Discrete Maths: proving that 2 variables are the same

[Mampac]

G'day,

In order to prove something, I need to show that if x² − y² = y² − x² and y² − x² = x² − y², then x = y. (proving a relation is antisymmetric)

I do realize this is VERY easy and evident, but i need to prove it algebraically. the thing is if i take a look at the first two statements, i get x² = y², which doesn't necessarily mean x = y, since if i take a square root it gives +-.

If i substitute right-hand side of second supposition for the right-hand side of the first one, i get x² − y² = x² − y² which will just cancel everything out and leave 0 = 0 which is true.

My question is: does this mean x = y? or what else I need to do?

 

[pka]

In order to prove something, I need to show that if x² − y² = y² − x² and y² − x² = x² − y², then x = y. (proving a relation is antisymmetric)
I do realize this is VERY easy and evident, but i need to prove it algebraically. the thing is if i take a look at the first two statements, i get x² = y², which doesn't necessarily mean x = y, since if i take a square root it gives +-.
If i substitute right-hand side of second supposition for the right-hand side of the first one, i get x² − y² = x² − y² which will just cancel everything out and leave 0 = 0 which is true.
My question is: does this mean x = y? or what else I need to do?

Please state the exact question so that we can see the point of your statements.
As is, if \(x =1~\&~y=-1\) then \(x^2-y^2=y^2-x^2\) BUT \(x\ne y\) ..................................................[edited]

 

[JeffM]

You are not going to get a valid proof of a false proposition.

[MATH]3^2 - (-3)^2 = 0 = (-3)^2 - 3^2 \not \implies 3 = - 3.[/MATH]
Of course if it is stipulated that x and y are non-negative, ...

 

[Mampac]

Please state the exact question so that we can see the point of your statements.
As is, if \(=1~\&~y=-1\) then \(x^2-y^2=y^2-x^2\) BUT \(x\ne y\)

I have a relation R which is defined on Z x Z, means both variables can be any integers.

R = {(a, b) : a² − b² = b² − a²}

So as a matter of fact, x and y can be negative.

So as I understood, I can disprove my proposition by bringing the counterexample Jeff provided us with?

 

[pka]

I have a relation R which is defined on Z x Z, means both variables can be any integers.
R = {(a, b) : a² − b² = b² − a²}
So as a matter of fact, x and y can be negative.

Thank you for the clarification. That relation is reflexive, symmetric and transitive.
Now is the question about antisymmetric?

 

[HallsofIvy]

Do you understand what "anti-symmetric" means?

 

[Steven G]

You have x^2 - y^2 = y^2 - x^2. The other equation is exactly the same. No need to use it (we already know that x^2 - y^2 = y^2 - x^2).

Now y^2 - x^2 = -(x^2 - y^2). So x^2 - y^2 = - (x^2 -y^2). If a number or expression equals a non zero multiple of itself then it must equal 0.

So x^2 - y^2 = 0. Now if you subtract two quantities and get 0 then the two quantities must be equal. So x^2 = y^2. Then x = +/- y.

 

[Mampac]

Thank you for the clarification. That relation is reflexive, symmetric and transitive.
Now is the question about antisymmetric?

yes, the question is about whether the relation is a partial order, so if i find out it's not anti-symmetric (noted in the very first message) then it's not a partial order.
thank you, i'll provide one of those as a counterexample to prove it's not a partial order

 

[pka]

yes, the question is about whether the relation is a partial order, so if i find out it's not anti-symmetric (noted in the very first message) then it's not a partial order. thank you, i'll provide one of those as a counterexample to prove it's not a partial order

If you had said that you objective was to prove or disprove a partial order, it would have really helped.

 

[Mampac]

If you had said that you objective was to prove or disprove a partial order, it would have really helped.

ok, will take this into consideration, thank you again!
by the way, am I posting these questions in the right section (Beginning Algebra)? I'm planning on creating another thread (hoping the last one) about Discrete Maths and am not sure whether Discrete Maths is a part of Beginning Algebra or of something else, because I attended school & high school not in U.S., thus I'm not familiar with this curricula and stuff.

 

[Steven G]

This section is fine