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Negating a statement: is there indeed a quantifier missing?
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There is this homework assignment that I seem to keep getting wrong. The question is:
Negate the following statement:
"For every positive number $epsilon$, there is a positive number $delta$ such that |x-a| < $delta$ implies |f(x)-f(a)| < $epsilon$".
My answer was:
"There exists a positive number $epsilon$, such that for every positive number $delta$, there exists an x such that |x-a| < $delta$ and |f(x)-f(a)| $geqslant epsilon$".
In math symbols this is
$exists epsilon > 0, forall delta >0, exists x, (|x-a|<delta) wedge(|f(x)-f(a)|geqslantepsilon)$
But then the answer I got back from my teacher, was "for which x,a? Is this true for all x,a such that |x-a|<$delta$ or just for one set of x.a?"
I just can't seems to figure out exactly what I'm missing. A quantifier for a?
Can anybody help me, please?
logic
asked 1 hour ago
Caroline
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up vote
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down vote
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There is this homework assignment that I seem to keep getting wrong. The question is:
Negate the following statement:
"For every positive number $epsilon$, there is a positive number $delta$ such that |x-a| < $delta$ implies |f(x)-f(a)| < $epsilon$".
My answer was:
"There exists a positive number $epsilon$, such that for every positive number $delta$, there exists an x such that |x-a| < $delta$ and |f(x)-f(a)| $geqslant epsilon$".
In math symbols this is
$exists epsilon > 0, forall delta >0, exists x, (|x-a|<delta) wedge(|f(x)-f(a)|geqslantepsilon)$
But then the answer I got back from my teacher, was "for which x,a? Is this true for all x,a such that |x-a|<$delta$ or just for one set of x.a?"
I just can't seems to figure out exactly what I'm missing. A quantifier for a?
Can anybody help me, please?
logic
asked 1 hour ago
Caroline
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There is this homework assignment that I seem to keep getting wrong. The question is:
Negate the following statement:
"For every positive number $epsilon$, there is a positive number $delta$ such that |x-a| < $delta$ implies |f(x)-f(a)| < $epsilon$".
My answer was:
"There exists a positive number $epsilon$, such that for every positive number $delta$, there exists an x such that |x-a| < $delta$ and |f(x)-f(a)| $geqslant epsilon$".
In math symbols this is
$exists epsilon > 0, forall delta >0, exists x, (|x-a|<delta) wedge(|f(x)-f(a)|geqslantepsilon)$
But then the answer I got back from my teacher, was "for which x,a? Is this true for all x,a such that |x-a|<$delta$ or just for one set of x.a?"
I just can't seems to figure out exactly what I'm missing. A quantifier for a?
Can anybody help me, please?
logic
asked 1 hour ago
Caroline
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There is this homework assignment that I seem to keep getting wrong. The question is:
Negate the following statement:
"For every positive number $epsilon$, there is a positive number $delta$ such that |x-a| < $delta$ implies |f(x)-f(a)| < $epsilon$".
My answer was:
"There exists a positive number $epsilon$, such that for every positive number $delta$, there exists an x such that |x-a| < $delta$ and |f(x)-f(a)| $geqslant epsilon$".
In math symbols this is
$exists epsilon > 0, forall delta >0, exists x, (|x-a|<delta) wedge(|f(x)-f(a)|geqslantepsilon)$
But then the answer I got back from my teacher, was "for which x,a? Is this true for all x,a such that |x-a|<$delta$ or just for one set of x.a?"
I just can't seems to figure out exactly what I'm missing. A quantifier for a?
Can anybody help me, please?
logic
logic
asked 1 hour ago
Caroline
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Caroline
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Caroline
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Caroline
161
asked 1 hour ago
Caroline
161
161
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Caroline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago
add a comment |Â
In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago
In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
The original statement is
$$forall epsilon > 0quad exists delta >0 :quad P(x,a)$$
and the negation is
$$existsepsilon > 0quad forall delta >0: quad lnot P(x,a)$$
answered 54 mins ago
gimusi
77.5k73889
add a comment |Â
up vote
2
down vote
A function $f:DtoBbb R$ is continuous at $ain D$ if $$forall epsiloninBbb R^+~exists deltainBbb R^+~forall xin D~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 1$$
So a function $f:DtoBbb R$ is discontinuous at $ain D$ if $$exists epsiloninBbb R^+~forall deltainBbb R^+~exists xin D~~(lvert x-arvert<delta~land~lvert f(x)-f(a)rvert geqslantepsilon)tag 2$$
However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$forall epsiloninBbb R^+~exists deltainBbb R^+~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 3$$
$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.
answered 50 mins ago
Graham Kemp
82.6k43378
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The original statement is
$$forall epsilon > 0quad exists delta >0 :quad P(x,a)$$
and the negation is
$$existsepsilon > 0quad forall delta >0: quad lnot P(x,a)$$
answered 54 mins ago
gimusi
77.5k73889
add a comment |Â
up vote
3
down vote
The original statement is
$$forall epsilon > 0quad exists delta >0 :quad P(x,a)$$
and the negation is
$$existsepsilon > 0quad forall delta >0: quad lnot P(x,a)$$
answered 54 mins ago
gimusi
77.5k73889
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The original statement is
$$forall epsilon > 0quad exists delta >0 :quad P(x,a)$$
and the negation is
$$existsepsilon > 0quad forall delta >0: quad lnot P(x,a)$$
answered 54 mins ago
gimusi
77.5k73889
The original statement is
$$forall epsilon > 0quad exists delta >0 :quad P(x,a)$$
and the negation is
$$existsepsilon > 0quad forall delta >0: quad lnot P(x,a)$$
answered 54 mins ago
gimusi
77.5k73889
answered 54 mins ago
gimusi
77.5k73889
answered 54 mins ago
gimusi
77.5k73889
answered 54 mins ago
gimusi
77.5k73889
77.5k73889
add a comment |Â
add a comment |Â
up vote
2
down vote
A function $f:DtoBbb R$ is continuous at $ain D$ if $$forall epsiloninBbb R^+~exists deltainBbb R^+~forall xin D~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 1$$
So a function $f:DtoBbb R$ is discontinuous at $ain D$ if $$exists epsiloninBbb R^+~forall deltainBbb R^+~exists xin D~~(lvert x-arvert<delta~land~lvert f(x)-f(a)rvert geqslantepsilon)tag 2$$
However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$forall epsiloninBbb R^+~exists deltainBbb R^+~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 3$$
$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.
answered 50 mins ago
Graham Kemp
82.6k43378
add a comment |Â
up vote
2
down vote
A function $f:DtoBbb R$ is continuous at $ain D$ if $$forall epsiloninBbb R^+~exists deltainBbb R^+~forall xin D~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 1$$
So a function $f:DtoBbb R$ is discontinuous at $ain D$ if $$exists epsiloninBbb R^+~forall deltainBbb R^+~exists xin D~~(lvert x-arvert<delta~land~lvert f(x)-f(a)rvert geqslantepsilon)tag 2$$
However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$forall epsiloninBbb R^+~exists deltainBbb R^+~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 3$$
$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.
answered 50 mins ago
Graham Kemp
82.6k43378
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A function $f:DtoBbb R$ is continuous at $ain D$ if $$forall epsiloninBbb R^+~exists deltainBbb R^+~forall xin D~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 1$$
So a function $f:DtoBbb R$ is discontinuous at $ain D$ if $$exists epsiloninBbb R^+~forall deltainBbb R^+~exists xin D~~(lvert x-arvert<delta~land~lvert f(x)-f(a)rvert geqslantepsilon)tag 2$$
However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$forall epsiloninBbb R^+~exists deltainBbb R^+~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 3$$
$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.
answered 50 mins ago
Graham Kemp
82.6k43378
A function $f:DtoBbb R$ is continuous at $ain D$ if $$forall epsiloninBbb R^+~exists deltainBbb R^+~forall xin D~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 1$$
So a function $f:DtoBbb R$ is discontinuous at $ain D$ if $$exists epsiloninBbb R^+~forall deltainBbb R^+~exists xin D~~(lvert x-arvert<delta~land~lvert f(x)-f(a)rvert geqslantepsilon)tag 2$$
However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$forall epsiloninBbb R^+~exists deltainBbb R^+~~(lvert x-arvert<delta~to~lvert f(x)-f(a)rvert <epsilon)tag 3$$
$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.
answered 50 mins ago
Graham Kemp
82.6k43378
answered 50 mins ago
Graham Kemp
82.6k43378
answered 50 mins ago
Graham Kemp
82.6k43378
answered 50 mins ago
Graham Kemp
82.6k43378
82.6k43378
add a comment |Â
add a comment |Â
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In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable.
– ajotatxe
1 hour ago
@Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was?
– AnyAD
1 hour ago
Why did you write $exists x$ but not $exists x,a$ in your negation? What made you treat $x,a$ differently?
– Michal Adamaszek
1 hour ago
See Continuous function : definition.
– Mauro ALLEGRANZA
42 mins ago