Mixing
How to read $exists x forall y exists z((x + y)z = 1)$?
$begingroup$
$$exists x forall y exists z((x + y)z = 1)$$
How can I translate this expression to English? And is the statement true or false? (for numbers in $Bbb R$).
$z=1$ is the part that confuses me.
predicate-logic quantifiers
edited Jan 24 at 10:20
Christoph
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
$endgroup$
add a comment |
$begingroup$
$$exists x forall y exists z((x + y)z = 1)$$
How can I translate this expression to English? And is the statement true or false? (for numbers in $Bbb R$).
$z=1$ is the part that confuses me.
predicate-logic quantifiers
edited Jan 24 at 10:20
Christoph
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
$endgroup$
1
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23
add a comment |
$begingroup$
$$exists x forall y exists z((x + y)z = 1)$$
How can I translate this expression to English? And is the statement true or false? (for numbers in $Bbb R$).
$z=1$ is the part that confuses me.
predicate-logic quantifiers
edited Jan 24 at 10:20
Christoph
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
$endgroup$
$$exists x forall y exists z((x + y)z = 1)$$
How can I translate this expression to English? And is the statement true or false? (for numbers in $Bbb R$).
$z=1$ is the part that confuses me.
predicate-logic quantifiers
predicate-logic quantifiers
edited Jan 24 at 10:20
Christoph
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
edited Jan 24 at 10:20
Christoph
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
edited Jan 24 at 10:20
Christoph
12.5k1642
edited Jan 24 at 10:20
Christoph
12.5k1642
edited Jan 24 at 10:20
Christoph
12.5k1642
12.5k1642
asked Jan 24 at 10:00
user635758user635758
164
asked Jan 24 at 10:00
user635758user635758
164
asked Jan 24 at 10:00
user635758user635758
164
164
1
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23
add a comment |
1
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23
1
1
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It translates as:
There exists an $x$ such that whatever $y$ you choose you can find an $z$ such that $x+y$ multiplied by $z$ equals $1$.
It is false because no matter what value $x$ has, whenever $y=-x$, $x+y$ multiplied by $z$ is zero.
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
$endgroup$
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
$begingroup$
It translates as:
There exists an $x$ such that whatever $y$ you choose you can find an $z$ such that $x+y$ multiplied by $z$ equals $1$.
It is false because no matter what value $x$ has, whenever $y=-x$, $x+y$ multiplied by $z$ is zero.
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
$endgroup$
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
add a comment |
$begingroup$
It translates as:
There exists an $x$ such that whatever $y$ you choose you can find an $z$ such that $x+y$ multiplied by $z$ equals $1$.
It is false because no matter what value $x$ has, whenever $y=-x$, $x+y$ multiplied by $z$ is zero.
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
$endgroup$
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
add a comment |
$begingroup$
It translates as:
There exists an $x$ such that whatever $y$ you choose you can find an $z$ such that $x+y$ multiplied by $z$ equals $1$.
It is false because no matter what value $x$ has, whenever $y=-x$, $x+y$ multiplied by $z$ is zero.
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
$endgroup$
It translates as:
There exists an $x$ such that whatever $y$ you choose you can find an $z$ such that $x+y$ multiplied by $z$ equals $1$.
It is false because no matter what value $x$ has, whenever $y=-x$, $x+y$ multiplied by $z$ is zero.
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
answered Jan 24 at 10:11
Bernard HurleyBernard Hurley
1787
1787
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
add a comment |
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
1
1
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
$begingroup$
You can't for $y=0$.
$endgroup$
– Bernard Hurley
Jan 24 at 10:21
add a comment |
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1
$begingroup$
It's not $z=1$, but $(x+y)z=1$. Hint: the statement is false.
$endgroup$
– Michael Burr
Jan 24 at 10:06
$begingroup$
Think of it as: There is an $x$ such that for all $y$, the equation $(x+y)z=1$ has a solution in $z$.
$endgroup$
– Chrystomath
Jan 24 at 10:08
$begingroup$
Do you mean it says z multiplied with (x+y) should be 1?
$endgroup$
– user635758
Jan 24 at 10:09
$begingroup$
Yes, of course.
$endgroup$
– TonyK
Jan 24 at 10:23