Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to convert this sentence to logic notation. "there is an integer less than or equal to all other integers greater than 0". "An integer exists that is less than or equal to all other integers greater than 0". So far I have -

L(x,y): x is less than or equal to y
∃y ∈ Z.(∀x ∈ Z.L(x,0))

Does this translate correctly? Any help would be appreciated.

share|improve this question
    
Are you free to define your own predicates, or must you use this particular $L$? –  Brian M. Scott Oct 27 '13 at 11:42
    
No I have to use the predicate L. Thanks. –  Darren Findlay Oct 27 '13 at 11:47
add comment

1 Answer

up vote 3 down vote accepted

Your sentence says that there is an integer $y$ such that all integers are less than or equal to $0$, which is clearly not what you wanted. Let’s try to sneak up on the correct expression.

How can we say ‘$x$ is an integer less than or equal to each integers that is greater than $0$’?

First, $y>0$ has to be translated. One way is $L(0,y)\land y\ne 0$, and another is $\neg L(y,0)$; you can use either. Now we want to say that if $y$ is an integer greater than $0$, then $x\le y$: $$\forall y\in\Bbb Z\Big(\big(L(0,y)\land y\ne 0\big)\to L(x,y)\Big)\;,$$ or

$$\forall y\in\Bbb Z\big(\neg L(y,0)\to L(x,y)\big)\;.$$

Finally, we have to say that there is such an integer $x$:

$$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\Big(\big(L(0,y)\land y\ne 0\big)\to L(x,y)\Big)\;,$$

or

$$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\big(\neg L(y,0)\to L(x,y)\big)\;.$$

But that’s still not quite right, because the other in all other integers greater than $0$ implies that $x$ is supposed to be greater than $0$, and we have to say so. For convenience I’m going to introduce an abbreviation: I’ll write $L'(x,y)$ for $L(x,y)\land x\ne y$ or $\neg L(y,x)$. Then what we want is

$$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\left(L'(0,x)\land\Big(\big(x\ne y\land L'(0,y)\big)\to L(x,y)\Big)\right)\;.$$

You can get rid of the new predicate $L'$ simply by expanding each instance using the definition of $L'$.

share|improve this answer
    
Thank you Brian. I decided to throw mine away! This looks great. I do need to look at it in the morning again; not exactly sure why I didn't see the first conjunct. –  Hunan Rostomyan Oct 27 '13 at 12:16
1  
@HunanRostomyan: You’re welcome. It’s actually a little tricky, since one has to realize the implications of the word other. –  Brian M. Scott Oct 27 '13 at 12:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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