Additive Inverse and integer modulo

Question

I am not completely sure how inverses work with sets of integer modulo. I have just started to learn about them. I have tried some practice problems, though I am not sure if my approach is correct in solving them.

Currently, this is what I am doing when solving for additive inverses:

Example Problem: Which element in $\mathbb{Z}_8$ is an additive inverse of $[13]$?

My solution:

$13\equiv 5(\text{mod }8 )$, and so we need to find the inverse of $5(\text{mod }8)$. The additive inverse of $x$ is simply the number which when added to $x$ yields the additive identity and the additive identity is $0$.

So what $y$ should we add to $x=5$ to give $x+y\equiv 0(\text{mod }8)$?

Say $y\equiv−x\equiv−5$ and ask what is equivalent to $−5$ modulo $8$? The answer will be $y\equiv 3(\text{mod }8)$.

Or, the solution to $5+y\equiv 8\equiv0(\text{mod }8)$. Again you’ll get $y\equiv 3(\text{mod }8)$.

Thus, the element $[3]$ is an additive inverse of $[13]$.

Can someone check if this is done correctly? If anyone has any other ways to solve this or advice on how to better understand integer modulo, I would appreciate it! Thanks!

You should justify the claim $, x+y\equiv 0\Rightarrow x\equiv -y,,$ e.g. (conceptually) by the Congruence Sum Rule, or directly by the definition of congruence, etc. — Bill Dubuque, Mar 25 '21 at 20:33
What is your definition of $,\Bbb Z_8?\ $ You are using both equivalence classes and congruences so it is not at all clear. — Bill Dubuque, Mar 25 '21 at 20:38

score 0 · Accepted Answer · edited Mar 25 '21 at 20:09

It's correct. Don't worry modular arithmetic is easy and is meant to be easy.

$a \equiv b \mod n \iff n\mid a-b$.

So $a + x \equiv b + x\pmod n \iff n\mid (a+x) - (b+x)$ but $(a+x) - (b+x) = a-b$.

So $a + x \equiv b + x \pmod n \iff a \equiv b \pmod n$.

So... you can always add to the same thing to both sides. Just liked you'd hope you could.

So $a + x \equiv 0\pmod n \iff$

$x \equiv - a \pmod n$.

And if you don't like negatives as $n \equiv 0 \pmod n$ (Because $n\mid n-0= n$)

We can add $n$.

$x \equiv -a \pmod n$ so

$x\equiv x + 0 \equiv -a + n \equiv n-a \pmod {n}$.

(That works because $n \equiv 0 \pmod n \implies n + (-a) \equiv 0 + (-a) \pmod n$).

It's meant to be easy.

....

So $13 + x \equiv 0 \pmod {8}$ so

$5 + x \equiv 0 \pmod 8$ so

$x \equiv -5 \equiv -5 + 8 \equiv 3\pmod 8$.

That's all you have to do and that is enough.

The key law used above is the Congruence Sum Rule $,a\equiv \bar a,\ b\equiv \bar b\Rightarrow a+b\equiv \bar a +\bar b.,$ That linked post has proofs of all of the basic congruence laws. — Bill Dubuque, Mar 25 '21 at 20:27

Additive Inverse and integer modulo

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