# Thread: A seemingly simple proof

1. ## 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 = -y-1
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

2. q-->p

x+y=-2

x+y=-1-1

x+1=-1-y

(x+1)^2=(-1-y)^2=(1+y)^2

3. 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.

4. @ 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