Determining Statement Truth or Falsity?
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I have to determine if these are true or false, for "∀x∃y" I read it as "for all X there is a Y". The domain
consists of pairs x and y, where x is -1, 0, or 1 and y is -1, 0, or 1. My answers are on the right in italics, however I dont know if they are correct, could someone explain?
(a) ∀x∃y(x + y = 0) true
(b) ∃x∀y(x + y = y) true
(c) ∀x∃y(xy ≥ y) true
(d) ∃x∀y(x ≤ y) false
discrete-mathematics logic quantifiers
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
$endgroup$
add a comment |
$begingroup$
I have to determine if these are true or false, for "∀x∃y" I read it as "for all X there is a Y". The domain
consists of pairs x and y, where x is -1, 0, or 1 and y is -1, 0, or 1. My answers are on the right in italics, however I dont know if they are correct, could someone explain?
(a) ∀x∃y(x + y = 0) true
(b) ∃x∀y(x + y = y) true
(c) ∀x∃y(xy ≥ y) true
(d) ∃x∀y(x ≤ y) false
discrete-mathematics logic quantifiers
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
$endgroup$
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@DanielSchepler sorry forgot to include that, just edited the post.
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– Usama Ghawji
Jan 24 at 22:17
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If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
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– fleablood
Jan 24 at 22:41
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That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
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– Daniel Schepler
Jan 24 at 23:11
add a comment |
$begingroup$
I have to determine if these are true or false, for "∀x∃y" I read it as "for all X there is a Y". The domain
consists of pairs x and y, where x is -1, 0, or 1 and y is -1, 0, or 1. My answers are on the right in italics, however I dont know if they are correct, could someone explain?
(a) ∀x∃y(x + y = 0) true
(b) ∃x∀y(x + y = y) true
(c) ∀x∃y(xy ≥ y) true
(d) ∃x∀y(x ≤ y) false
discrete-mathematics logic quantifiers
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
$endgroup$
I have to determine if these are true or false, for "∀x∃y" I read it as "for all X there is a Y". The domain
consists of pairs x and y, where x is -1, 0, or 1 and y is -1, 0, or 1. My answers are on the right in italics, however I dont know if they are correct, could someone explain?
(a) ∀x∃y(x + y = 0) true
(b) ∃x∀y(x + y = y) true
(c) ∀x∃y(xy ≥ y) true
(d) ∃x∀y(x ≤ y) false
discrete-mathematics logic quantifiers
discrete-mathematics logic quantifiers
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
edited Jan 24 at 22:17
Usama Ghawji
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
asked Jan 24 at 22:08
Usama GhawjiUsama Ghawji
525
525
$begingroup$
@DanielSchepler sorry forgot to include that, just edited the post.
$endgroup$
– Usama Ghawji
Jan 24 at 22:17
$begingroup$
If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
$endgroup$
– fleablood
Jan 24 at 22:41
$begingroup$
That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
$endgroup$
– Daniel Schepler
Jan 24 at 23:11
add a comment |
$begingroup$
@DanielSchepler sorry forgot to include that, just edited the post.
$endgroup$
– Usama Ghawji
Jan 24 at 22:17
$begingroup$
If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
$endgroup$
– fleablood
Jan 24 at 22:41
$begingroup$
That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
$endgroup$
– Daniel Schepler
Jan 24 at 23:11
$begingroup$
@DanielSchepler sorry forgot to include that, just edited the post.
$endgroup$
– Usama Ghawji
Jan 24 at 22:17
$begingroup$
@DanielSchepler sorry forgot to include that, just edited the post.
$endgroup$
– Usama Ghawji
Jan 24 at 22:17
$begingroup$
If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
$endgroup$
– fleablood
Jan 24 at 22:41
$begingroup$
If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
$endgroup$
– fleablood
Jan 24 at 22:41
$begingroup$
That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
$endgroup$
– Daniel Schepler
Jan 24 at 23:11
$begingroup$
That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
$endgroup$
– Daniel Schepler
Jan 24 at 23:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It looks like all your answers are correct.
The qualifier $forall x exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $forall (x in mathbb{N}) exists (y in mathbb{N}) (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y in {-1, 0, 1}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$forall x exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $exists x forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $exists x forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$exists x forall y (x + y = y)$ is true, because $x = 0$ works for all values of $y$.
answered Jan 24 at 22:37
ConManConMan
7,9021324
$endgroup$
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
add a comment |
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1 Answer
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$begingroup$
It looks like all your answers are correct.
The qualifier $forall x exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $forall (x in mathbb{N}) exists (y in mathbb{N}) (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y in {-1, 0, 1}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$forall x exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $exists x forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $exists x forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$exists x forall y (x + y = y)$ is true, because $x = 0$ works for all values of $y$.
answered Jan 24 at 22:37
ConManConMan
7,9021324
$endgroup$
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
add a comment |
$begingroup$
It looks like all your answers are correct.
The qualifier $forall x exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $forall (x in mathbb{N}) exists (y in mathbb{N}) (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y in {-1, 0, 1}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$forall x exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $exists x forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $exists x forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$exists x forall y (x + y = y)$ is true, because $x = 0$ works for all values of $y$.
answered Jan 24 at 22:37
ConManConMan
7,9021324
$endgroup$
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
add a comment |
$begingroup$
It looks like all your answers are correct.
The qualifier $forall x exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $forall (x in mathbb{N}) exists (y in mathbb{N}) (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y in {-1, 0, 1}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$forall x exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $exists x forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $exists x forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$exists x forall y (x + y = y)$ is true, because $x = 0$ works for all values of $y$.
answered Jan 24 at 22:37
ConManConMan
7,9021324
$endgroup$
It looks like all your answers are correct.
The qualifier $forall x exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $forall (x in mathbb{N}) exists (y in mathbb{N}) (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y in {-1, 0, 1}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$forall x exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $exists x forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $exists x forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$exists x forall y (x + y = y)$ is true, because $x = 0$ works for all values of $y$.
answered Jan 24 at 22:37
ConManConMan
7,9021324
answered Jan 24 at 22:37
ConManConMan
7,9021324
answered Jan 24 at 22:37
ConManConMan
7,9021324
answered Jan 24 at 22:37
ConManConMan
7,9021324
7,9021324
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
add a comment |
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I appreciate the feedback, great answer!
$endgroup$
– Usama Ghawji
Jan 24 at 22:52
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
I'd think (d) would be true because $x=-1$ works.
$endgroup$
– Daniel Schepler
Jan 24 at 23:53
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
$begingroup$
Sorry, you're right. I think I saw something different in there before, or just confused myself.
$endgroup$
– ConMan
Jan 25 at 1:00
add a comment |
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$begingroup$
@DanielSchepler sorry forgot to include that, just edited the post.
$endgroup$
– Usama Ghawji
Jan 24 at 22:17
$begingroup$
If you can explain why you got those answers, we can tell you if your reasoning is correct. For example a) is true because for each $x = -1,0,1$ there is a $y = 1,0,-1$ respectively so that $x + y = 0; (-1+1=0+0=1+(-1) = 0$. If that (or something similar to it) was your reasoning. But if your reasoning was weird and you got "true" for weird reasons, you aren't correct.
$endgroup$
– fleablood
Jan 24 at 22:41
$begingroup$
That domain is a bit problematic, because then the natural interpretation of the binary function symbol $+$ does not always take values which are in that domain of discourse.
$endgroup$
– Daniel Schepler
Jan 24 at 23:11