Simple proof that if x^2 = y^2 then x=-y or x=y

  • Thread starter Jose Bouza
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  • #1

Homework Statement


Hi, I'm a high school student currently working through Spivak for the first time and this is my first proof based introduction to mathematics so I just wanted to make sure that I was doing the following correctly:

Prove that if [itex] x^2 = y^2[/itex] then [itex]x=y[/itex] or [itex]x=-y[/itex]

The Attempt at a Solution


[itex] (-y)^2 = (-y)*(-y) = (-1*y)*(-1*y) = (-1)*(-1)*y*y = 1*y*y = y*y = y^2 [/itex]
so...
[itex] (-y)^2 = y^2 [/itex]
or...
[itex] x^2 = y^2 [/itex] when [itex] x= -y [/itex]

Is this proof logical assuming the 12 basic properties of multiplication and addition that spivak lists in chapter 1?

For the fist case ([itex] x= y [/itex]) I'm kind of confused. I could do this:

if [itex] y*y = y^2 [/itex]
then [itex] y^2 = y^2 [/itex] and [itex]x^2 = y^2[/itex] if [itex] x=y[/itex]

but that seems far too easy, am I missing something or is that part just that simple?

Thanks for any help :)
 

Answers and Replies

  • #2
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Homework Statement


Hi, I'm a high school student currently working through Spivak for the first time and this is my first proof based introduction to mathematics so I just wanted to make sure that I was doing the following correctly:

Prove that if [itex] x^2 = y^2[/itex] then [itex]x=y[/itex] or [itex]x=-y[/itex]

The Attempt at a Solution


[itex] (-y)^2 = (-y)*(-y) = (-1*y)*(-1*y) = (-1)*(-1)*y*y = 1*y*y = y*y = y^2 [/itex]
so...
[itex] (-y)^2 = y^2 [/itex]
or...
[itex] x^2 = y^2 [/itex] when [itex] x= -y [/itex]

Is this proof logical assuming the 12 basic properties of multiplication and addition that spivak lists in chapter 1?

For the fist case ([itex] x= y [/itex]) I'm kind of confused. I could do this:

if [itex] y*y = y^2 [/itex]
then [itex] y^2 = y^2 [/itex] and [itex]x^2 = y^2[/itex] if [itex] x=y[/itex]

but that seems far too easy, am I missing something or is that part just that simple?

Thanks for any help :)
I don't have Spivak, so I don't know the 12 properties he lists.

Starting with x2 = y2, you can write an equivalent equation, x2 - y2 = 0. The left side can be factored...
 
  • #3
I don't have Spivak, so I don't know the 12 properties he lists.

Starting with x2 = y2, you can write an equivalent equation, x2 - y2 = 0. The left side can be factored...

Basically just the associative, commutative and distributive properties (as in ##a(x+y) = ax + ay##, not at the level of factoring of ##x^2-y^2##) . Factoring is a higher level algebraic property and must first be proved using those three if you wish to use it here as far as I know.
I know this property is super elementary but the basic idea of the problem set is to prove all the common algebraic properties using the most fundamental ones.
 
  • #4
HallsofIvy
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Homework Statement


Hi, I'm a high school student currently working through Spivak for the first time and this is my first proof based introduction to mathematics so I just wanted to make sure that I was doing the following correctly:

Prove that if [itex] x^2 = y^2[/itex] then [itex]x=y[/itex] or [itex]x=-y[/itex]

The Attempt at a Solution


[itex] (-y)^2 = (-y)*(-y) = (-1*y)*(-1*y) = (-1)*(-1)*y*y = 1*y*y = y*y = y^2 [/itex]
so...
[itex] (-y)^2 = y^2 [/itex]
or...
[itex] x^2 = y^2 [/itex] when [itex] x= -y [/itex]

Is this proof logical assuming the 12 basic properties of multiplication and addition that spivak lists in chapter 1?

For the fist case ([itex] x= y [/itex]) I'm kind of confused. I could do this:

if [itex] y*y = y^2 [/itex]
then [itex] y^2 = y^2 [/itex] and [itex]x^2 = y^2[/itex] if [itex] x=y[/itex]
The problem is that this is NOT what you want to prove!
You want to prove that if [itex]x^2= y^2[/itex] then x= y or x= -y, NOT that "if x= y then x^2= y^2.

Instead, I suggest that you go from [itex]x^2= y^2[/itex] to [itex]x^2- y^2= 0[/itex] and factor the left side.

but that seems far too easy, am I missing something or is that part just that simple?

Thanks for any help :)
 
  • #5
Dick
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Basically just the associative, commutative and distributive properties (as in ##a(x+y) = ax + ay##, not at the level of factoring of ##x^2-y^2##) . Factoring is a higher level algebraic property and must first be proved using those three if you wish to use it here as far as I know.
I know this property is super elementary but the basic idea of the problem set is to prove all the common algebraic properties using the most fundamental ones.

"Factoring" isn't a separate property. You should be able to show that ##(x+y)(x-y)=x^2-y^2## using Spivak's properties. So you can use that to prove what you want.
 
  • #6
"Factoring" isn't a separate property. You should be able to show that ##(x+y)(x-y)=x^2-y^2## using Spivak's properties. So you can use that to prove what you want.
Are you saying I can just assume that you can factor ##x^2 - y^2## to ## (x+y)(x-y)## using the property ##a(x+y) = ax + ay## or are you suggesting I prove it using the three given properties?
If its the latter I don't think that the case since we actually have to prove what ##x^2 - y^2## factors into in a later problem so I'm assuming the author didn't intend that property to be used yet (the general structure involves using proofs from previous problems to solve new ones). Thats why I'm asking if the "proof" I provided is correct.
But thanks for the help anyways!
 
  • #7
Dick
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Are you saying I can just assume that you can factor ##x^2 - y^2## to ## (x+y)(x-y)## using the property ##a(x+y) = ax + ay## or are you suggesting I prove it using the three given properties?
If its the latter I don't think that the case since we actually have to prove what ##x^2 - y^2## factors into in a later problem so I'm assuming the author didn't intend that property to be used yet (the general structure involves using proofs from previous problems to solve new ones). Thats why I'm asking if the "proof" I provided is correct.
But thanks for the help anyways!

Yes, your proof that if ##x=(-y)## then ##x^2=y^2## and that if ##x=y## then ##x^2=y^2## (and that is as obvious as you think - maybe even more so). What your approach is missing is showing that those are the ONLY two solutions. You can do that with the factorization. Multiply ##(x+y)(x-y)## out and show the result is ##x^2-y^2## using the Spivak properties. Then use the factorization to show that there are only two solutions.
 
  • #8
SammyS
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How about adding zero ( in the form of ##\ xy-xy\ ##) to ##\ x^2-y^2\ ## ?
 
  • #9
Yes, your proof that if ##x=(-y)## then ##x^2=y^2## and that if ##x=y## then ##x^2=y^2## (and that is as obvious as you think - maybe even more so). What your approach is missing is showing that those are the ONLY two solutions. You can do that with the factorization. Multiply ##(x+y)(x-y)## out and show the result is ##x^2-y^2## using the Spivak properties. Then use the factorization to show that there are only two solutions.
Ahh okay I see, thanks again.
 
  • #10
WWGD
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To nitpick, notice that you need to assume commutativity when you show ## x^2-y^2= (x+y)(x-y) ##.
 
  • #11
How about adding zero ( in the form of ##\ xy-xy\ ##) to ##\ x^2-y^2\ ## ?
To then get to ##(x+y)(x-y)##? I think that going from ##x(x+y) - y(x+y)## to ##(x+y)(x-y)## must be proved to be correct in of itself no?
 
  • #12
Dick
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To then get to ##(x+y)(x-y)##? I think that going from ##x(x+y) - y(x+y)## to ##(x+y)(x-y)## must be proved to be correct in of itself no?

Of course, but isn't that easy??
 
  • #13
SammyS
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To then get to ##(x+y)(x-y)##? I think that going from ##x(x+y) - y(x+y)## to ##(x+y)(x-y)## must be proved to be correct in of itself no?
What? No distributive law?
 
  • #14
Of course, but isn't that easy??
Oh okay I see. At this point the inverse of the distributive property can be used since ##x+y## is equal to the ##a## in ##a(x-y) = ax - ay## right?
 
  • #15
Dick
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Oh okay I see. At this point the inverse of the distributive property can be used since ##x+y## is equal to the ##a## in ##a(x-y) = ax - ay## right?

Yes!
 
  • #16
What? No distributive law?
Yeah I didn't see how to directly apply the distributive property at first but I got it now, thank you both!
 

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