Transitive relation help

**Danneedshelp** · December 2nd 2009, 06:56 PM

How do I go about proving the relation $\rho$ defined such that $u\rho\\v$ $\Leftrightarrow$ $g(u)$ and $g(v)$ have the same sign ( $u,v\in{\mathbb{R}}$ ) is transitive?

proof: Suppose $u\rho\\w$ and $w\rho\\v$ , then $g(u)$ , $g(w)$ , and $g(v)$ have the same sign. Without loss of generality, suppose $g(u)\leq$ $g(w)$ $\leq\\g(v)$ are all positive. Then...

Am I on the right track? Can I assume that the outputs are ordered as I did?

Thanks

**Drexel28** · December 2nd 2009, 08:32 PM

Originally Posted by Danneedshelp

How do I go about proving the relation $\rho$ defined such that $u\rho\\v$ $\Leftrightarrow$ $g(u)$ and $g(v)$ have the same sign ( $u,v\in{\mathbb{R}}$ ) is transitive?

proof: Suppose $u\rho\\w$ and $w\rho\\v$ , then $g(u)$ , $g(w)$ , and $g(v)$ have the same sign. Without loss of generality, suppose $g(u)\leq$ $g(w)$ $\leq\\g(v)$ are all positive. Then...

Am I on the right track? Can I assume that the outputs are ordered as I did?

Thanks

I don't understand what the diffculty is. Suppose that $x\sim y\Longleftrightarrow \text{sgn}\text{ }g(x)=\text{sgn }g(y)$ . This clearly implies though that since $\text{sgn}$ is never zero that the above is the same as $\frac{\text{sgn }g(x)}{\text{sgn }g(y)}=1$ . So assume then that $y\sim z\implies \frac{\text{sgn }g(y)}{\text{sgn }g(z)}=\frac{\text{sgn }g(z)}{\text{sgn }g(y)}=1$ . Equate the two and solve. Am I misinterpreting the question?

Remark This actually isn't true in general, only if $g(x)\ne0\quad \forall x\in\mathbb{R}$ , otherwise the signum function is undefined.

**Danneedshelp** · December 3rd 2009, 07:55 AM

Originally Posted by Drexel28

I don't understand what the diffculty is. Suppose that $x\sim y\Longleftrightarrow \text{sgn}\text{ }g(x)=\text{sgn }g(y)$ . This clearly implies though that since $\text{sgn}$ is never zero that the above is the same as $\frac{\text{sgn }g(x)}{\text{sgn }g(y)}=1$ . So assume then that $y\sim z\implies \frac{\text{sgn }g(y)}{\text{sgn }g(z)}=\frac{\text{sgn }g(z)}{\text{sgn }g(y)}=1$ . Equate the two and solve. Am I misinterpreting the question?

Remark This actually isn't true in general, only if $g(x)\ne0\quad \forall x\in\mathbb{R}$ , otherwise the signum function is undefined.

Yes, I am sorry for not being more clear about the function g(x). I have defined g(x) such that g(x)=f(x)-L where g(x) never takes on the value zero, because for any x in my domian, f(x) does not equal L.

I have only seen the signum function once in a homework problem. As I understand it, the sgn function allows us just deal with the sign of a value?

Thanks for your help

**Drexel28** · December 3rd 2009, 12:01 PM

Originally Posted by Danneedshelp

Yes, I am sorry for not being more clear about the function g(x). I have defined g(x) such that g(x)=f(x)-L where g(x) never takes on the value zero, because for any x in my domian, f(x) does not equal L.

I have only seen the signum function once in a homework problem. As I understand it, the sgn function allows us just deal with the sign of a value?

Thanks for your help

I am not entirely sure if what you're saying invalidates what I have done. Does it? If so, let me know. And yes if we define $f:\mathbb{R}^*\mapsto\{-1,1\}$ with $x\mapsto\begin{cases} -1 & \mbox{if}\quad x<0 \\ 1 & \mbox{if}\quad x>0\end{cases}$ then we are really describing the signum function.

**emakarov** · December 3rd 2009, 12:53 PM

In my opinion, it does not matter whether g or sgn may return 0. Almost any relation R of the form "x and y have the same something" is transitive (as well as reflexive and symmetric). More precisely, if something(x) is a function and the relation "same" is itself transitive, then so is R.

In this case, if g(x) and g(y) have the same sign, and g(y) and g(z) have the same sign, then it is clear that all three have the same sign.

**Drexel28** · December 3rd 2009, 01:09 PM

Originally Posted by emakarov

In my opinion, it does not matter whether g or sgn may return 0. Almost any relation R of the form "x and y have the same something" is transitive (as well as reflexive and symmetric). More precisely, if something(x) is a function and the relation "same" is itself transitive, then so is R.

In this case, if g(x) and g(y) have the same sign, and g(y) and g(z) have the same sign, then it is clear that all three have the same sign.

What is the sign of $\sin(0)=-\sin(0)$ ?

**emakarov** · December 3rd 2009, 03:23 PM

What is the sign of

?

Well, this depends on the definition of the sgn function. I would say $\text{sgn}(0)$ should be $0$ , so $\text{sgn}(\sin(0))=\text{sgn}(-\sin(0))=0$ .

However, I don't see how this changes anything. If, for some function $g$ and some transitive relation $\equiv$ , the relation $R(x,y)$ is defined as $g(x)\equiv g(y)$ , then $R$ is transitive. If we have some unusual case, e.g., when $g(x)$ is not defined, then we cannot say $R(x,y)$ for any $y$ . If $g(x)$ and $g(y)$ are defined but $g(x)\equiv g(y)$ is not, then again $x$ and $y$ are not related, so the premise of the transitivity property is false. What is important is for $g$ to be a function, i.e., return exactly one value for each $x$ .

Do you have any counterexample in mind?

**Drexel28** · December 3rd 2009, 03:38 PM

Originally Posted by emakarov

Well, this depends on the definition of the sgn function. I would say $\text{sgn}(0)$ should be $0$ , so $\text{sgn}(\sin(0))=\text{sgn}(-\sin(0))=0$ .

However, I don't see how this changes anything. If, for some function $g$ and some transitive relation $\equiv$ , the relation $R(x,y)$ is defined as $g(x)\equiv g(y)$ , then $R$ is transitive. If we have some unusual case, e.g., when $g(x)$ is not defined, then we cannot say $R(x,y)$ for any $y$ . If $g(x)$ and $g(y)$ are defined but $g(x)\equiv g(y)$ is not, then again $x$ and $y$ are not related, so the premise of the transitivity property is false. What is important is for $g$ to be a function, i.e., return exactly one value for each $x$ .

Do you have any counterexample in mind?

No, I just merely had in mind a different definition of sign than you. Clearly though, according to Mathworld.com, yours is the correct. If the sign of zero was undefined then clearly this relation would not be well defined. For what we are saying is not when $g(x)$ is undefined but when it's sign is indeterminable.

**Plato** · December 3rd 2009, 03:40 PM

I have tried to stay away from commentating on this thread.
However, I think any misunderstandings have resulted from a badly worded question.
Testing companies pay good money to have possible test questions reviewed.
The original post would have failed that process.
The phrase “same sign” would have sent up red flags.
It is so easy to say $g(u)~\&~g(v)$ are either both negative or both non-negative.

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