Proving $exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $?











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$$exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $$



I have no idea how to approach this problem? If true, prove it, and if false prove the the negation. With its negation being:



$$forall k in mathbb{Z}:exists l in mathbb{Z}: k mid l$$



My intuition is telling me the negation is true, because for any k we can just choose k for l and then we get $k mid k$, which is always true for the integers.



But then, I believe the orginal statement is also true. Because clearly, $ 0 mid l $ only for $l = 0$. So, $$forall l in mathbb{Z}: lnot(0 mid l)$$



Any suggestions of where I am going wrong?






elementary-number-theory proof-verification propositional-calculus predicate-logic







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  • 1




    You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
    – user3482749
    Nov 16 at 23:28










  • Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
    – Fefnir Wilhelm
    Nov 16 at 23:31















up vote
0
down vote

favorite












$$exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $$



I have no idea how to approach this problem? If true, prove it, and if false prove the the negation. With its negation being:



$$forall k in mathbb{Z}:exists l in mathbb{Z}: k mid l$$



My intuition is telling me the negation is true, because for any k we can just choose k for l and then we get $k mid k$, which is always true for the integers.



But then, I believe the orginal statement is also true. Because clearly, $ 0 mid l $ only for $l = 0$. So, $$forall l in mathbb{Z}: lnot(0 mid l)$$



Any suggestions of where I am going wrong?






elementary-number-theory proof-verification propositional-calculus predicate-logic







share|cite|improve this question


















  • 1




    You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
    – user3482749
    Nov 16 at 23:28










  • Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
    – Fefnir Wilhelm
    Nov 16 at 23:31













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $$



I have no idea how to approach this problem? If true, prove it, and if false prove the the negation. With its negation being:



$$forall k in mathbb{Z}:exists l in mathbb{Z}: k mid l$$



My intuition is telling me the negation is true, because for any k we can just choose k for l and then we get $k mid k$, which is always true for the integers.



But then, I believe the orginal statement is also true. Because clearly, $ 0 mid l $ only for $l = 0$. So, $$forall l in mathbb{Z}: lnot(0 mid l)$$



Any suggestions of where I am going wrong?






elementary-number-theory proof-verification propositional-calculus predicate-logic







share|cite|improve this question













$$exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $$



I have no idea how to approach this problem? If true, prove it, and if false prove the the negation. With its negation being:



$$forall k in mathbb{Z}:exists l in mathbb{Z}: k mid l$$



My intuition is telling me the negation is true, because for any k we can just choose k for l and then we get $k mid k$, which is always true for the integers.



But then, I believe the orginal statement is also true. Because clearly, $ 0 mid l $ only for $l = 0$. So, $$forall l in mathbb{Z}: lnot(0 mid l)$$



Any suggestions of where I am going wrong?






elementary-number-theory proof-verification propositional-calculus predicate-logic




elementary-number-theory proof-verification propositional-calculus predicate-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 16 at 23:26









Fefnir Wilhelm

62




62








  • 1




    You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
    – user3482749
    Nov 16 at 23:28










  • Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
    – Fefnir Wilhelm
    Nov 16 at 23:31














  • 1




    You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
    – user3482749
    Nov 16 at 23:28










  • Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
    – Fefnir Wilhelm
    Nov 16 at 23:31








1




1




You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
– user3482749
Nov 16 at 23:28




You go wrong after "But then". You first say that $0 | 0$, but then say that $forall l in mathbb{Z}: lnot(0|l)$. These statements are not compatible.
– user3482749
Nov 16 at 23:28












Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
– Fefnir Wilhelm
Nov 16 at 23:31




Oh so your saying that since $exists lin mathbb{Z}: 0 mid l.$ We cannot conclude $forall l in Z: lnot (0 mid l)$
– Fefnir Wilhelm
Nov 16 at 23:31










1 Answer
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According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.






share|cite|improve this answer























  • If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
    – Fefnir Wilhelm
    Nov 16 at 23:50












  • @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
    – gimusi
    Nov 17 at 8:15











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According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.






share|cite|improve this answer























  • If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
    – Fefnir Wilhelm
    Nov 16 at 23:50












  • @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
    – gimusi
    Nov 17 at 8:15















up vote
0
down vote













According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.






share|cite|improve this answer























  • If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
    – Fefnir Wilhelm
    Nov 16 at 23:50












  • @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
    – gimusi
    Nov 17 at 8:15













up vote
0
down vote










up vote
0
down vote









According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.






share|cite|improve this answer














According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 17 at 8:14

























answered Nov 16 at 23:33









gimusi

86.9k74393




86.9k74393












  • If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
    – Fefnir Wilhelm
    Nov 16 at 23:50












  • @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
    – gimusi
    Nov 17 at 8:15


















  • If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
    – Fefnir Wilhelm
    Nov 16 at 23:50












  • @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
    – gimusi
    Nov 17 at 8:15
















If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
– Fefnir Wilhelm
Nov 16 at 23:50






If we let the definition of "k divides l" be: $$exists n in mathbb{Z} : l = k times n$$ then the negation statement works for $0$, no?
– Fefnir Wilhelm
Nov 16 at 23:50














@FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
– gimusi
Nov 17 at 8:15




@FefnirWilhelm Yes you are right, with reference to the general definition the negation is true.
– gimusi
Nov 17 at 8:15


















 

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