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Math Help - baffled...bijective arithmetic &congruence

  1. #1
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    baffled...bijective arithmetic &congruence

    consider arithmetic modulo 14. Determine which of the following functions are bijective.If it is bijective determine the inverse function.

    i.)  f: \mathbb{Z}_{14} \rightarrow \mathbb{Z}_{14}:<br />
f([a]_{14})=[a]_{14}\cdot[3]_{14};

    ii.) f: \mathbb{Z}_{14} \rightarrow \mathbb{Z}_{14}:<br />
f([a]_{14})=[a]_{14}\cdot[4]_{14};

    confused need help.
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    Quote Originally Posted by skystar View Post
    consider arithmetic modulo 14. Determine which of the following functions are bijective.If it is bijective determine the inverse function.

    i.)  f: \mathbb{Z}_{14} \rightarrow \mathbb{Z}_{14}:<br />
f([a]_{14})=[a]_{14}\cdot[3]_{14};

    ii.) f: \mathbb{Z}_{14} \rightarrow \mathbb{Z}_{14}:<br />
f([a]_{14})=[a]_{14}\cdot[4]_{14};

    confused need help.
    \mathbb{Z}_{14} is a cyclic group with respect to addition and 3 and 14 relatively prime. That means the 3 is a generator of the group. So F is both 1-1 and onto. 3^{-1}=5 in \mathbb{Z}_{14}

    So f^{-1}([a]_{14})=[a]_{14} \cdot [5]_{14}

    This can be verified by function composition.

    For ii) The range of f is the even elements so the function is not onto.

    I hope this helps.

    Good luck.
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    i dont understand how  3^{-1}=5 though? thanks
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  4. #4
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    Hello

    Quote Originally Posted by skystar View Post
    i dont understand how  3^{-1}=5 though? thanks
    The inverse of 3 modulo 14

    The inverse d of 3 is such that 3d = 1 \mod 14

    Use the Euclidian algorithm to find it :

    \textbf{14}=\textbf{3} \times 4+{\color{red}2} \implies {\color{red}2}=\textbf{14}-\textbf{3} \times 4

    \textbf{3}={\color{red}2}+1 \implies 1=\textbf{3}-{\color{red}2}=\textbf{3}-(\textbf{14}-\textbf{3} \times 4)=-\textbf{14}+{\color{blue}5} \times \textbf{3}

    ---> 5 \times 3=1+14=1 \mod 14
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