# Additive Inverse and integer modulo

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!

• It's quite correct for me. – Bernard Mar 25 at 20:08
• 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 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 at 20:38

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 at 20:27
• Thanks so much! It really clarifies everything! – Eugene Mar 25 at 21:58