1. 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.) $f: \mathbb{Z}_{14} \rightarrow \mathbb{Z}_{14}:
f([a]_{14})=[a]_{14}\cdot[3]_{14};$

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

confused need help.

2. Originally Posted by skystar
consider arithmetic modulo 14. Determine which of the following functions are bijective.If it is bijective determine the inverse function.

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

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

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.

3. i dont understand how $3^{-1}=5$ though? thanks

4. Hello

Originally Posted by skystar
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 :

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

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

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