Let $ A ={ninmathbb{Z}: 2 | n}$ and $B={ninmathbb{Z}: 4 | n}$. Prove that $nin (A - B)$ if and only if $n=2k$...
$begingroup$
Let $A = {ninmathbb{Z}: 2, |, n}$ and B={$ninmathbb{Z}: 4 , | , n$}. Prove that $nin(A - B)$ if and only if $n=2k$ for some odd integer $k$.
I'm not sure how to prove this 'correctly'. Any help would be greatly appreciated, thanks!
discrete-mathematics elementary-set-theory proof-writing
edited Dec 6 '18 at 21:41
user1101010
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
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add a comment |
$begingroup$
Let $A = {ninmathbb{Z}: 2, |, n}$ and B={$ninmathbb{Z}: 4 , | , n$}. Prove that $nin(A - B)$ if and only if $n=2k$ for some odd integer $k$.
I'm not sure how to prove this 'correctly'. Any help would be greatly appreciated, thanks!
discrete-mathematics elementary-set-theory proof-writing
edited Dec 6 '18 at 21:41
user1101010
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
$endgroup$
add a comment |
$begingroup$
Let $A = {ninmathbb{Z}: 2, |, n}$ and B={$ninmathbb{Z}: 4 , | , n$}. Prove that $nin(A - B)$ if and only if $n=2k$ for some odd integer $k$.
I'm not sure how to prove this 'correctly'. Any help would be greatly appreciated, thanks!
discrete-mathematics elementary-set-theory proof-writing
edited Dec 6 '18 at 21:41
user1101010
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
$endgroup$
Let $A = {ninmathbb{Z}: 2, |, n}$ and B={$ninmathbb{Z}: 4 , | , n$}. Prove that $nin(A - B)$ if and only if $n=2k$ for some odd integer $k$.
I'm not sure how to prove this 'correctly'. Any help would be greatly appreciated, thanks!
discrete-mathematics elementary-set-theory proof-writing
discrete-mathematics elementary-set-theory proof-writing
edited Dec 6 '18 at 21:41
user1101010
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
edited Dec 6 '18 at 21:41
user1101010
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
edited Dec 6 '18 at 21:41
user1101010
7391730
edited Dec 6 '18 at 21:41
user1101010
7391730
edited Dec 6 '18 at 21:41
user1101010
7391730
7391730
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
asked Dec 6 '18 at 21:34
hunnybunshunnybuns
15
15
add a comment |
add a comment |
2 Answers
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$begingroup$
Suppose $nin A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.
We proceed to prove that $k$ is odd by contradiction.
Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $nnotin B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.
Putting this together, $n = 2k$ for an odd integer $k$, as required.
answered Dec 6 '18 at 21:42
ODFODF
1,486510
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$begingroup$
Say $nin A-B$. Clearly, $A-B={...,-6,-2,2,6,...}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $minmathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $ forall minmathbb Z$.
Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, ainmathbb Zimplies n=2(2a+1)=4a+2implies2|n, 4nmid nimplies nin A-B$
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
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2 Answers
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2 Answers
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$begingroup$
Suppose $nin A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.
We proceed to prove that $k$ is odd by contradiction.
Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $nnotin B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.
Putting this together, $n = 2k$ for an odd integer $k$, as required.
answered Dec 6 '18 at 21:42
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Suppose $nin A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.
We proceed to prove that $k$ is odd by contradiction.
Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $nnotin B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.
Putting this together, $n = 2k$ for an odd integer $k$, as required.
answered Dec 6 '18 at 21:42
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Suppose $nin A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.
We proceed to prove that $k$ is odd by contradiction.
Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $nnotin B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.
Putting this together, $n = 2k$ for an odd integer $k$, as required.
answered Dec 6 '18 at 21:42
ODFODF
1,486510
$endgroup$
Suppose $nin A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.
We proceed to prove that $k$ is odd by contradiction.
Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $nnotin B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.
Putting this together, $n = 2k$ for an odd integer $k$, as required.
answered Dec 6 '18 at 21:42
ODFODF
1,486510
answered Dec 6 '18 at 21:42
ODFODF
1,486510
answered Dec 6 '18 at 21:42
ODFODF
1,486510
answered Dec 6 '18 at 21:42
ODFODF
1,486510
1,486510
add a comment |
add a comment |
$begingroup$
Say $nin A-B$. Clearly, $A-B={...,-6,-2,2,6,...}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $minmathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $ forall minmathbb Z$.
Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, ainmathbb Zimplies n=2(2a+1)=4a+2implies2|n, 4nmid nimplies nin A-B$
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
Say $nin A-B$. Clearly, $A-B={...,-6,-2,2,6,...}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $minmathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $ forall minmathbb Z$.
Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, ainmathbb Zimplies n=2(2a+1)=4a+2implies2|n, 4nmid nimplies nin A-B$
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
Say $nin A-B$. Clearly, $A-B={...,-6,-2,2,6,...}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $minmathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $ forall minmathbb Z$.
Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, ainmathbb Zimplies n=2(2a+1)=4a+2implies2|n, 4nmid nimplies nin A-B$
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
$endgroup$
Say $nin A-B$. Clearly, $A-B={...,-6,-2,2,6,...}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $minmathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $ forall minmathbb Z$.
Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, ainmathbb Zimplies n=2(2a+1)=4a+2implies2|n, 4nmid nimplies nin A-B$
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
edited Dec 6 '18 at 21:45
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
answered Dec 6 '18 at 21:40
Shubham JohriShubham Johri
5,172717
5,172717
add a comment |
add a comment |
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