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In an old exam it's asked to say which elements in $\mathbb{Z}_{14}$ have multiplicative inverses. Furthermore, it's asked to calculate the multiplicative inverse of $9 \text { mod } 14$ in $\mathbb{Z}_{14}$.

Now my question is how can you do this? Because it's not explained in our readings, also no task in any of our homework, so I don't know how this could be solved because I have no source / example : /

I have read some on the internet for the first part, how to get multiplicative inverse elements in $\mathbb{Z}_{14}$.

I think for this we need $\text{gcd(x, 14)=1}$? If so these would be inverse: $1,3,5,9,11,13$

I hope this is correctly done at least? But I have no idea for the second question..?

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    Hint: $3\times 5\equiv 1\pmod {14}$. So the multiplicative inverse of $3$ is $5$. – 2017-02-11
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    @lulu "$\times$" stands for multiplication sign here right? – 2017-02-11
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    General note: there are systematic ways to find multiplicative inverses...see [Bezout's Identity](https://en.wikipedia.org/wiki/B%C3%A9zout's_identity) But for small numbers, as here, it is often easier to just play around until you see it. – 2017-02-11
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    Yes, $\times$ means multiply. – 2017-02-11
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    Any elements that are relatively prime to 14 will have an inverse mod 14, so you indeed found them all. I'm not sure why you're asked to calculate the inverse of 9 mod 14 when you just showed there isn't one. – 2017-02-11
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    The general method to find a modular inverse is to find a *Bézout's relation* between the modulus and the invertible number. This is done by the *Extended Euclidean algorithm*. You can see an example in my answer to [this question](http://math.stackexchange.com/questions/2100531/how-to-solve-this-linear-congruence-46x-equiv-6-pmod-202/2100598#2100598). – 2017-02-11
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    @Kaynex there is indeed an inverse of $9 \bmod 14$. – 2017-02-12

2 Answers 2

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You have to search through the elements of $\Bbb Z_{14}$ those who satisfy $gcd(x,14)=1$.

For instance, $11 \in \Bbb Z_{14}$ is such that $11\cdot9\equiv 1$ $(\mathsf {mod}$ $14)$.

Why we don't search those $y \in \Bbb Z_{14}$ such that $gcd(y, 14) \neq 1$? Because otherwise we can find a $t \in \Bbb Z_{14}$, $t\neq0$ such that $yt \equiv0$ $(\mathsf {mod}$ $14)$, thus nor $y$ nor $t$ can be invertible in $\Bbb Z_{14}$.

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In order for an inverse to exist, you need the value and the modulus to be coprime. Here $\gcd(9,14)=1$ so we're OK there.

To get the idea of a multiplicative inverse with these small numbers, you can just explore the multiples of $9 \bmod 14$:
$2\cdot 9 \equiv 18\equiv 4$
$3\cdot 9 \equiv 4+9\equiv 13$
$4\cdot 9 \equiv 13+9\equiv 8$
$5\cdot 9 \equiv 8+9\equiv 3$
$6\cdot 9 \equiv 3+9\equiv 12$
$7\cdot 9 \equiv 12+9\equiv 7$
$8\cdot 9 \equiv 7+9\equiv 2$
$9\cdot 9 \equiv 2+9\equiv 11$
$10\cdot 9 \equiv 11+9\equiv 6$
$11\cdot 9 \equiv 6+9\equiv 1$ as required

Of course you can also just multiply the numbers directly and find the next lowest multiple of $14$ too, to do the same exploration.

It's worth getting comfortable with the idea of using the negative versions of congruences, because you can see there that $3\cdot 9 \equiv 13 \bmod 14$, and $13\equiv -1 \bmod 14$. So since $3\cdot 9 \equiv -1$, $-3\cdot 9 \equiv 1 \bmod 14$ and this gives the same answer since $-3 \equiv 11 \bmod 14$.

Also, for other shortcuts, we could look at "the negative side" of $9$:
$9\equiv -5 \bmod 14$ and
$-3\cdot -5 = 15\equiv 1 \bmod 14$ so
the inverse of $9$ is $-3\equiv 11 \bmod 14$.