Classifying Relations: Reflexive, Irreflexive, or Neither?

[scorpius1782]

Homework Statement

xRy iff (x+y)²=1
determine if reflexive, irreflexive or neither

Homework Equations

The Attempt at a Solution

I'm having trouble understanding the definitions. I know that reflexive means that ever element is related to itself and that irreflexive means that no element is related to itself but I don't really understand what that means.

For this problem it is only true y=-x-1 or y=1-x.

But I don't understand what to do from here. Does this mean that since x and y are dependent on one another then they are reflexive? But how would I know if it an element in 'x' (say 10) is dependent onitself?

 

[LCKurtz]

Homework Statement

xRy iff (x+y)²=1
determine if reflexive, irreflexive or neither

Homework Equations

The Attempt at a Solution

I'm having trouble understanding the definitions. I know that reflexive means that ever element is related to itself and that irreflexive means that no element is related to itself but I don't really understand what that means.

For this problem it is only true y=-x-1 or y=1-x.

But I don't understand what to do from here. Does this mean that since x and y are dependent on one another then they are reflexive? But how would I know if it an element in 'x' (say 10) is dependent onitself?

You haven't told us what set $x$ and $y$ are from. I will assume the real numbers. As you say, your relation is reflexive if for all $x \in \mathbb R$ you have $xRx$ or $(x+x)^2 = 1$. Is that true or false?

Now write carefully what it means for your relation to be irreflexive and decide whether or not it is true.

 

[scorpius1782]

Sorry yes, x,y are Real. I would say that your example is irreflexive because x can only be $\frac{1}{2}$ or $\frac{-1}{2}$.

However, I stumble on the idea of "related". What does that even mean, specifically? Because, in the original question there is an equation that specifically relates x and y. They are related via that equation but is that what the definition is referring to? They can never be equal or divisible by one another but does that mean they can't be related? These sort of things are glossed over in my text and it is quite frustrating. I still don't understand how a number can be related to itself.

 

[LCKurtz]

You haven't told us what set $x$ and $y$ are from. I will assume the real numbers. As you say, your relation is reflexive if for all $x \in \mathbb R$ you have $xRx$ or $(x+x)^2 = 1$. Is that true or false?

You didn't answer that question.

Now write carefully what it means for your relation to be irreflexive

You didn't do that.

Sorry yes, x,y are Real. I would say that your example is irreflexive because x can only be $\frac{1}{2}$ or $\frac{-1}{2}$.

Until you write the definition of what it means to be irreflexive in this problem, you are just guessing.

However, I stumble on the idea of "related". What does that even mean, specifically?

In this problem $x$ and $y$ are related, written $xRy$, if $(x+y)^2=1$. That's the definition. Nothing more nor less. So try again to answer the questions above.

 

[scorpius1782]

I guess I didn't understand what you were saying in the question (I didn't really get the 'or') part. So, this must be false because if xRx is true then x=x but this is clearly not the case in the equation.

Irreflexive-no element can be related to itself. So, for xRy it must be that the conditional equation demands that x=y. But it clearly can't and never can. They are related but not reflexive(ly) related (this part actually confused me for some reason).

Have I got the idea now?

 

[LCKurtz]

If you want to know if you have an equivalence relation, here is what you have to prove:

$R$ is refexive if for all $x \in \mathbb R$ you have $xRx$, which means $(x+x)^2 = 1$.

I guess I didn't understand what you were saying in the question (I didn't really get the 'or') part. So, this must be false because if xRx is true then x=x but this is clearly not the case in the equation.

I re-worded it without the "or". Now can you well me whether $R$ is reflexive and why? All you have to do is explain whether the statement is true or false and why.

Irreflexive-no element can be related to itself. So, for xRy it must be that the conditional equation demands that x=y. But it clearly can't and never can. They are related but not reflexive(ly) related (this part actually confused me for some reason).

Have I got the idea now?

No. Once you get the reflexive question figured out you need to write a careful definition for irreflexive and work it similarly.

 

[scorpius1782]

x is reflexive because it can solve that equation as $\frac{1}{2}$ or $\frac{−1}{2}$ (I was still thinking x,y before for some reason).

This is as careful as I can be without actually understanding it. I'm not trying to be a pain but I can't define something I don't seem to understand.
Irreflexive-no element can be related to itself. So, for xRy it must be that the conditional equation demands that $x \neq y$ and is then irreflexive. No other solution can have x=y or else it would be neither reflexive or irreflexive.

 

[LCKurtz]

x is reflexive because it can solve that equation as $\frac{1}{2}$ or $\frac{−1}{2}$ (I was still thinking x,y before for some reason).

It isn't $x$ that is or isn't reflexive. It is $R$. At the risk of answering the question for you:

Is it true or false that for all $x\in \mathbb R$, $(x+x)^2=1$? Why or why not? That is what $xRx$ means and that's how you answer the question.

 

[scorpius1782]

Then false because there is only 2 real numbers that solve the equation while there is an infinite number of reals that do not.

 

[LCKurtz]

It isn't $x$ that is or isn't reflexive. It is $R$. At the risk of answering the question for you:

Is it true or false that for all $x\in \mathbb R$, $(x+x)^2=1$? Why or why not? That is what $xRx$ means and that's how you answer the question.

Then false because there is only 2 real numbers that solve the equation while there is an infinite number of reals that do not.

Yes. Instead of stating a sentence like that, give an equation. Just say, for example, that it is false because $(0+0)^2\ne 1$.

Now here's your problem. State carefully, with equations, similar to what is above and using the definition of $R$ what you have to prove true or false to say whether $R$ is irreflexive. Don't give me a paragraph of explanation. Give equations.

 

[scorpius1782]

I don't want to post the final answer since it is not homework that we're not suppose to share. However, I think I fully understand it now. Thank you very much for the help.