Transitive Math Help: xRy if (x+y)2 = 1

[scorpius1782]

Homework Statement

R is the relation of the set R; xRy if and only if (x + y)² = 1.
Is it transitive.

Homework Equations

The Attempt at a Solution

Every description of transitive involves 3 elements. Here, I just have x and y. It is defined as xRy so there is a relation but is it transitive all by itself with just those two elements?

 

[LCKurtz]

Homework Statement

R is the relation of the set R; xRy if and only if (x + y)² = 1.
Is it transitive.

Homework Equations

The Attempt at a Solution

Every description of transitive involves 3 elements. Here, I just have x and y. It is defined as xRy so there is a relation but is it transitive all by itself with just those two elements?

Write down what xRy means, what yRz means and what xRz means using your given equation. Then decide whether you can prove the third from the first two or find a counterexample.

 

[scorpius1782]

This confuses me because there is no z. What kind of relationship can y or x have with something that doesn't exist? I can't describe what it means for yRz at all because of this.

I can say that xRy stems from the fact that x and y are a pair that satisfies the given equation. R contains the pair (x,y). Since there is no z so: (x,y)$\bar{R}$z. (Can't find a way to strikeout R.. so \bar = strikeout)

But like I said, does it matter? If I just have the two elements is that enough to say that it is transitive?

 

[LCKurtz]

This confuses me because there is no z. What kind of relationship can y or x have with something that doesn't exist? I can't describe what it means for yRz at all because of this.

I can say that xRy stems from the fact that x and y are a pair that satisfies the given equation. R contains the pair (x,y). Since there is no z so: (x,y)$\bar{R}$z. (Can't find a way to strikeout R.. so \bar = strikeout)

But like I said, does it matter? If I just have the two elements is that enough to say that it is transitive?

No, you aren't understanding this at all. R is a relation between numbers. Any two numbers may or may not be related. Is 1R3? To answer you have to ask whether (1+3)^2=1. It doesn't make any sense to talk about whether (x,y)Rz because (x,y) is not a number. If w and z are numbers it makes sense to say wRz if $(w+z)^2 = 1$. It will be true for some choices of w and z and not for others.

So I ask you again: What are you given and what are you trying to prove?

 

[scorpius1782]

I shouldn't have said (x,y) but rather x or y. I was trying to save myself time.

Are x,y,z suppose to be elements in the same set then? I've been thinking of them as in different real sets.

 

[vela]

What is the definition of transitivity? Does it ever involve R where there's an instance with more than two elements?

 

[scorpius1782]

What is the definition of transitivity? Does it ever involve R where there's an instance with more than two elements?

I thought I already defined it. If I did so incorrectly please tell me how/why! I'm not sure I understand your question but R is only a reltionship between 2 elements as far as I've seen. But my question is still are they all in the same set? In the end I think it probably doesn't matter. z could be in there but it just isn't used. My gut tells me it is transitive but I have no basis for the assumption.

Sorry grammar, on phone.

 

[vela]

No, you haven't defined what it means for R to be transitive, and it seems pretty clear from your questions that you don't know the definition. Even if you do know it, just humor us and tell us what the precise definition of transitivity is.

 

[scorpius1782]

I'm sorry, I thought I already had but couldn't look because I couldn't scroll up on my phone. As my book explains it:if xRy and yRz then xRz. This makes perfect sense to me. But in my case, what is z? That is the part I don't get.

 

[vela]

I'm not sure I understand your confusion here. Suppose two elements x and y are related if x=y. Transitivity would mean if a=b and b=c, then a=c. Your question is asking "what is c?"

 

[scorpius1782]

Yes, because z is not in my equation anywhere. Do I just say that z exists but x and y are not related? Do I even need to mention z?

 

[vela]

What equation are you referring to? There are no equations here.

 

[scorpius1782]

The $(x+y)^2=1$ at the top

 

[vela]

You do realize that x and y are dummy variables, right? Reread what LCKurtz said in post #4.

 

[scorpius1782]

okay so given that y=1-x from the equation. If x=2 then y=-1 then 2R-1. Then if z=1-y then -1R2. So x=z. Am I on the right track now?

 

[vela]

Not really. You shouldn't be solving for y or z. You need to prove the statement for any x, y, and z that satisfy the given conditions — or find a counterexample to show R isn't transitive.

xRy means that (x+y)²=1.
yRz means that (y+z)²=1.

That's all you know about x, y, and z. Given that, you want to show that (x+z)²=1 from which you can conclude that xRz. Then you will have shown that xRy and yRz implies xRz, i.e. that R is transitive.

Or you might decide R is not transitive. To show that, you want to find a counterexample. Find specific values of x, y, and z such that x and y are related and y and z are related, but x and z are not related.

 

[HallsofIvy]

okay so given that y=1-x from the equation.

No, that isn't given. If $(x+ y)^2= 1$ then either $x+ y= 1$, whence $y= 1- x$ or $x+ y= -1$, whence $y= -1- x$.

If x=2 then y=-1 then 2R-1. Then if z=1-y then -1R2. So x=z. Am I on the right track now?