Compute the Cardinality of the set ${z in Bbb{Z}|z>-10, z^3<0}$
$begingroup$
Compute the cardinality of the set ${z in Bbb{Z} mid z>-10, z^3<0}$
We will also write this with the notation: compute $bigl|{z in mathbb{Z} mid z>-10, z^3<0}bigr|$. (To show your work, makesure you show the set explicitly before telling us how many elements it has.)
If we're talking about all integers ($Bbb{Z}$) including negatives. For the first part of $z > -10$. $z$ can only be up to $-9$ otherwise it would be greater then or equal to $-10$. But what would the cardinality be? Since the set would have ${-9,-8,-7,-6...$ and so on till infinity?$}$
And for the second part $z^3<0$. $z$ can only be $0$ otherwise anything else for $z^3$ would be grater then $0$. So the set would be ${0}$ with cardinality $1$?
discrete-mathematics elementary-set-theory logic
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
$endgroup$
add a comment |
$begingroup$
Compute the cardinality of the set ${z in Bbb{Z} mid z>-10, z^3<0}$
We will also write this with the notation: compute $bigl|{z in mathbb{Z} mid z>-10, z^3<0}bigr|$. (To show your work, makesure you show the set explicitly before telling us how many elements it has.)
If we're talking about all integers ($Bbb{Z}$) including negatives. For the first part of $z > -10$. $z$ can only be up to $-9$ otherwise it would be greater then or equal to $-10$. But what would the cardinality be? Since the set would have ${-9,-8,-7,-6...$ and so on till infinity?$}$
And for the second part $z^3<0$. $z$ can only be $0$ otherwise anything else for $z^3$ would be grater then $0$. So the set would be ${0}$ with cardinality $1$?
discrete-mathematics elementary-set-theory logic
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
$endgroup$
3
$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30
add a comment |
$begingroup$
Compute the cardinality of the set ${z in Bbb{Z} mid z>-10, z^3<0}$
We will also write this with the notation: compute $bigl|{z in mathbb{Z} mid z>-10, z^3<0}bigr|$. (To show your work, makesure you show the set explicitly before telling us how many elements it has.)
If we're talking about all integers ($Bbb{Z}$) including negatives. For the first part of $z > -10$. $z$ can only be up to $-9$ otherwise it would be greater then or equal to $-10$. But what would the cardinality be? Since the set would have ${-9,-8,-7,-6...$ and so on till infinity?$}$
And for the second part $z^3<0$. $z$ can only be $0$ otherwise anything else for $z^3$ would be grater then $0$. So the set would be ${0}$ with cardinality $1$?
discrete-mathematics elementary-set-theory logic
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
$endgroup$
Compute the cardinality of the set ${z in Bbb{Z} mid z>-10, z^3<0}$
We will also write this with the notation: compute $bigl|{z in mathbb{Z} mid z>-10, z^3<0}bigr|$. (To show your work, makesure you show the set explicitly before telling us how many elements it has.)
If we're talking about all integers ($Bbb{Z}$) including negatives. For the first part of $z > -10$. $z$ can only be up to $-9$ otherwise it would be greater then or equal to $-10$. But what would the cardinality be? Since the set would have ${-9,-8,-7,-6...$ and so on till infinity?$}$
And for the second part $z^3<0$. $z$ can only be $0$ otherwise anything else for $z^3$ would be grater then $0$. So the set would be ${0}$ with cardinality $1$?
discrete-mathematics elementary-set-theory logic
discrete-mathematics elementary-set-theory logic
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
edited Jan 24 at 6:44
Taroccoesbrocco
5,64271840
5,64271840
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
asked Jan 24 at 2:22
Brad GuzmanBrad Guzman
63
63
3
$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30
add a comment |
3
$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30
3
3
$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30
$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The integers satisfying $z>-10$ form the set ${z in mathbb{Z} | z>-10}={-9,-8,-7,...}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set ${z in mathbb{Z}|z>-10, z^3<0}$, you want both of these conditions to be true, so $${z in mathbb{Z}|z>-10, z^3<0}={-9,-8,...,-2,-1}$$ which has $9$ elements. That is, the cardinality is $9$.
answered Jan 24 at 2:33
greeliousgreelious
472112
$endgroup$
add a comment |
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$begingroup$
The integers satisfying $z>-10$ form the set ${z in mathbb{Z} | z>-10}={-9,-8,-7,...}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set ${z in mathbb{Z}|z>-10, z^3<0}$, you want both of these conditions to be true, so $${z in mathbb{Z}|z>-10, z^3<0}={-9,-8,...,-2,-1}$$ which has $9$ elements. That is, the cardinality is $9$.
answered Jan 24 at 2:33
greeliousgreelious
472112
$endgroup$
add a comment |
$begingroup$
The integers satisfying $z>-10$ form the set ${z in mathbb{Z} | z>-10}={-9,-8,-7,...}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set ${z in mathbb{Z}|z>-10, z^3<0}$, you want both of these conditions to be true, so $${z in mathbb{Z}|z>-10, z^3<0}={-9,-8,...,-2,-1}$$ which has $9$ elements. That is, the cardinality is $9$.
answered Jan 24 at 2:33
greeliousgreelious
472112
$endgroup$
add a comment |
$begingroup$
The integers satisfying $z>-10$ form the set ${z in mathbb{Z} | z>-10}={-9,-8,-7,...}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set ${z in mathbb{Z}|z>-10, z^3<0}$, you want both of these conditions to be true, so $${z in mathbb{Z}|z>-10, z^3<0}={-9,-8,...,-2,-1}$$ which has $9$ elements. That is, the cardinality is $9$.
answered Jan 24 at 2:33
greeliousgreelious
472112
$endgroup$
The integers satisfying $z>-10$ form the set ${z in mathbb{Z} | z>-10}={-9,-8,-7,...}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set ${z in mathbb{Z}|z>-10, z^3<0}$, you want both of these conditions to be true, so $${z in mathbb{Z}|z>-10, z^3<0}={-9,-8,...,-2,-1}$$ which has $9$ elements. That is, the cardinality is $9$.
answered Jan 24 at 2:33
greeliousgreelious
472112
answered Jan 24 at 2:33
greeliousgreelious
472112
answered Jan 24 at 2:33
greeliousgreelious
472112
answered Jan 24 at 2:33
greeliousgreelious
472112
472112
add a comment |
add a comment |
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$begingroup$
Don’t all negative integers have negative cubes? So $z^3<0$ is equivalent to $z<0$, right?
$endgroup$
– MPW
Jan 24 at 2:30