4
$\begingroup$

"Express the following sentence symbollically, using only quantifiers for real numbers, logical connectives, the order relation < and the symbol Q having the meaning 'x is rational'"

I have to translate the sentence "There is a rational number between any two unequal real numbers". I worked a bit on it and eventually deduced the following:

$$(\forall x,y\in \mathbb{R})[x> y](\exists q\in \mathbb{Q})[q>y \wedge q< x]$$

In light of some comments a correct version of my incorrect statement should be: $$(\forall x,y\in \mathbb{R})[x≠ y \Rightarrow (\exists q\in \mathbb{Q})[q>y \wedge x> q]\vee[y>q \; \wedge \;q>x]]$$

Can you help me understand why my answer is wrong?

$\endgroup$
12
  • 1
    $\begingroup$ Well, first, $x>y$ is not right. $x\ne y$ would be more appropriate. Also I believe you could shorten down $q>y∧q<x$ to $x<q<y$. $\endgroup$
    – user122283
    Mar 4 '14 at 1:23
  • 1
    $\begingroup$ Even though the statements with $x\neq y$ and $x>y$ are equivalent, they don't say the same. I don't know the notation you're using, maybe it is correct in the context you're in, but more common would be $$(\forall x,y\in \mathbb R)(x>y\to \exists q\in \mathbb Q(x>q\land q>y)).$$ Correct would be $$(\forall x,y\in \mathbb R)(x\neq y\to \exists q\in \mathbb Q(x>q\land q>y)).$$ $\endgroup$
    – Git Gud
    Mar 4 '14 at 1:25
  • $\begingroup$ @SanathDevalapurkar I corrected the statement, thanks for your input! $\endgroup$
    – torr
    Mar 4 '14 at 1:26
  • $\begingroup$ @GitGud It is in the context of a pre-college courses about Mathematical thinking. I understand the nuance and will correct that thanks! $\endgroup$
    – torr
    Mar 4 '14 at 1:27
  • 2
    $\begingroup$ @torr As pointed out, what I said was wrong, please see the correct answer below. Correct is $$(\forall x,y\in \mathbb R)(x\neq y\to \exists q\in \mathbb Q((x>q\land q>y)\lor (y>q \land q> x))).$$ $\endgroup$
    – Git Gud
    Mar 4 '14 at 1:32
2
$\begingroup$

In your first formula, following @Danul G's advise, I assume $\mathbb R$ as the domain of the variables, and I'll rewrite it with the predicate $Q(x)$ (for $x$ is rational) to stay with the original formulation of the problem :

$\forall x \forall y ( x > y ) \exists q (Q(q) \land [q>y \land q<x])$

there are basically two mistakes :

(i) one substantial: as pointed in the above comments, in this way you are assuming that $x > y$, but your problem says $x \ne y$; so you must correct the formula accordingly.

(ii) the other one is connected to the way you have written the formula; juxtaposition of the two sub-formulas is not formally correct: at most you can read it as a conjunction.

But in this way you are saying that :

$\forall x \forall y ( x > y )$ and $\exists q (Q(q) \land [q>y \land q<x])$

i.e.that for two real numbers whatever, the first is greater than the second (false ! : take $0$ as $x$ and $1$ as $y$) and ... and the resulting statement is false, being the conjunction of two formulas, one of which is false.

$\endgroup$
3
  • $\begingroup$ The question is slightly ill posed, I think. OP says that: "Express the following sentence symbollically, using only quantifiers for real numbers, logical connectives, the order relation < and the symbol Q having the meaning x is rational" which leads me to believe that he does not actually have $\in$ in his language to phrase is this way. $\endgroup$
    – UserB1234
    Mar 4 '14 at 12:30
  • $\begingroup$ @Danul G - I agree with you that is better to introduce two predicates $Q$ and $R$. $\endgroup$ Mar 4 '14 at 12:40
  • $\begingroup$ I just assumed that the domain of discourse was the reals in my answer. I think that is what is meant by "quantifiers for real numbers". But I'm not sure. $\endgroup$
    – UserB1234
    Mar 4 '14 at 12:43
2
$\begingroup$

I think that it should actually be $\forall{x,y}(x\neq{y}\implies{\exists{q}(Q(q)\wedge(\neg(x<q)\wedge{x\neq{q}}\wedge{y<q})\vee(\neg{(y<q)}\wedge{y\neq{q}}\wedge{x<q}})))$.

This answer is written assuming that your variables range accross $\mathbb{R}$, since you technically don't have $\epsilon$ symbol in your language and $x\neq{y}$ is shorthand for $\neg(x=y)$

$\endgroup$
10
  • $\begingroup$ @Git Gud: I think the above addresses your issue about $\mathbb{Q}$ being used as a predicate. $\endgroup$
    – UserB1234
    Mar 4 '14 at 1:49
  • $\begingroup$ Can you explain why your answer is correct and why mine is not? Thanks!! :) $\endgroup$
    – torr
    Mar 4 '14 at 1:55
  • $\begingroup$ First it depends on what language you are using. I would actually like to see the question (as it is asked of you) before I give a final answer (For a start I don't even know if you have membership in your language). The error you are making for the most part is that you are assuming $x\neq{y}$ is logically equivalent to $x>y$. It is not. If $x=0$ and $y=1$, then $x\neq{y}$ is true. However there is no number $z$ (let alone a rational) such that $y<z$ and $z<x$. $\endgroup$
    – UserB1234
    Mar 4 '14 at 2:02
  • $\begingroup$ What you are saying is if $x>y$, then there is a rational between them. But what happens if $x<y$? After all if $x\neq{y}$, then that is a huge possibility $\endgroup$
    – UserB1234
    Mar 4 '14 at 2:08
  • 1
    $\begingroup$ @Danul_G math.stackexchange.com/questions/698347/… The solution of the book use existence for x,y and forall for q. $\endgroup$
    – torr
    Mar 4 '14 at 2:10
0
$\begingroup$

Try the next. Using $\sim$ for denying: translate Trichomoty Law, use $\forall x,y\in\mathbb{R}\colon \phi(x,y)$ and $\phi(x,y)\equiv\sim(x=y)\rightarrow[[x<y]\vee[x>y]]$. Then we also have $\forall x\in\mathbb{R}\forall{q}\in Q\colon \phi(x,q)$ and as the antecedent is fullfilled we have $[x<q]\vee[x>q]$. The same holds for $\phi(q,y)$. Now the existence of q. Take p.e. $x<q$ and $q<y$, fix x and y; then you will have two sets of rationals for which the cut is not empty.

$\endgroup$
-1
$\begingroup$

$$\forall x,y\in\mathbb{R} x\not=y \implies \exists q\in\mathbb{Q} (y-q)(q-x) > 0 $$

$\endgroup$
11
  • $\begingroup$ Can you explain why is true and why am I wrong? Thanks! $\endgroup$
    – torr
    Mar 4 '14 at 1:30
  • $\begingroup$ You have not made a stipulation about the order of $x$ and $y$. You could say $\forall x, y, x < y \implies \cdots$. $\endgroup$ Mar 4 '14 at 1:34
  • $\begingroup$ @ncmathsadist I believe it is not nitpicking in saying that this is incorrect. Despite being equivalent to what is asked, it is not what is asked. $\endgroup$
    – Git Gud
    Mar 4 '14 at 1:38
  • 1
    $\begingroup$ This may be equivalent but it is certainly not a direct translation into logical notation. I would suggest just "$x<q<y$ or $y<q<x$" $\endgroup$
    – MPW
    Mar 4 '14 at 1:38
  • 1
    $\begingroup$ You are assuming that multiplication is there in the language. I believe that it isn't. $\endgroup$
    – UserB1234
    Mar 4 '14 at 1:39

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.