0
$\begingroup$

Universe of Discourse: When we are examining two predicate statements that have different universes of discourse that intersect one another, should we expand the universe of discourse to union of the two universes, or should we contract the universe of discourse to include only the intersection of the two? I know this is a super long post but, I am very confused and will appreciate all of the guidance I can get.

I came across the following exercise the other day, and I don't know what to conclude; it is not the algebra but it is the fundamentals of the logic that have me confused; I don't know what to conclude.

Prove the following: If $x,y,z\in\mathbb{R}$ , then $x\leqslant y\leqslant z$ iff $\mid x-y\mid+\mid y-z\mid=\mid x-z\mid$.

This statement can be proved both correct and incorrect depending on what universe of discourse you establish. For example let $P(x,y,z,R)=\space'xRyRz'$, and let $Q(x,y,z)='\mid x-y\mid+\mid y-z\mid=\mid x-z\mid'$ both be predicate statements, where $x,y,z\in\mathbb{R}$, and let $R$ belong to the set $A=\{<,\leqslant\}$ or the set $B=\{<,\leqslant,\geqslant,>\}$, where the relations in both $A$ and $B$ are the relations that we might be familiar with.

Now the predicate $Q$ exists in the real number where all of the relation in set $B$ are well defined, however $x,y,z$ also happily lives in a world that only considers the relations in set $A$ as well. Although, the predicate $Q$ doesn't take any relations it does however provoke set $B$ through the definition of the absolute value. On the other hand the predicate $P$ only provokes set $A$ however it can peacefully coincide with the set $B$. As I was writing this it became apparent that, the predicate $Q$ can not exist in a world without the set $B$ because of its absolute values functions that require the relation $\geqslant$ on $\mathbb{R}$. So in order for us to consider this relationship between the predicates $Q$ and $P$ we must choose our universe of discourse to be $A\cup B$. However the question still remains is this always the case? Might it be possible that we could consider an equality that does not depend on the set $B$ (even though it could be natural to assume that it is in a world that contains the set $B$ ) in which case we could have chosen the universe of discourse to be $A\cap B$ or $A\cup B$, which would be the correct universe to choose? Or is it somewhat subjective?

$\endgroup$
  • 1
    $\begingroup$ I personally think that your "universe of discourse" question should be asked as a separate question, perhaps including an example where this has come up. The two question here are quite independent, and really shouldn't be asked together. $\endgroup$ – user642796 Sep 21 '13 at 3:27
  • $\begingroup$ I guess I am seeing this very differently;consider the following $x,y,z\in \mathbb{R}$ and we have the statement $xRyRz$ where $R$ is a relation on the real numbers. Now consider the case that $R\in\{<,\leqslant\}$, then the original question really is an if and only if statement. Now consider another case if $R\in \{<, \leqslant, \geqslant, >\}$, Then, we cannot establish an if and only if statement. At any rate the first predicate can live easily in the world of the first set of relations, while the equality provokes the second set of relations. Which set should we settle on? $\endgroup$ – JimmyJackson Sep 21 '13 at 3:39
  • $\begingroup$ Then perhaps I misunderstand what you mean by universe of discourse. More commonly it refers to just the underlying objects under consideration (in this case the real numbers) and not the relations on those objects. $\endgroup$ – user642796 Sep 21 '13 at 4:05
  • 1
    $\begingroup$ By the way, don't use \mid for absolute value signs, just use |. With \mid you get weird spacing, e.g. $\mid x-y\mid=\mid y-x\mid$, instead of the correct $|x-y|=|y-x|$ with |. $\endgroup$ – Rahul Sep 21 '13 at 4:10
  • $\begingroup$ @RahulNarain Thanks for the tip $\endgroup$ – JimmyJackson Sep 21 '13 at 4:43
3
$\begingroup$

You can simplify your first proof.

Suppose $x\leq y\leq z$ then $|x-y|+|y-z|=(y-x)+(z-y)=z-x=|x-z|$. On the other hand, the same argument works for $x>y>z$ (which is what you show above).

The correct statement is $|x-y|+|y-z|=|x-z|$ if and only if $x\leq y\leq z$ or $z\leq y\leq x$.

$\endgroup$
  • $\begingroup$ @azerel Thanks, for this. But i am still unsure about the universe of discourse thing. See if I wanted to show this with a truth table i would naturally be looking at arguments of different sizes. Namely, $x\leqslant y\leqslant z$ entails only for possible situations for $x,y,z$ while the equality naturally brings about more arguments. Also, I think you still have to show the case when $x=y=z$ because of the way that the absolute value is traditionally defined, and then could simplify down to the line above. $\endgroup$ – JimmyJackson Sep 21 '13 at 2:58
  • $\begingroup$ @JimmyJackson $|x|= x$ if $x\geq 0$ and $-x$ otherwise. So there is no need to consider the case of equality separately. To prove the statement suppose $x\leq z$ but $z\notin [x,y]$ which splits in two cases. $\endgroup$ – azarel Sep 21 '13 at 3:07
  • $\begingroup$ But the original statement makes makes it OK for $z$ to belong to the interval $[x,y]$. I guess it is OK to ignore the case that $x=y=z$, because if you compare the equality $(x-y)+(y-z)=(y-x)+(z-y)$ which leads to $x=z$ which is true in the case that $x=y=z$, but I feel it is worth considering this case for safety. $\endgroup$ – JimmyJackson Sep 21 '13 at 3:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.