
A seemingly simple proof
The question: Prove that if x and y are distinct real numbers, then (x+1)^2=(y+1)^2 iff x+y=2. How does the conclusion change if we allow x=y.
(The professor's instruction require that we assume a^2=b^2 implies a=b or a=b)
This is what I have down so far:
pf  Let x,y be distinct real numbers, then (x+1)^2=(y+1)^2 iff x+y=2
assume a^2=b^2 implies a=b or a=b
Let P be the statement that (x+1)^2=(y+1)^2, and Q be the statement that x+y=2
We will show that 1. P implies Q, and 2. Q implies P
1. Showing P implies Q
(x+1)^2=(y+1)^2
thus,
x+1=y+1
x=y, but since x and y are distinct numbers, we ignore this case
OR
x+1 =  (y+1)
x+1 = y1
x + y = 2
(Did what I just do there show that P implies Q?)
2. Q implies P
To show this, do I just solve x+y=2, plug it back into P, and then confirm that P is in fact true?
And then, since I have shown that both P implies Q and Q implies P, does that mean my proof is finished?
Also, if we allow x=y then does the conclusion change at all? Seems like it wouldn't because if I used the instructors instructions (assume a^2=b^2 implies a=b or a=b), then the statements still hold true (because x=x or x=x, while x=x isn't true, x=x is, and we just need one of those statements to be true).
Please let me be on track...I'm SO lost in this course...
Thank You

q>p
x+y=2
x+y=11
x+1=1y
(x+1)^2=(1y)^2=(1+y)^2

For if x=y, the conclusion does change quite a bit. You lose the "only if" portion. For instance, allow x=y=2, then your P holds trivially, however, x+y=2+2=4≠2, so clearly P does not imply Q.

@ also sprach: simple enough. thanks for the explanation :
@ mattman: thanks for that! I couldn't get it straight on paper. That makes perfect sense though.
thanks guys so much. i am so glad God put capable mathematicians on this planet