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I have the following statement:

"The open interval (0,1) contains no 'largest' number"

Which I am asked to "formalize" (I hope I am using the correct mathematical jargon here).

I think the most correct way to formalize it is:

$$\forall a (((a>0) \wedge (a<1)) \rightarrow \exists b ((a<b) \wedge (b<1)))$$

I would like to hear your opinion about it.

But is the following also correct? If not, why?

$$\forall a \exists b((a>0) \wedge (a<1) \wedge (a<b) \wedge (b<1))$$

Mauro ALLEGRANZA
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MC From Scratch
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  • It depends on the domain you are quantifying over when you write "for all $a$". For all $a$ where? – Michal Adamaszek Mar 10 '22 at 15:15
  • It doesn't specify in the question. The question just gives different formalizations of the statement I have given, and I am asked to mark which ones are correct. @MichalAdamaszek – MC From Scratch Mar 10 '22 at 15:18

2 Answers2

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If you translate this to human language the first expression says:

  • "for all $a$ in $(0,1)$ there exists a $b$ between $a$ and one" (perfect).

The second expression sounds odd:

  • "for all $a$ [which ones?] there exists a $b$ such that $a$ is in $(0,1)$ and $b$ is between $a$ and one."

It is certainly not true that any $a$ is in $(0,1)$.

Kurt G.
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  • But is it even correct to say "For all a, if a is in (0,1) ...... " because then it's not for all a. It's like saying "For every colour, if it is white ......" . It's not for every colour in that case. It sounds a bit stupid to say. – MC From Scratch Mar 10 '22 at 16:58
  • You are right. "For all $a$, if a is in $(0,1)$ ... " is another human language version of your first logical statement. The second one is bogus. – Kurt G. Mar 10 '22 at 17:51
  • Second (borrowing the legend in my Answer) is syntactically correct and semantically meaningful, and in fact, the form $∀x{\in}(2,3);P(x)$ that you're used to is actually an abbreviation of $∀x;\big(x{\in}(2,3){\implies}P(x)\big).$ This statement asserts that $P(2.7)$ is true while making no claim about the truth value of $P(4.5).$ Here is a longer explanation. @MCFromScratch – ryang Mar 11 '22 at 13:36
  • @ryang . Thanks for pointing this out. If I now understand correctly : $\forall x \exists y;(((x>0) \wedge (x<1)) \color{green}{\to} ((x<y) \wedge (y<1)))$ is a correct version of the first statement and $\forall x \exists y;(((x>0) \wedge(x<1)) \color{red}{\wedge} ((x<y) \wedge (y<1)))$ is the wrong second statement. The difference lies in $\color{green}{\to}$ versus $\color{red}{\wedge} $. – Kurt G. Mar 11 '22 at 13:48
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$$\forall a\;(((a>0) \wedge (a<1)) \rightarrow \exists b\;((a<b) \wedge (b<1)))$$ $$\forall a \exists b\:\:((a>0) \wedge (a<1) \wedge (a<b) \wedge (b<1))$$

Let's call the two statements First and Second. Their domain of discourse is $\mathbb R.$

  1. First accurately formalises the given verbal statement, and is a true statement.
  2. Second claims that whatever $a$ is, we can find some $b$ such that $a∈(0,1)$ and $a<b<1;$ the counterexample $a=100$ clearly shows that Second is a false statement.
  3. First is logically equivalent to $\forall x \exists y\;(((x>0) \wedge (x<1)) \to ((x<y) \wedge (y<1))).$ (Why?)
  4. Second logically implies First (why?), but not vice versa (why?); that is, Second is a stronger statement than First.
ryang
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