# Prove that if x and y are

1. Jun 29, 2010

### Jamin2112

Prove that if x and y are ....

1. The problem statement, all variables and given/known data

Prove that if x and y are distinct real numbers, then (x+1)2=(y+1)2 if and only if x+y=-2. How does the conclusion change if we allow x=y?

2. Relevant equations

...

3. The attempt at a solution

Suppose x and y are real numbers. If x≠y then

x+2≠y+2
----> (x+2)/(y+2)≠1
----> (x+2)/(y+2)≠ x/y or y/x (since x/y=y/x=1)
----> (x+2)x=(y+2)y
or
(x+2)/x=(y+2)/y
----> (x+2)x=(y+2)y (since (x+2)/x=(y+2)/y ---> x=y)
----> x2+2x=y2+2y
---->x2+2x+1=y2+2y+1
---->(x+1)2=(y+1)2
----> x+1 = (y+1) or (-y-1)
----> x+1=-y-1 (since x+1=y+1 ---> x=y)
----> x+y=-2

?????

But I'm not sure if I've done everything it asked. I know for "P if and only Q," it needs to be proven that P--->Q and Q--->P, but it seems here that if P is (x+1)2=(y+1)2 and Q is x+y=-2, I'm just kind of going in circles by proving the "if and only if" part. See what I'm saying?

2. Jun 29, 2010

### Office_Shredder

Staff Emeritus
Re: Prove that if x and y are ....

It seems like your not equals sign turns into an equals sign here

To prove an if and only if statement you should really do two proofs. One is: assume P, and prove Q. The other is: assume Q, and then prove P. Here, x not equal to y is neither P nor Q, it's just an added assumption to be made at all times. We need to do two things: if (x+1)2=(y+1)2, then x+y=-2.

As a separate proof, you need to show also, if x+y=-2, then (x+1)2=(y+1)2.

3. Jun 30, 2010

### Staff: Mentor

Re: Prove that if x and y are ....

Or you can add x $\neq$ y to the hypothesis for each statement you're trying to prove.

I.e. x $\neq$ y and (x + 1)2 = (y + 1)2 ==> x + y = -2
for the one direction, and

x $\neq$ y and x + y = -2 ==> (x + 1)2 = (y + 1)2
for the other.

Hint: for the first direction, solve for x in the equation (x + 1)2 = (y + 1)2.

4. Jun 30, 2010

### Jamin2112

Re: Prove that if x and y are ....

Ha-ha! It was supposed to stay an "not equals" sign.

Anyhow, it's easy to prove (x+1)2=(y+1)2 ---> x+y=-2

-------------------------------------------------------------------

Proof:

Suppose x,y are real numbers and x≠y. If (x+1)2=(y+1)2, then taking the square root of both sides yields

x+1= (y+1) or (-y-1)

x+1=y+1 ---> x=y,

which doesn't meet the condition x≠y. We need only consider

x+1=-y-1 ---> x+y=-22.

-------------------------------------------------------------------

Right? And then I just go backwards to prove x+y=-2 ---> (x+1)2=(y+1)2 ?

5. Jun 30, 2010

### Staff: Mentor

Re: Prove that if x and y are ....

You have a typo in the last line of the first part.
It's pretty straightforward to go from x + y = -2 to (x+1)2=(y+1)2.

6. Jun 30, 2010

### Jamin2112

Re: Prove that if x and y are ....

x+y=-2

Adding 1 to both sides and subtracting y from both sides yields x+1=-(y+1). Squaring both sides yields (x+1)2=(y+1)2.

7. Jun 30, 2010

### Staff: Mentor

Re: Prove that if x and y are ....

There you go!

8. Jun 30, 2010

### hunt_mat

Re: Prove that if x and y are ....

Suppose
$$(x+1)^{2}=(y+1)^{2}$$
Then the following must be true:
$$0=(x+1)^{2}-(y+1)^{2}=(x+y+2)(x-y)$$
As we are told that x and y are distinct, it must be the case that:
$$x+y=-2$$