Mixing
Quantifier difference
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What s the difference between $ n in Z implies n(n+1) =2k $ such that $k in Z$ and $ forall n in Z implies n(n+1) =2k $ such that $k in Z$
Is this true:
$ (n in Z implies n(n+1) =2k $ such that $k in Z) implies forall n in Z; n(n+1) =2k $ such that$ k in Z$
quantifiers
asked 5 hours ago
J.Moh
395
add a comment |
up vote
0
down vote
favorite
What s the difference between $ n in Z implies n(n+1) =2k $ such that $k in Z$ and $ forall n in Z implies n(n+1) =2k $ such that $k in Z$
Is this true:
$ (n in Z implies n(n+1) =2k $ such that $k in Z) implies forall n in Z; n(n+1) =2k $ such that$ k in Z$
quantifiers
asked 5 hours ago
J.Moh
395
None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
/ means such that in my case
– J.Moh
3 hours ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What s the difference between $ n in Z implies n(n+1) =2k $ such that $k in Z$ and $ forall n in Z implies n(n+1) =2k $ such that $k in Z$
Is this true:
$ (n in Z implies n(n+1) =2k $ such that $k in Z) implies forall n in Z; n(n+1) =2k $ such that$ k in Z$
quantifiers
asked 5 hours ago
J.Moh
395
What s the difference between $ n in Z implies n(n+1) =2k $ such that $k in Z$ and $ forall n in Z implies n(n+1) =2k $ such that $k in Z$
Is this true:
$ (n in Z implies n(n+1) =2k $ such that $k in Z) implies forall n in Z; n(n+1) =2k $ such that$ k in Z$
quantifiers
quantifiers
asked 5 hours ago
J.Moh
395
asked 5 hours ago
J.Moh
395
edited 1 hour ago
asked 5 hours ago
J.Moh
395
asked 5 hours ago
J.Moh
395
asked 5 hours ago
J.Moh
395
395
None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
/ means such that in my case
– J.Moh
3 hours ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago
add a comment |
None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
/ means such that in my case
– J.Moh
3 hours ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago
None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
/ means such that in my case
– J.Moh
3 hours ago
/ means such that in my case
– J.Moh
3 hours ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Is this true that :
$(n ∈ mathbb Z ⟹ dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k text { , for } k ∈ mathbb Z$ ?
The formula :
$dfrac {n}{(n+1)} = 2k$
is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.
Specifically, the formula is true only for $n=k=0$.
This means that the consequent : $∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k$ is false.
Thus the original formula is false for $n=k=0$ (because in that case we have $text T to text F$, which is $text F$) and true in all other cases (because $text F to text F$ is $text T$).
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
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
Is this true that :
$(n ∈ mathbb Z ⟹ dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k text { , for } k ∈ mathbb Z$ ?
The formula :
$dfrac {n}{(n+1)} = 2k$
is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.
Specifically, the formula is true only for $n=k=0$.
This means that the consequent : $∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k$ is false.
Thus the original formula is false for $n=k=0$ (because in that case we have $text T to text F$, which is $text F$) and true in all other cases (because $text F to text F$ is $text T$).
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
add a comment |
up vote
0
down vote
Is this true that :
$(n ∈ mathbb Z ⟹ dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k text { , for } k ∈ mathbb Z$ ?
The formula :
$dfrac {n}{(n+1)} = 2k$
is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.
Specifically, the formula is true only for $n=k=0$.
This means that the consequent : $∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k$ is false.
Thus the original formula is false for $n=k=0$ (because in that case we have $text T to text F$, which is $text F$) and true in all other cases (because $text F to text F$ is $text T$).
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Is this true that :
$(n ∈ mathbb Z ⟹ dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k text { , for } k ∈ mathbb Z$ ?
The formula :
$dfrac {n}{(n+1)} = 2k$
is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.
Specifically, the formula is true only for $n=k=0$.
This means that the consequent : $∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k$ is false.
Thus the original formula is false for $n=k=0$ (because in that case we have $text T to text F$, which is $text F$) and true in all other cases (because $text F to text F$ is $text T$).
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
Is this true that :
$(n ∈ mathbb Z ⟹ dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k text { , for } k ∈ mathbb Z$ ?
The formula :
$dfrac {n}{(n+1)} = 2k$
is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.
Specifically, the formula is true only for $n=k=0$.
This means that the consequent : $∀n ∈ mathbb Z dfrac {n}{(n+1)} = 2k$ is false.
Thus the original formula is false for $n=k=0$ (because in that case we have $text T to text F$, which is $text F$) and true in all other cases (because $text F to text F$ is $text T$).
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
edited 1 hour ago
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
answered 4 hours ago
Mauro ALLEGRANZA
63.3k448110
63.3k448110
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
add a comment |
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
/ doesn t mean division in my case, it means such that
– J.Moh
3 hours ago
add a comment |
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None, in that I cannot understand either: why have you introduced a $k$, only to use it in the expression $2k/k$ which merely simplifies to $2$?
– Lord Shark the Unknown
4 hours ago
/ means such that in my case
– J.Moh
3 hours ago
Impossible to understand ... The first one is an equation : $dfrac {n}{n+1}=2k$.
– Mauro ALLEGRANZA
1 hour ago
The second one is a formula : $(dfrac {n}{n+1}=2k) to forall n (dfrac {n}{n+1}=2k)$ whose truth value depends on $n$ and $k$.
– Mauro ALLEGRANZA
1 hour ago
Rewritten in an unambiguous way
– J.Moh
1 hour ago