle n=ms$ that is, $\displaystyle \prod p_i^{\alpha_i}=\prod p_i^{\beta_i+\gamma_i}$ . By the unique factorization, $\displaystyle \alpha_i=\beta_i+\gamma_i$ , hence $\displaystyle \beta_i\leq \alpha_i$ for all $\displaystyle i$ .
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