Mixing
How to write this $(exists (x,y) in R, y- 3x-2 wedge y = 1-x^2) $in words?
$begingroup$
I am translating the negation of the statement $(exists (x,y) in R, y= 3x-2 wedge y = 1-x^2) $
$(forall (x,y) in R, yne 3x-2 vee y ne 1-x^2) $
So far I have, "For every $(x,y)$ there exists a real number such that $yne 3x-2 $ or $ y ne 1-x^2$ "
The reason I am second guessing myself (even if i am correct) is that it doesn't really make sense to me in the form of a sentence. Am I right? If not, any suggestions?
proof-verification proof-writing predicate-logic
asked Jan 15 at 1:39
ForextraderForextrader
717
$endgroup$
add a comment |
$begingroup$
I am translating the negation of the statement $(exists (x,y) in R, y= 3x-2 wedge y = 1-x^2) $
$(forall (x,y) in R, yne 3x-2 vee y ne 1-x^2) $
So far I have, "For every $(x,y)$ there exists a real number such that $yne 3x-2 $ or $ y ne 1-x^2$ "
The reason I am second guessing myself (even if i am correct) is that it doesn't really make sense to me in the form of a sentence. Am I right? If not, any suggestions?
proof-verification proof-writing predicate-logic
asked Jan 15 at 1:39
ForextraderForextrader
717
$endgroup$
$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18
add a comment |
$begingroup$
I am translating the negation of the statement $(exists (x,y) in R, y= 3x-2 wedge y = 1-x^2) $
$(forall (x,y) in R, yne 3x-2 vee y ne 1-x^2) $
So far I have, "For every $(x,y)$ there exists a real number such that $yne 3x-2 $ or $ y ne 1-x^2$ "
The reason I am second guessing myself (even if i am correct) is that it doesn't really make sense to me in the form of a sentence. Am I right? If not, any suggestions?
proof-verification proof-writing predicate-logic
asked Jan 15 at 1:39
ForextraderForextrader
717
$endgroup$
I am translating the negation of the statement $(exists (x,y) in R, y= 3x-2 wedge y = 1-x^2) $
$(forall (x,y) in R, yne 3x-2 vee y ne 1-x^2) $
So far I have, "For every $(x,y)$ there exists a real number such that $yne 3x-2 $ or $ y ne 1-x^2$ "
The reason I am second guessing myself (even if i am correct) is that it doesn't really make sense to me in the form of a sentence. Am I right? If not, any suggestions?
proof-verification proof-writing predicate-logic
proof-verification proof-writing predicate-logic
asked Jan 15 at 1:39
ForextraderForextrader
717
asked Jan 15 at 1:39
ForextraderForextrader
717
edited Jan 15 at 2:18
Forextrader
asked Jan 15 at 1:39
ForextraderForextrader
717
asked Jan 15 at 1:39
ForextraderForextrader
717
asked Jan 15 at 1:39
ForextraderForextrader
717
717
$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18
add a comment |
$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18
$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18
add a comment |
1 Answer
1
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oldest
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$begingroup$
Note:
$$ lnot (forall x in A, phi(x) wedge psi(x))$$
is the same as
$$ exists x in A, (lnotphi(x)) vee (lnot psi(x))$$
If $wedge$ and $vee$ swapped places in the above, you still get a true statement.
If $exists$ and $forall$ are swapped places in the above, you still get a true statement.
Your correct if you take out the "the exists part". That is, the correct statement is: "For every $(x,y), y ne 3x - 2 vee y ne 1 - x^2$".
Note that in your case, $A = mathbb{R}^2$, so elements in $A$ are of the form $(x,y)$ where $x in mathbb{R}$ and $y in mathbb{R}$.
You asked about readability. For the first sentence, you have "There exists a pair $(x,y)$ such Eq1 is true and Eq2 is true". The negation of this statement is that "There doesn't exist a pair $(x,y)$ such that Eq1 is true and Eq2 is true". That is, "For every pair $(x,y)$, either Eq1 is false or Eq2 is false".
answered Jan 15 at 1:50
MetricMetric
1,23649
$endgroup$
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
Note:
$$ lnot (forall x in A, phi(x) wedge psi(x))$$
is the same as
$$ exists x in A, (lnotphi(x)) vee (lnot psi(x))$$
If $wedge$ and $vee$ swapped places in the above, you still get a true statement.
If $exists$ and $forall$ are swapped places in the above, you still get a true statement.
Your correct if you take out the "the exists part". That is, the correct statement is: "For every $(x,y), y ne 3x - 2 vee y ne 1 - x^2$".
Note that in your case, $A = mathbb{R}^2$, so elements in $A$ are of the form $(x,y)$ where $x in mathbb{R}$ and $y in mathbb{R}$.
You asked about readability. For the first sentence, you have "There exists a pair $(x,y)$ such Eq1 is true and Eq2 is true". The negation of this statement is that "There doesn't exist a pair $(x,y)$ such that Eq1 is true and Eq2 is true". That is, "For every pair $(x,y)$, either Eq1 is false or Eq2 is false".
answered Jan 15 at 1:50
MetricMetric
1,23649
$endgroup$
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
add a comment |
$begingroup$
Note:
$$ lnot (forall x in A, phi(x) wedge psi(x))$$
is the same as
$$ exists x in A, (lnotphi(x)) vee (lnot psi(x))$$
If $wedge$ and $vee$ swapped places in the above, you still get a true statement.
If $exists$ and $forall$ are swapped places in the above, you still get a true statement.
Your correct if you take out the "the exists part". That is, the correct statement is: "For every $(x,y), y ne 3x - 2 vee y ne 1 - x^2$".
Note that in your case, $A = mathbb{R}^2$, so elements in $A$ are of the form $(x,y)$ where $x in mathbb{R}$ and $y in mathbb{R}$.
You asked about readability. For the first sentence, you have "There exists a pair $(x,y)$ such Eq1 is true and Eq2 is true". The negation of this statement is that "There doesn't exist a pair $(x,y)$ such that Eq1 is true and Eq2 is true". That is, "For every pair $(x,y)$, either Eq1 is false or Eq2 is false".
answered Jan 15 at 1:50
MetricMetric
1,23649
$endgroup$
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
add a comment |
$begingroup$
Note:
$$ lnot (forall x in A, phi(x) wedge psi(x))$$
is the same as
$$ exists x in A, (lnotphi(x)) vee (lnot psi(x))$$
If $wedge$ and $vee$ swapped places in the above, you still get a true statement.
If $exists$ and $forall$ are swapped places in the above, you still get a true statement.
Your correct if you take out the "the exists part". That is, the correct statement is: "For every $(x,y), y ne 3x - 2 vee y ne 1 - x^2$".
Note that in your case, $A = mathbb{R}^2$, so elements in $A$ are of the form $(x,y)$ where $x in mathbb{R}$ and $y in mathbb{R}$.
You asked about readability. For the first sentence, you have "There exists a pair $(x,y)$ such Eq1 is true and Eq2 is true". The negation of this statement is that "There doesn't exist a pair $(x,y)$ such that Eq1 is true and Eq2 is true". That is, "For every pair $(x,y)$, either Eq1 is false or Eq2 is false".
answered Jan 15 at 1:50
MetricMetric
1,23649
$endgroup$
Note:
$$ lnot (forall x in A, phi(x) wedge psi(x))$$
is the same as
$$ exists x in A, (lnotphi(x)) vee (lnot psi(x))$$
If $wedge$ and $vee$ swapped places in the above, you still get a true statement.
If $exists$ and $forall$ are swapped places in the above, you still get a true statement.
Your correct if you take out the "the exists part". That is, the correct statement is: "For every $(x,y), y ne 3x - 2 vee y ne 1 - x^2$".
Note that in your case, $A = mathbb{R}^2$, so elements in $A$ are of the form $(x,y)$ where $x in mathbb{R}$ and $y in mathbb{R}$.
You asked about readability. For the first sentence, you have "There exists a pair $(x,y)$ such Eq1 is true and Eq2 is true". The negation of this statement is that "There doesn't exist a pair $(x,y)$ such that Eq1 is true and Eq2 is true". That is, "For every pair $(x,y)$, either Eq1 is false or Eq2 is false".
answered Jan 15 at 1:50
MetricMetric
1,23649
edited Jan 15 at 1:55
answered Jan 15 at 1:50
MetricMetric
1,23649
answered Jan 15 at 1:50
MetricMetric
1,23649
answered Jan 15 at 1:50
MetricMetric
1,23649
1,23649
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
add a comment |
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Ok so my negation statement is correct then from you're saying
$endgroup$
– Forextrader
Jan 15 at 1:56
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
Well you said "For every (x,y) there exists real numbers ...". What real numbers are you referring to? It is sufficient to say "For every (x,y), Eq1 is not true or Eq2 is not true".
$endgroup$
– Metric
Jan 15 at 1:58
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
$begingroup$
When you say "For every (x,y)...", what is implied is "For every pair of real numbers (x,y)...".
$endgroup$
– Metric
Jan 15 at 2:00
add a comment |
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$begingroup$
"For all real numbers $x$ and $y$, $ynot =$..."
$endgroup$
– T. Fo
Jan 15 at 1:42
$begingroup$
Ok, thank you. so except for the "for all real numbers $x$ and $y$" mine was correct? It just seems like not a natural sentence
$endgroup$
– Forextrader
Jan 15 at 1:50
$begingroup$
Is $y-3x-2$ a typo for $y=3x-2$?
$endgroup$
– bof
Jan 15 at 2:17
$begingroup$
Yes, thank you, corrected
$endgroup$
– Forextrader
Jan 15 at 2:18