Given $A \subseteq \mathbf{Z}$ and $x\in \mathbf{Z}$, we say that $x$ is $A$-mirrored if and only if $−x\in A$. We also define...

Question

Sorry if this question seems kind of long but I am confused for part C.

My proof for part C that $M_a$ is closed under addition is as follows: The set $M_a$ is closed.

Let $x$ be in $M_a$ and $Y$ be in $M_a$ then $(X+y)$ exists in $M_a$. Then $-x$ is in $A$ and $-y$ is in $A$ then $-(x+y) =-x-y= (-x)+(-y)$. Because we know that $-x$ exists in $A$ and $(-y)$ exists in A then $-(x+y)$ exists in $M_a$.

Therefore Ma is also closed under addition.

Sorry again, I am just really curious on how to approach this and whether my approach is right or wrong.

Answer 1

$a)$ it is $-x\not\in A$

$b)$ The elements are $\{0,-1,6,7,-7,100\}$

$c)$ take $a,b\in M_b$.Then $-a,-b\in B$ so $-a-b\in B\implies a+b\in M_b$

$d)$ when $B$ satisfies $a\in B\implies -a\in B$

Answer 2

In short: The approach is correct but the answer is not.

Longer version: You have to be careful about what you write in proofs. Always keep in mind what you are trying to prove. The exercise says:

Show that $M_A$ is closed under addition.

Going straight to the definition, you would have to show that for all $x,y \in $$M_A$, $x+y\in$$M_A$. So as with almost every problem of this nature, we start by taking arbitrary $x,y \in M_A$. Now, we want to show that $x+y\in M_A$ as well.

You said: Let $x$ be in $M_a$ and $Y$ be in $M_a$ then (X+y) exists in $M_a$. That is what you want to prove, so you can't just state it. Moreover, be careful with the uppercases, as $X$ and $x$ are not the same thing (same with $y$ and $a$).

The next thing you do is right, writing that $-x\in A$ and $-y\in A$. But then you mix up again, saying that this implies that * ... then $-(x+y)$ exists in $M_a$*. This again, is not true. This is not even what we want ... we want to show that $(x+y)\in M_A$. The argument here would be: ... then, since $A$ is closed under addition, $-(x+y) \in$ $A$.

Finally we finish the proof: By definition, $-(x+y) \in A$ if and only if $(x+y) \in M_A$, since $-(x+y) \in A$, we are done.