I am taking real analysis, but that is really not important.....the important part is that my professor says that one of my methods of proof is invalid.
Essentially he is asking me to prove something of the form:
$$ X \iff Y $$
So basically I break the world into 2 cases: $A$ and $B$ (That is at least one of these is true and in fact only one is true)
I then prove the following:
$A\Rightarrow(X\iff Y)$
$B\Rightarrow(X\iff Y)$
Then I conclude:
$X\iff Y$
The problem my professor has is that my proof of $A\Rightarrow(X\iff Y)$ looks like this:
Assume $A$
...
$\therefore X \space and \space Y$
$\therefore X \iff Y$
And my proof of $B\Rightarrow(X\iff Y)$ is similar:
Assume $B$
...
$\therefore \tilde X \space and \space \tilde Y$
$\therefore X \iff Y$
My question is this......Is my method of proof valid, or am I missing something?
My professor's argument is with my statement that:
$ X \space and \space Y \Rightarrow (X \iff Y)$
He says, "Well you haven't shown that $X$ and $Y$ are related in any way"....his example of a counter example is:
Let $X=$ Milwaukee is by Lake Michigan and let $Y=$ All squares are rectangles....."Clearly" $X$ does not imply $Y$......
Edit:
Here is my theorem and proof:
Thm. $abs(a+b)=abs(a)+abs(b) \iff ab \ge 0$ where $abs(x)$ is the absolute value of $x$.
Proof:
There are four cases:
Case 1:
$$a\gt 0 \space and \space b\gt 0$$
Obviously $ab \ge 0$.
$$abs(a)=a \space and \space abs(b)=b$$
$$abs(a+b)=a+b$$
$$\therefore abs(a+b)=abs(a)+abs(b)$$
$$\therefore abs(a+b)=abs(a)+abs(b) \iff ab \ge 0$$
Case 2:
$$abs(a) < 0 \space and \space abs(b) < 0$$
Clearly $ab\ge 0$
$$abs(a)=-a$$
$$abs(b)=-b$$
$$-a-b=-(a+b)$$
But clearly $a+b\lt 0$
So
$$abs(a+b)=-a-b$$
$$\therefore abs(a+b)=abs(a)+abs(b)$$
$$\therefore abs(a+b)=abs(a)+abs(b) \iff ab \ge 0$$
Case 3:
$ a \gt 0 \space and \space b \lt 0$
Clearly $ab \lt 0$
$$abs(a)=a$$
$$abs(b)=-b$$
$$a-b=abs(a)+abs(b)$$
$$a+b=abs(a)-abs(b)$$
$$a-b\ne a+b \space and \space a-b\ne -a-b$$
$$\therefore abs(a+b)=abs(a)+abs(b)\iff ab\ge 0$$
$$\therefore abs(a+b)=abs(a)+abs(b)\iff ab\ge 0 \space in \space general$$
Case 4:
$a=0$
Clearly $ab\ge 0$ because it is zero.
By inspection we see:
$abs(b)=abs(b)$
$$\therefore abs(a+b)=abs(a)+abs(b)\iff ab\ge 0$$
$$\therefore abs(a+b)=abs(a)+abs(b)\iff ab\ge 0 \space in \space general$$
We technically have two more cases where in 3 and 4 we just swap $a$ and $b$ but those are symmetric and follow the same structure.