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"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?

torr
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    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 $x<q<y$. –  Mar 04 '14 at 01:23
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    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 $$(\forall x,y\in \mathbb R)(x\neq y\to \exists q\in \mathbb Q(x>q\land q>y)).$$ – Git Gud Mar 04 '14 at 01:25
  • @SanathDevalapurkar I corrected the statement, thanks for your input! – torr Mar 04 '14 at 01:26
  • @GitGud It is in the context of a pre-college courses about Mathematical thinking. I understand the nuance and will correct that thanks! – torr Mar 04 '14 at 01:27
  • @GitGud: Your suggested correction is flawed. It actually assumes $y<q<x$ – MPW Mar 04 '14 at 01:28
  • @torr That's not enough context. Maybe you are following a text in which that notation is used. – Git Gud Mar 04 '14 at 01:28
  • @MPW Thanks for spotting it. – Git Gud Mar 04 '14 at 01:31
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    @torr As pointed out, what I said was wrong, please see the correct answer below.

    Correct is $$(\forall x,y\in \mathbb R)(x\neq y\to \exists q\in \mathbb Q((x>q\land q>y)\lor (y>q \land q> x))).$$

     – Git Gud Mar 04 '14 at 01:32
  • "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'. – torr Mar 04 '14 at 01:33
  • @torr In light of the above comment, it should be: $$(\forall x,y\in \mathbb R)(x\neq y\to \exists q\in \mathbb R(\mathbb Q(q) \land (x<q\land q<y)\lor (y<q \land q< x))).$$ Apparently $\mathbb Q$ should be looked at as a predicate and not as a set. But this is weird, so be careful with I'm suggesting. – Git Gud Mar 04 '14 at 01:35
  • It's really confusing that the question changes all the time. The formula you started asking "why is this wrong" about is not the statement you're now asking "why is this wrong" about, which makes some of the answers and comments meaningless :-( – hmakholm left over Monica Mar 04 '14 at 01:45
  • @HenningMakholm I am going to fix this now. – torr Mar 04 '14 at 01:49

4 Answers4

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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 $\neg(x=y)$

UserB1234
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  • @Git Gud: I think the above addresses your issue about $\mathbb{Q}$ being used as a predicate. – UserB1234 Mar 04 '14 at 01:49
  • Can you explain why your answer is correct and why mine is not? Thanks!! :) – torr Mar 04 '14 at 01:55
  • 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 $z<x$. – UserB1234 Mar 04 '14 at 02:02
  • 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 – UserB1234 Mar 04 '14 at 02:08
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    @Danul_G http://math.stackexchange.com/questions/698347/why-is-my-logical-statement-wrong/698366?noredirect=1#comment1462896_698347 The solution of the book use existence for x,y and forall for q. – torr Mar 04 '14 at 02:10
  • Then I believe the above is correct. But I cannot guarantee it. I would probably need to read the book for the buildup of the question before I can answer. For example: You are using the symbol $\in$. However it is not there in your posting of the question. – UserB1234 Mar 04 '14 at 02:17
  • You do not mention the domain of x,y ! Can you do this? Why? – torr Mar 04 '14 at 04:05
  • As I said, this depends on what you are assuming about your language. You still have not told me properly whether you are allowed to use the $\mathbb{in}$.

    However I am assuming that my variables range over $\mathbb{R}$. So in a way, yes, I can do this. But in order to give you a definite answer I would probably need read the book.

     – UserB1234 Mar 04 '14 at 12:35
  • Thanks for your assistance Danul! – torr Mar 04 '14 at 17:56
  • Rather sloppy! (i) $Q(x)$ should be $Q(q)$. (ii) $(\neg{y<q})$ should be $\neg({y<q})$. (iii) The predicate after $\exists{q}$ has the form $(A \wedge B \vee C)$. Is this $(A \wedge (B \vee C))$ or $((A \wedge B) \vee C)$? – TonyK Nov 13 '15 at 11:48
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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.

Mauro ALLEGRANZA
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  • 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.

     – UserB1234 Mar 04 '14 at 12:30
  • @Danul G - I agree with you that is better to introduce two predicates $Q$ and $R$. – Mauro ALLEGRANZA Mar 04 '14 at 12:40
  • 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. – UserB1234 Mar 04 '14 at 12:43
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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.

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

ncmathsadist
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  • Can you explain why is true and why am I wrong? Thanks! – torr Mar 04 '14 at 01:30
  • You have not made a stipulation about the order of $x$ and $y$. You could say $\forall x, y, x < y \implies \cdots$. – ncmathsadist Mar 04 '14 at 01:34
  • @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. – Git Gud Mar 04 '14 at 01:38
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    This may be equivalent but it is certainly not a direct translation into logical notation. I would suggest just "$x<q<y$ or $y<q<x$" – MPW Mar 04 '14 at 01:38
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    You are assuming that multiplication is there in the language. I believe that it isn't. – UserB1234 Mar 04 '14 at 01:39
  • The reals come equipped with multiplication. This is defined. – ncmathsadist Mar 04 '14 at 01:58
  • Need a bit of explanation... – vonbrand Mar 04 '14 at 02:03
  • What about min and max? $(y-q)*(q-x) \iff (x\wedge y) < q < (x\vee y)$. – ncmathsadist Mar 04 '14 at 02:06
  • I did not say that it isn't or cannot be defined. I just said that OP might not be able to talk about multiplication since he doesn't have a symbol for it.However OP hasn't posted what his original language was, so I'm not sure. – UserB1234 Mar 04 '14 at 02:14
  • same as $((y > q)\wedge(q > x)) \vee ((y < q)\wedge(q < x))$ – ncmathsadist Mar 04 '14 at 02:15
  • I'm confused. I agree that what you have written down is correct, provided OP has the necessary vocabulary to talk about it. However he can't talk about multiplication (you can't define it using the symbols he has) – UserB1234 Mar 04 '14 at 02:24