Mixing
Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }
$begingroup$
Let
A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
and
B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}.
Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.
So I'm preparing for an exam in Discrete math, I came up across this question and can't seem to get the answer. I've tried using some of the set identities but I don't know how to get the answer.
discrete-mathematics logic proof-writing
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
$endgroup$
add a comment |
$begingroup$
Let
A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
and
B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}.
Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.
So I'm preparing for an exam in Discrete math, I came up across this question and can't seem to get the answer. I've tried using some of the set identities but I don't know how to get the answer.
discrete-mathematics logic proof-writing
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
$endgroup$
add a comment |
$begingroup$
Let
A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
and
B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}.
Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.
So I'm preparing for an exam in Discrete math, I came up across this question and can't seem to get the answer. I've tried using some of the set identities but I don't know how to get the answer.
discrete-mathematics logic proof-writing
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
$endgroup$
Let
A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
and
B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}.
Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.
So I'm preparing for an exam in Discrete math, I came up across this question and can't seem to get the answer. I've tried using some of the set identities but I don't know how to get the answer.
discrete-mathematics logic proof-writing
discrete-mathematics logic proof-writing
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
asked Jan 27 at 4:36
Usama GhawjiUsama Ghawji
666
666
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Clearly $A$ is non-empty, so take some arbitrary $nin A$. Then $n=4j-3$ for some $j>0$ (convince yourself that if $j=0$ then $n$ could not be in $A$). It follows that:
$$n=4j-3=4(j-1)+1=2(2j-2)+1$$
Since $j>0$, $(2j-2)in N$, so $nin B$. Since $n$ was arbitrary, $Asubseteq B$.
answered Jan 27 at 4:40
Harry QuHarry Qu
986
$endgroup$
add a comment |
$begingroup$
In order to show $Asubseteq B$ we have to show that every element of $A$ is an element of $B$.
Let $n=4j-3$ be an arbitrary element of $A$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $k=2j-2$ then we have $n=2k+1 in B$
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
votes
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votes
$begingroup$
Clearly $A$ is non-empty, so take some arbitrary $nin A$. Then $n=4j-3$ for some $j>0$ (convince yourself that if $j=0$ then $n$ could not be in $A$). It follows that:
$$n=4j-3=4(j-1)+1=2(2j-2)+1$$
Since $j>0$, $(2j-2)in N$, so $nin B$. Since $n$ was arbitrary, $Asubseteq B$.
answered Jan 27 at 4:40
Harry QuHarry Qu
986
$endgroup$
add a comment |
$begingroup$
Clearly $A$ is non-empty, so take some arbitrary $nin A$. Then $n=4j-3$ for some $j>0$ (convince yourself that if $j=0$ then $n$ could not be in $A$). It follows that:
$$n=4j-3=4(j-1)+1=2(2j-2)+1$$
Since $j>0$, $(2j-2)in N$, so $nin B$. Since $n$ was arbitrary, $Asubseteq B$.
answered Jan 27 at 4:40
Harry QuHarry Qu
986
$endgroup$
add a comment |
$begingroup$
Clearly $A$ is non-empty, so take some arbitrary $nin A$. Then $n=4j-3$ for some $j>0$ (convince yourself that if $j=0$ then $n$ could not be in $A$). It follows that:
$$n=4j-3=4(j-1)+1=2(2j-2)+1$$
Since $j>0$, $(2j-2)in N$, so $nin B$. Since $n$ was arbitrary, $Asubseteq B$.
answered Jan 27 at 4:40
Harry QuHarry Qu
986
$endgroup$
Clearly $A$ is non-empty, so take some arbitrary $nin A$. Then $n=4j-3$ for some $j>0$ (convince yourself that if $j=0$ then $n$ could not be in $A$). It follows that:
$$n=4j-3=4(j-1)+1=2(2j-2)+1$$
Since $j>0$, $(2j-2)in N$, so $nin B$. Since $n$ was arbitrary, $Asubseteq B$.
answered Jan 27 at 4:40
Harry QuHarry Qu
986
answered Jan 27 at 4:40
Harry QuHarry Qu
986
answered Jan 27 at 4:40
Harry QuHarry Qu
986
answered Jan 27 at 4:40
Harry QuHarry Qu
986
986
add a comment |
add a comment |
$begingroup$
In order to show $Asubseteq B$ we have to show that every element of $A$ is an element of $B$.
Let $n=4j-3$ be an arbitrary element of $A$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $k=2j-2$ then we have $n=2k+1 in B$
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
In order to show $Asubseteq B$ we have to show that every element of $A$ is an element of $B$.
Let $n=4j-3$ be an arbitrary element of $A$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $k=2j-2$ then we have $n=2k+1 in B$
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
In order to show $Asubseteq B$ we have to show that every element of $A$ is an element of $B$.
Let $n=4j-3$ be an arbitrary element of $A$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $k=2j-2$ then we have $n=2k+1 in B$
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
In order to show $Asubseteq B$ we have to show that every element of $A$ is an element of $B$.
Let $n=4j-3$ be an arbitrary element of $A$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $k=2j-2$ then we have $n=2k+1 in B$
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
edited Jan 28 at 21:28
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 27 at 4:45
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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