Mixing
Discrete Math - Question on Sets
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Been trying for a while to figure out how to solve this question.
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, . . . }.
discrete-mathematics
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$begingroup$
Been trying for a while to figure out how to solve this question.
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, . . . }.
discrete-mathematics
$endgroup$
add a comment |
$begingroup$
Been trying for a while to figure out how to solve this question.
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, . . . }.
discrete-mathematics
$endgroup$
Been trying for a while to figure out how to solve this question.
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, . . . }.
discrete-mathematics
discrete-mathematics
asked Jan 27 at 2:15
user638615
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$begingroup$
Take an arbitrary element $x in A$. Then, by the definition of set $A$, we can
$$x = 4j - 3$$
for some $j in mathbb{N}$. Moreover, since $4j - 3 geq 1$ and $x equiv 1 pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.
Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A subseteq B$. So we're done.
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
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1 Answer
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1 Answer
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$begingroup$
Take an arbitrary element $x in A$. Then, by the definition of set $A$, we can
$$x = 4j - 3$$
for some $j in mathbb{N}$. Moreover, since $4j - 3 geq 1$ and $x equiv 1 pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.
Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A subseteq B$. So we're done.
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
$endgroup$
add a comment |
$begingroup$
Take an arbitrary element $x in A$. Then, by the definition of set $A$, we can
$$x = 4j - 3$$
for some $j in mathbb{N}$. Moreover, since $4j - 3 geq 1$ and $x equiv 1 pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.
Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A subseteq B$. So we're done.
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
$endgroup$
add a comment |
$begingroup$
Take an arbitrary element $x in A$. Then, by the definition of set $A$, we can
$$x = 4j - 3$$
for some $j in mathbb{N}$. Moreover, since $4j - 3 geq 1$ and $x equiv 1 pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.
Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A subseteq B$. So we're done.
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
$endgroup$
Take an arbitrary element $x in A$. Then, by the definition of set $A$, we can
$$x = 4j - 3$$
for some $j in mathbb{N}$. Moreover, since $4j - 3 geq 1$ and $x equiv 1 pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.
Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A subseteq B$. So we're done.
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
answered Jan 27 at 2:18
Ekesh KumarEkesh Kumar
1,07428
1,07428
add a comment |
add a comment |
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