Mixing
Find the number(s) $k$ such that $kmathbb{Z} = mathbb{Z}$
$begingroup$
I am just starting out with Discrete Math and there is a question in my book that is
Find the number(s) $k$ such that $k mathbb{Z} = mathbb{Z}$.
The answer is -1 and 1. I understand why it is 1 because any number multiplied by 1 is itself, what I don't understand is why the other answer is -1.
For example, if $k = -1$ then $(-1)mathbb{Z}$ would be all the values of $mathbb{Z}$, but if that is the case, then why can't $k$ could also equal 2 or any other integer?
discrete-mathematics integers
asked Jan 27 at 16:47
SamSam
43718
$endgroup$
add a comment |
$begingroup$
I am just starting out with Discrete Math and there is a question in my book that is
Find the number(s) $k$ such that $k mathbb{Z} = mathbb{Z}$.
The answer is -1 and 1. I understand why it is 1 because any number multiplied by 1 is itself, what I don't understand is why the other answer is -1.
For example, if $k = -1$ then $(-1)mathbb{Z}$ would be all the values of $mathbb{Z}$, but if that is the case, then why can't $k$ could also equal 2 or any other integer?
discrete-mathematics integers
asked Jan 27 at 16:47
SamSam
43718
$endgroup$
add a comment |
$begingroup$
I am just starting out with Discrete Math and there is a question in my book that is
Find the number(s) $k$ such that $k mathbb{Z} = mathbb{Z}$.
The answer is -1 and 1. I understand why it is 1 because any number multiplied by 1 is itself, what I don't understand is why the other answer is -1.
For example, if $k = -1$ then $(-1)mathbb{Z}$ would be all the values of $mathbb{Z}$, but if that is the case, then why can't $k$ could also equal 2 or any other integer?
discrete-mathematics integers
asked Jan 27 at 16:47
SamSam
43718
$endgroup$
I am just starting out with Discrete Math and there is a question in my book that is
Find the number(s) $k$ such that $k mathbb{Z} = mathbb{Z}$.
The answer is -1 and 1. I understand why it is 1 because any number multiplied by 1 is itself, what I don't understand is why the other answer is -1.
For example, if $k = -1$ then $(-1)mathbb{Z}$ would be all the values of $mathbb{Z}$, but if that is the case, then why can't $k$ could also equal 2 or any other integer?
discrete-mathematics integers
discrete-mathematics integers
asked Jan 27 at 16:47
SamSam
43718
asked Jan 27 at 16:47
SamSam
43718
asked Jan 27 at 16:47
SamSam
43718
asked Jan 27 at 16:47
SamSam
43718
asked Jan 27 at 16:47
SamSam
43718
43718
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)mathbb{Z} = mathbb{Z}$ but unsure whether, for example, $2mathbb{Z} ne mathbb{Z}$. This makes me think that $kmathbb{Z}$ is slightly confusing to you.
So by $kmathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$kmathbb{Z} = { kn mid n in mathbb{Z} } = {dots,-2k,-k,0,k,2k,dots}$$
Thus,
$$1mathbb{Z} = { 1n mid n in mathbb{Z} } = {dots,-2,-1,0,1,2,dots} = mathbb{Z} tag 1$$
$$(-1)mathbb{Z} = { (-1)n mid n in mathbb{Z} } = {dots,2,1,0,-1,-2,dots} = mathbb{Z} tag 2$$
$$2mathbb{Z} = { 2n mid n in mathbb{Z} } = {dots,-4,-2,0,2,4,dots} ne mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer.
answered Jan 27 at 17:41
MetricMetric
1,22659
$endgroup$
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
|
show 1 more comment
$begingroup$
If $kmathbb{Z} = mathbb{Z}$ then $1in kmathbb{Z}$ (in fact this is an if and only if) but then $frac{1}{k}inmathbb{Z}$. This means that $kin{-1,1}$.
answered Jan 27 at 16:49
YankoYanko
8,0832830
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)mathbb{Z} = mathbb{Z}$ but unsure whether, for example, $2mathbb{Z} ne mathbb{Z}$. This makes me think that $kmathbb{Z}$ is slightly confusing to you.
So by $kmathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$kmathbb{Z} = { kn mid n in mathbb{Z} } = {dots,-2k,-k,0,k,2k,dots}$$
Thus,
$$1mathbb{Z} = { 1n mid n in mathbb{Z} } = {dots,-2,-1,0,1,2,dots} = mathbb{Z} tag 1$$
$$(-1)mathbb{Z} = { (-1)n mid n in mathbb{Z} } = {dots,2,1,0,-1,-2,dots} = mathbb{Z} tag 2$$
$$2mathbb{Z} = { 2n mid n in mathbb{Z} } = {dots,-4,-2,0,2,4,dots} ne mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer.
answered Jan 27 at 17:41
MetricMetric
1,22659
$endgroup$
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
|
show 1 more comment
$begingroup$
Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)mathbb{Z} = mathbb{Z}$ but unsure whether, for example, $2mathbb{Z} ne mathbb{Z}$. This makes me think that $kmathbb{Z}$ is slightly confusing to you.
So by $kmathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$kmathbb{Z} = { kn mid n in mathbb{Z} } = {dots,-2k,-k,0,k,2k,dots}$$
Thus,
$$1mathbb{Z} = { 1n mid n in mathbb{Z} } = {dots,-2,-1,0,1,2,dots} = mathbb{Z} tag 1$$
$$(-1)mathbb{Z} = { (-1)n mid n in mathbb{Z} } = {dots,2,1,0,-1,-2,dots} = mathbb{Z} tag 2$$
$$2mathbb{Z} = { 2n mid n in mathbb{Z} } = {dots,-4,-2,0,2,4,dots} ne mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer.
answered Jan 27 at 17:41
MetricMetric
1,22659
$endgroup$
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
|
show 1 more comment
$begingroup$
Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)mathbb{Z} = mathbb{Z}$ but unsure whether, for example, $2mathbb{Z} ne mathbb{Z}$. This makes me think that $kmathbb{Z}$ is slightly confusing to you.
So by $kmathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$kmathbb{Z} = { kn mid n in mathbb{Z} } = {dots,-2k,-k,0,k,2k,dots}$$
Thus,
$$1mathbb{Z} = { 1n mid n in mathbb{Z} } = {dots,-2,-1,0,1,2,dots} = mathbb{Z} tag 1$$
$$(-1)mathbb{Z} = { (-1)n mid n in mathbb{Z} } = {dots,2,1,0,-1,-2,dots} = mathbb{Z} tag 2$$
$$2mathbb{Z} = { 2n mid n in mathbb{Z} } = {dots,-4,-2,0,2,4,dots} ne mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer.
answered Jan 27 at 17:41
MetricMetric
1,22659
$endgroup$
Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)mathbb{Z} = mathbb{Z}$ but unsure whether, for example, $2mathbb{Z} ne mathbb{Z}$. This makes me think that $kmathbb{Z}$ is slightly confusing to you.
So by $kmathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$kmathbb{Z} = { kn mid n in mathbb{Z} } = {dots,-2k,-k,0,k,2k,dots}$$
Thus,
$$1mathbb{Z} = { 1n mid n in mathbb{Z} } = {dots,-2,-1,0,1,2,dots} = mathbb{Z} tag 1$$
$$(-1)mathbb{Z} = { (-1)n mid n in mathbb{Z} } = {dots,2,1,0,-1,-2,dots} = mathbb{Z} tag 2$$
$$2mathbb{Z} = { 2n mid n in mathbb{Z} } = {dots,-4,-2,0,2,4,dots} ne mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer.
answered Jan 27 at 17:41
MetricMetric
1,22659
answered Jan 27 at 17:41
MetricMetric
1,22659
answered Jan 27 at 17:41
MetricMetric
1,22659
answered Jan 27 at 17:41
MetricMetric
1,22659
1,22659
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
|
show 1 more comment
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Is $2mathbb{Z}$ is invalid because numbers get skipped in the set that exist in $mathbb{Z}$? If that is the case then is $7mathbb{Q}= mathbb{Q}$ valid?
$endgroup$
– Sam
Jan 27 at 18:08
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Yes, $2mathbb{Z}$ is just the even integers, meaning that it doesn't contain any odd integers! For the second question, what do you think? Have you tried enumerating some example elements yet?
$endgroup$
– Metric
Jan 27 at 18:11
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Any rational number multiplied by 7 still returns a rational number, but I am not sure if skips any numbers as with $2mathbb{Z}$
$endgroup$
– Sam
Jan 27 at 18:14
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
$begingroup$
Right, so $7mathbb{Q} subseteq mathbb{Q}$. Is it possible to prove the other direction, that is, $mathbb{Q} subseteq 7mathbb{Q}$?
$endgroup$
– Metric
Jan 27 at 18:18
1
1
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
$begingroup$
To help a little bit more, can you prove formally that $(-1)mathbb{Z} = mathbb{Z}$ by showing that they are subsets of each other? I think you came to conclusion of $(-1)mathbb{Z} = mathbb{Z}$ by having an intuitive feel for what each set look like, but proving it formally (i.e., showing that each set is a subset of the other) will help you more in the long run, especially with this problem concerning the rationals.
$endgroup$
– Metric
Jan 27 at 18:23
|
show 1 more comment
$begingroup$
If $kmathbb{Z} = mathbb{Z}$ then $1in kmathbb{Z}$ (in fact this is an if and only if) but then $frac{1}{k}inmathbb{Z}$. This means that $kin{-1,1}$.
answered Jan 27 at 16:49
YankoYanko
8,0832830
$endgroup$
add a comment |
$begingroup$
If $kmathbb{Z} = mathbb{Z}$ then $1in kmathbb{Z}$ (in fact this is an if and only if) but then $frac{1}{k}inmathbb{Z}$. This means that $kin{-1,1}$.
answered Jan 27 at 16:49
YankoYanko
8,0832830
$endgroup$
add a comment |
$begingroup$
If $kmathbb{Z} = mathbb{Z}$ then $1in kmathbb{Z}$ (in fact this is an if and only if) but then $frac{1}{k}inmathbb{Z}$. This means that $kin{-1,1}$.
answered Jan 27 at 16:49
YankoYanko
8,0832830
$endgroup$
If $kmathbb{Z} = mathbb{Z}$ then $1in kmathbb{Z}$ (in fact this is an if and only if) but then $frac{1}{k}inmathbb{Z}$. This means that $kin{-1,1}$.
answered Jan 27 at 16:49
YankoYanko
8,0832830
answered Jan 27 at 16:49
YankoYanko
8,0832830
answered Jan 27 at 16:49
YankoYanko
8,0832830
answered Jan 27 at 16:49
YankoYanko
8,0832830
8,0832830
add a comment |
add a comment |
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