Mixing
Help with quantifiers: $forall yinmathbb{Z},exists xinmathbb{Z},(x^2+ygeq1)$
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As the title suggests, I am revisiting quantifiers and am having trouble deciphering the proper meaning. I know it is one of these two:
Does it mean that for each individual $yinmathbb{Z}$ we can find an $xinmathbb{Z}$ pertaining to that individual $yinmathbb{Z}$ such that the statement is true
or
Does it mean that for all $yinmathbb{Z}$ collectively, there must exist a single unique $xinmathbb{Z}$ that satisfies the above proposition for any $yinmathbb{Z}$.
I'm leaning towards the latter statement. Thank you.
logic quantifiers
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
$endgroup$
add a comment |
$begingroup$
As the title suggests, I am revisiting quantifiers and am having trouble deciphering the proper meaning. I know it is one of these two:
Does it mean that for each individual $yinmathbb{Z}$ we can find an $xinmathbb{Z}$ pertaining to that individual $yinmathbb{Z}$ such that the statement is true
or
Does it mean that for all $yinmathbb{Z}$ collectively, there must exist a single unique $xinmathbb{Z}$ that satisfies the above proposition for any $yinmathbb{Z}$.
I'm leaning towards the latter statement. Thank you.
logic quantifiers
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
$endgroup$
add a comment |
$begingroup$
As the title suggests, I am revisiting quantifiers and am having trouble deciphering the proper meaning. I know it is one of these two:
Does it mean that for each individual $yinmathbb{Z}$ we can find an $xinmathbb{Z}$ pertaining to that individual $yinmathbb{Z}$ such that the statement is true
or
Does it mean that for all $yinmathbb{Z}$ collectively, there must exist a single unique $xinmathbb{Z}$ that satisfies the above proposition for any $yinmathbb{Z}$.
I'm leaning towards the latter statement. Thank you.
logic quantifiers
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
$endgroup$
As the title suggests, I am revisiting quantifiers and am having trouble deciphering the proper meaning. I know it is one of these two:
Does it mean that for each individual $yinmathbb{Z}$ we can find an $xinmathbb{Z}$ pertaining to that individual $yinmathbb{Z}$ such that the statement is true
or
Does it mean that for all $yinmathbb{Z}$ collectively, there must exist a single unique $xinmathbb{Z}$ that satisfies the above proposition for any $yinmathbb{Z}$.
I'm leaning towards the latter statement. Thank you.
logic quantifiers
logic quantifiers
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
edited Jan 23 at 21:36
J. W. Tanner
3,2401320
3,2401320
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
asked Jan 23 at 21:05
Albert DiazAlbert Diaz
1157
1157
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It means the first. Think of it as $forall y in mathbb{Z}: (exists x in mathbb{Z}: (x^2+y^2 geq 1))$ -- that is, each quantifier "quantifies" a single expression and you can nest them to build more complex statements.
To obtain the latter meaning, you'd invert the order of the quantifiers: $exists x in mathbb{Z}: (forall y in mathbb{Z}: (x^2+y^2 geq 1))$ means that at least one $x in mathbb{Z}$ works for each $y$.
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
The first interpretation is correct: $forall y, exists x$ means that for any single $y$ there is some $x$ which works. If asserting that there is a single $x$ that works simultaneously for all $y$, it would be $exists x,forall y$.
answered Jan 23 at 21:09
ArthurArthur
118k7117200
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add a comment |
$begingroup$
It means that if you fix an integer $y$ you can find an integer $x$ such that $x^2+y geq 1$. There is no uniqueness of $x$ understood.
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It means the first. Think of it as $forall y in mathbb{Z}: (exists x in mathbb{Z}: (x^2+y^2 geq 1))$ -- that is, each quantifier "quantifies" a single expression and you can nest them to build more complex statements.
To obtain the latter meaning, you'd invert the order of the quantifiers: $exists x in mathbb{Z}: (forall y in mathbb{Z}: (x^2+y^2 geq 1))$ means that at least one $x in mathbb{Z}$ works for each $y$.
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
It means the first. Think of it as $forall y in mathbb{Z}: (exists x in mathbb{Z}: (x^2+y^2 geq 1))$ -- that is, each quantifier "quantifies" a single expression and you can nest them to build more complex statements.
To obtain the latter meaning, you'd invert the order of the quantifiers: $exists x in mathbb{Z}: (forall y in mathbb{Z}: (x^2+y^2 geq 1))$ means that at least one $x in mathbb{Z}$ works for each $y$.
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
It means the first. Think of it as $forall y in mathbb{Z}: (exists x in mathbb{Z}: (x^2+y^2 geq 1))$ -- that is, each quantifier "quantifies" a single expression and you can nest them to build more complex statements.
To obtain the latter meaning, you'd invert the order of the quantifiers: $exists x in mathbb{Z}: (forall y in mathbb{Z}: (x^2+y^2 geq 1))$ means that at least one $x in mathbb{Z}$ works for each $y$.
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
$endgroup$
It means the first. Think of it as $forall y in mathbb{Z}: (exists x in mathbb{Z}: (x^2+y^2 geq 1))$ -- that is, each quantifier "quantifies" a single expression and you can nest them to build more complex statements.
To obtain the latter meaning, you'd invert the order of the quantifiers: $exists x in mathbb{Z}: (forall y in mathbb{Z}: (x^2+y^2 geq 1))$ means that at least one $x in mathbb{Z}$ works for each $y$.
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
answered Jan 23 at 21:10
Bastián NúñezBastián Núñez
1765
1765
add a comment |
add a comment |
$begingroup$
The first interpretation is correct: $forall y, exists x$ means that for any single $y$ there is some $x$ which works. If asserting that there is a single $x$ that works simultaneously for all $y$, it would be $exists x,forall y$.
answered Jan 23 at 21:09
ArthurArthur
118k7117200
$endgroup$
add a comment |
$begingroup$
The first interpretation is correct: $forall y, exists x$ means that for any single $y$ there is some $x$ which works. If asserting that there is a single $x$ that works simultaneously for all $y$, it would be $exists x,forall y$.
answered Jan 23 at 21:09
ArthurArthur
118k7117200
$endgroup$
add a comment |
$begingroup$
The first interpretation is correct: $forall y, exists x$ means that for any single $y$ there is some $x$ which works. If asserting that there is a single $x$ that works simultaneously for all $y$, it would be $exists x,forall y$.
answered Jan 23 at 21:09
ArthurArthur
118k7117200
$endgroup$
The first interpretation is correct: $forall y, exists x$ means that for any single $y$ there is some $x$ which works. If asserting that there is a single $x$ that works simultaneously for all $y$, it would be $exists x,forall y$.
answered Jan 23 at 21:09
ArthurArthur
118k7117200
answered Jan 23 at 21:09
ArthurArthur
118k7117200
answered Jan 23 at 21:09
ArthurArthur
118k7117200
answered Jan 23 at 21:09
ArthurArthur
118k7117200
118k7117200
add a comment |
add a comment |
$begingroup$
It means that if you fix an integer $y$ you can find an integer $x$ such that $x^2+y geq 1$. There is no uniqueness of $x$ understood.
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
$endgroup$
add a comment |
$begingroup$
It means that if you fix an integer $y$ you can find an integer $x$ such that $x^2+y geq 1$. There is no uniqueness of $x$ understood.
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
$endgroup$
add a comment |
$begingroup$
It means that if you fix an integer $y$ you can find an integer $x$ such that $x^2+y geq 1$. There is no uniqueness of $x$ understood.
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
$endgroup$
It means that if you fix an integer $y$ you can find an integer $x$ such that $x^2+y geq 1$. There is no uniqueness of $x$ understood.
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
answered Jan 23 at 21:09
GibbsGibbs
5,4103827
5,4103827
add a comment |
add a comment |
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