What Happens When Distinct Real Numbers Satisfy (x+1)^2 = (y+1)^2?

[Jamin2112]

Prove that if x and y are ...

Homework Statement

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?

Homework Equations

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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?

 

[Office_Shredder]

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

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.

 

[Mark44]

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.

 

[Jamin2112]

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.

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

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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.

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

 

[Mark44]

You have a typo in the last line of the first part.

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

It's pretty straightforward to go from x + y = -2 to (x+1)²=(y+1)².

 

[Jamin2112]

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.

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.

 

[Mark44]

There you go!

 

[hunt_mat]

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:
$$x+y=-2$$