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Math Help - Congruence proof, did I do it somewhat correct?

  1. #1
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    Congruence proof, did I do it somewhat correct?

    I just want to see how well I did with this proof, and would like to know any corrections to make if I did it improperly, thanks!

    The proof is Show if a ≡ b(mod m) and n ∈ Z+ where n | m, then a ≡ b(mod n)

    My answer:

    suppose
    a ≡ b(mod m). Then n ∈ Z+ , and a,b,m,k ∈ Z. Now suppose a=b+km, then a-b=km and km=n. Thus n|m. Then a ≡ b(mod n). Hence, if a ≡ b(mod m) then a ≡ b(mod n).

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  2. #2
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    Re: Congruence proof, did I do it somewhat correct?

    Quote Originally Posted by michaelgg13 View Post
    I just want to see how well I did with this proof, and would like to know any corrections to make if I did it improperly, thanks!

    The proof is Show if a ≡ b(mod m) and n ∈ Z+ where n | m, then a ≡ b(mod n)

    My answer:

    suppose
    a ≡ b(mod m). Then n ∈ Z+ , and a,b,m,k ∈ Z. Now suppose a=b+km, then a-b=km and km=n. Thus n|m. Then a ≡ b(mod n). Hence, if a ≡ b(mod m) then a ≡ b(mod n).

    your use of "thus n|m" is odd because you are given that n|m.

    just do this

    $a \equiv b (\bmod~ m) \Rightarrow \exists k \in \mathbb{Z} \ni a = km + b$

    Now $n|m \Rightarrow \exists j \in \mathbb{Z} \ni m = jn$ so

    $a \equiv b (\bmod~ m) \Rightarrow a = jkn + b = (jk)n + b$ and thus

    $a \equiv b (\bmod ~n)$
    Thanks from michaelgg13
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  3. #3
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    Re: Congruence proof, did I do it somewhat correct?

    Quote Originally Posted by romsek View Post
    your use of "thus n|m" is odd because you are given that n|m.

    just do this

    $a \equiv b (\bmod~ m) \Rightarrow \exists k \in \mathbb{Z} \ni a = km + b$

    Now $n|m \Rightarrow \exists j \in \mathbb{Z} \ni m = jn$ so

    $a \equiv b (\bmod~ m) \Rightarrow a = jkn + b = (jk)n + b$ and thus

    $a \equiv b (\bmod ~n)$
    ahh ok, that makes more sense. thanks so much!
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  4. #4
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    Re: Congruence proof, did I do it somewhat correct?

    Quote Originally Posted by michaelgg13 View Post
    I just want to see how well I did with this proof, and would like to know any corrections to make if I did it improperly, thanks!

    The proof is Show if a ≡ b(mod m) and n ∈ Z+ where n | m, then a ≡ b(mod n)
    Here another way. You should know that if $m|n~\&~n|j$ then $m|j$.

    We are given that $m|(a-b) ~\&~n|m$ so $n|(a-b)$. qed
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