Give a counterexample, if possible, to these universally quantified statements.
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Give a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. That is, show a reason why the statement is NOT universally true when applied to the domain of integers.
a. $forall x (|x| > 0)$
b. $forall x exists y (x = 1/y) $
c. For each of the quantified statements in a-b above, give a domain for the variables for which each universally quantified statement a-b is true.
For part a I put $x=0$. For b. I'm not sure what it is asking. I put $y=0$ as a shot in the dark but I've no clue.
discrete-mathematics
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
$endgroup$
|
show 1 more comment
$begingroup$
Give a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. That is, show a reason why the statement is NOT universally true when applied to the domain of integers.
a. $forall x (|x| > 0)$
b. $forall x exists y (x = 1/y) $
c. For each of the quantified statements in a-b above, give a domain for the variables for which each universally quantified statement a-b is true.
For part a I put $x=0$. For b. I'm not sure what it is asking. I put $y=0$ as a shot in the dark but I've no clue.
discrete-mathematics
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
$endgroup$
$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39
|
show 1 more comment
$begingroup$
Give a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. That is, show a reason why the statement is NOT universally true when applied to the domain of integers.
a. $forall x (|x| > 0)$
b. $forall x exists y (x = 1/y) $
c. For each of the quantified statements in a-b above, give a domain for the variables for which each universally quantified statement a-b is true.
For part a I put $x=0$. For b. I'm not sure what it is asking. I put $y=0$ as a shot in the dark but I've no clue.
discrete-mathematics
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
$endgroup$
Give a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. That is, show a reason why the statement is NOT universally true when applied to the domain of integers.
a. $forall x (|x| > 0)$
b. $forall x exists y (x = 1/y) $
c. For each of the quantified statements in a-b above, give a domain for the variables for which each universally quantified statement a-b is true.
For part a I put $x=0$. For b. I'm not sure what it is asking. I put $y=0$ as a shot in the dark but I've no clue.
discrete-mathematics
discrete-mathematics
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
asked Sep 19 '15 at 3:30
theguy1991theguy1991
5616
5616
$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39
|
show 1 more comment
$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39
$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39
|
show 1 more comment
1 Answer
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Part $b$ is saying that every number has a reciprocal. Not so! $0$ doesn't, so you need to define your domain to exclude $0$.
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Part $b$ is saying that every number has a reciprocal. Not so! $0$ doesn't, so you need to define your domain to exclude $0$.
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
$endgroup$
add a comment |
$begingroup$
Part $b$ is saying that every number has a reciprocal. Not so! $0$ doesn't, so you need to define your domain to exclude $0$.
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
$endgroup$
add a comment |
$begingroup$
Part $b$ is saying that every number has a reciprocal. Not so! $0$ doesn't, so you need to define your domain to exclude $0$.
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
$endgroup$
Part $b$ is saying that every number has a reciprocal. Not so! $0$ doesn't, so you need to define your domain to exclude $0$.
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
answered Sep 19 '15 at 3:34
Adam HrankowskiAdam Hrankowski
2,098930
2,098930
add a comment |
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$begingroup$
$y$ isn't the one you can pick. It's $x$. But $0$ is correct.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:33
$begingroup$
So for where it is true in C. I could just use x = 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:34
$begingroup$
$x=1$ is not a domain. You could use ${1}$. That's not the one that's intended, but it works.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:35
$begingroup$
So like x > or equal to 1?
$endgroup$
– theguy1991
Sep 19 '15 at 3:38
$begingroup$
That is a description of another possible domain. But the actual domain is ${xinmathbb{Z}|xgeq 1}$. It's a set.
$endgroup$
– Matt Samuel
Sep 19 '15 at 3:39