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
asked Nov 16 at 23:26
Fefnir Wilhelm
62
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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
asked Nov 16 at 23:26
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
add a comment |
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
asked Nov 16 at 23:26
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$exists k in mathbb{Z}: forall l in mathbb{Z}: lnot(k mid l) $$
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
asked Nov 16 at 23:26
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 23:26
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 23:26
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 23:26
Fefnir Wilhelm
62
asked Nov 16 at 23:26
Fefnir Wilhelm
62
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Nov 16 at 23:33
gimusi
86k74292
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 16 at 23:33
gimusi
86k74292
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
add a comment |
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.
answered Nov 16 at 23:33
gimusi
86k74292
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
add a comment |
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.
answered Nov 16 at 23:33
gimusi
86k74292
According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.
answered Nov 16 at 23:33
gimusi
86k74292
edited Nov 17 at 8:14
answered Nov 16 at 23:33
gimusi
86k74292
answered Nov 16 at 23:33
gimusi
86k74292
answered Nov 16 at 23:33
gimusi
86k74292
86k74292
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
add a comment |
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
add a comment |
Fefnir Wilhelm is a new contributor. Be nice, and check out our Code of Conduct.
Fefnir Wilhelm is a new contributor. Be nice, and check out our Code of Conduct.
Fefnir Wilhelm is a new contributor. Be nice, and check out our Code of Conduct.
Fefnir Wilhelm is a new contributor. Be nice, and check out our Code of Conduct.
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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