Prove that if x and y are

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  • #1
Jamin2112
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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



...

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?
 

Answers and Replies

  • #2
Office_Shredder
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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.
 
  • #3
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Or you can add x [itex]\neq[/itex] y to the hypothesis for each statement you're trying to prove.

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

x [itex]\neq[/itex] 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
Jamin2112
986
12


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

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

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
36,437
8,412


You have a typo in the last line of the first part.
Jamin2112 said:
x+1=-y-1 ---> x+y=-22.

It's pretty straightforward to go from x + y = -2 to (x+1)2=(y+1)2.
 
  • #6
Jamin2112
986
12


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.
 
  • #8
hunt_mat
Homework Helper
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Suppose
[tex]
(x+1)^{2}=(y+1)^{2}
[/tex]
Then the following must be true:
[tex]
0=(x+1)^{2}-(y+1)^{2}=(x+y+2)(x-y)
[/tex]
As we are told that x and y are distinct, it must be the case that:
[tex]
x+y=-2
[/tex]
 

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