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Prove that Z(integers) and A = a = 4r + 2 for some r ∈Z have the same cardinality
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I'm having trouble coming up with a proof.
I know that to how an equal cardinality I must show each of the sets has the same numbers of elements (in this case infinite integers). I have begun my proof as follows:
$ a = 4r + 2$ for some $r in Z$
Because $a$ is an integer than $a$ means that some integer is equal to $4r+2$.
Because an integer can be positive or negative, we must show that $a$ is either $2k$ or $2k+1$
Case 1: Show $a$ is even
$a = 4r+2$
$a = 4(2k)+2$
$a = 8k+2$
$a = 2(4k+1)$
$a = 2(n)$
Case 2: show that $a$ is odd
$a = 4r+2$
$a = 4(2k+1)+2$
$a = 8k+6$
$a = 2(4k+2)+1$
$a = 2(n)+1$
Because we have proven $a$ is an even or odd integer then some $a$ can be represented by some even or odd integer. Thus for whatever $a$, there is an integer $r$ that can represent it and thus they have the same cardinality.
Is my proof proof correct and if not could you explain the correct process I should be taking?
discrete-mathematics elementary-set-theory
edited Aug 3 at 15:23
Jneven
440218
asked Aug 3 at 12:50
Jlee
342
add a comment |Â
up vote
1
down vote
favorite
I'm having trouble coming up with a proof.
I know that to how an equal cardinality I must show each of the sets has the same numbers of elements (in this case infinite integers). I have begun my proof as follows:
$ a = 4r + 2$ for some $r in Z$
Because $a$ is an integer than $a$ means that some integer is equal to $4r+2$.
Because an integer can be positive or negative, we must show that $a$ is either $2k$ or $2k+1$
Case 1: Show $a$ is even
$a = 4r+2$
$a = 4(2k)+2$
$a = 8k+2$
$a = 2(4k+1)$
$a = 2(n)$
Case 2: show that $a$ is odd
$a = 4r+2$
$a = 4(2k+1)+2$
$a = 8k+6$
$a = 2(4k+2)+1$
$a = 2(n)+1$
Because we have proven $a$ is an even or odd integer then some $a$ can be represented by some even or odd integer. Thus for whatever $a$, there is an integer $r$ that can represent it and thus they have the same cardinality.
Is my proof proof correct and if not could you explain the correct process I should be taking?
discrete-mathematics elementary-set-theory
edited Aug 3 at 15:23
Jneven
440218
asked Aug 3 at 12:50
Jlee
342
2
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having trouble coming up with a proof.
I know that to how an equal cardinality I must show each of the sets has the same numbers of elements (in this case infinite integers). I have begun my proof as follows:
$ a = 4r + 2$ for some $r in Z$
Because $a$ is an integer than $a$ means that some integer is equal to $4r+2$.
Because an integer can be positive or negative, we must show that $a$ is either $2k$ or $2k+1$
Case 1: Show $a$ is even
$a = 4r+2$
$a = 4(2k)+2$
$a = 8k+2$
$a = 2(4k+1)$
$a = 2(n)$
Case 2: show that $a$ is odd
$a = 4r+2$
$a = 4(2k+1)+2$
$a = 8k+6$
$a = 2(4k+2)+1$
$a = 2(n)+1$
Because we have proven $a$ is an even or odd integer then some $a$ can be represented by some even or odd integer. Thus for whatever $a$, there is an integer $r$ that can represent it and thus they have the same cardinality.
Is my proof proof correct and if not could you explain the correct process I should be taking?
discrete-mathematics elementary-set-theory
edited Aug 3 at 15:23
Jneven
440218
asked Aug 3 at 12:50
Jlee
342
I'm having trouble coming up with a proof.
I know that to how an equal cardinality I must show each of the sets has the same numbers of elements (in this case infinite integers). I have begun my proof as follows:
$ a = 4r + 2$ for some $r in Z$
Because $a$ is an integer than $a$ means that some integer is equal to $4r+2$.
Because an integer can be positive or negative, we must show that $a$ is either $2k$ or $2k+1$
Case 1: Show $a$ is even
$a = 4r+2$
$a = 4(2k)+2$
$a = 8k+2$
$a = 2(4k+1)$
$a = 2(n)$
Case 2: show that $a$ is odd
$a = 4r+2$
$a = 4(2k+1)+2$
$a = 8k+6$
$a = 2(4k+2)+1$
$a = 2(n)+1$
Because we have proven $a$ is an even or odd integer then some $a$ can be represented by some even or odd integer. Thus for whatever $a$, there is an integer $r$ that can represent it and thus they have the same cardinality.
Is my proof proof correct and if not could you explain the correct process I should be taking?
discrete-mathematics elementary-set-theory
edited Aug 3 at 15:23
Jneven
440218
asked Aug 3 at 12:50
Jlee
342
edited Aug 3 at 15:23
Jneven
440218
edited Aug 3 at 15:23
Jneven
440218
edited Aug 3 at 15:23
Jneven
440218
440218
asked Aug 3 at 12:50
Jlee
342
asked Aug 3 at 12:50
Jlee
342
asked Aug 3 at 12:50
Jlee
342
342
2
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44
add a comment |Â
2
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44
2
2
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44
add a comment |Â
1 Answer
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We say that $A$ and $B$ have the same cardinality if there exists a bijection between them, that is, a function $f : A to B$ such that:
- (Injective) Whenever $f(x) = f(y)$ for $x,y$ in $A$ we have $x=y$, and
- (Surjective) For any $y$ in $B$ there is an $x$ in $A$ with $f(x)=y$.
So in this case we'll take $B$ to be $mathbbZ$ (the integers), and $A$ the set $a in mathbbZ : a=4r+2 text for some r in mathbbZ $.
Let's consider the function $f(x) = 4x+2$.
Firstly, if $r$ is in $mathbbZ$, then $f(r) = 4r+2$ for an integer $r$, so $f(r) in A$. This tells us that $f$ is indeed a function from $mathbbZ$ to $A$. Now we want to show that it is a bijection.
- Suppose we have $f(x) = f(y)$ for integers $x$ and $y$. This means that $4x+2 = 4y+2$. Can you see how this implies that $x=y$?
- Let's pick any element $a$ in $A$. By the definition of $A$ we know that there is some integer $r$ with $a=4r+2$. So $f(r)=a$.
And now we've shown that $f$ is a bijection between $mathbbZ$ and $A$, therefore the two sets have the same cardinality.
answered Aug 3 at 15:08
Daniel Mroz
851314
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We say that $A$ and $B$ have the same cardinality if there exists a bijection between them, that is, a function $f : A to B$ such that:
- (Injective) Whenever $f(x) = f(y)$ for $x,y$ in $A$ we have $x=y$, and
- (Surjective) For any $y$ in $B$ there is an $x$ in $A$ with $f(x)=y$.
So in this case we'll take $B$ to be $mathbbZ$ (the integers), and $A$ the set $a in mathbbZ : a=4r+2 text for some r in mathbbZ $.
Let's consider the function $f(x) = 4x+2$.
Firstly, if $r$ is in $mathbbZ$, then $f(r) = 4r+2$ for an integer $r$, so $f(r) in A$. This tells us that $f$ is indeed a function from $mathbbZ$ to $A$. Now we want to show that it is a bijection.
- Suppose we have $f(x) = f(y)$ for integers $x$ and $y$. This means that $4x+2 = 4y+2$. Can you see how this implies that $x=y$?
- Let's pick any element $a$ in $A$. By the definition of $A$ we know that there is some integer $r$ with $a=4r+2$. So $f(r)=a$.
And now we've shown that $f$ is a bijection between $mathbbZ$ and $A$, therefore the two sets have the same cardinality.
answered Aug 3 at 15:08
Daniel Mroz
851314
add a comment |Â
up vote
0
down vote
We say that $A$ and $B$ have the same cardinality if there exists a bijection between them, that is, a function $f : A to B$ such that:
- (Injective) Whenever $f(x) = f(y)$ for $x,y$ in $A$ we have $x=y$, and
- (Surjective) For any $y$ in $B$ there is an $x$ in $A$ with $f(x)=y$.
So in this case we'll take $B$ to be $mathbbZ$ (the integers), and $A$ the set $a in mathbbZ : a=4r+2 text for some r in mathbbZ $.
Let's consider the function $f(x) = 4x+2$.
Firstly, if $r$ is in $mathbbZ$, then $f(r) = 4r+2$ for an integer $r$, so $f(r) in A$. This tells us that $f$ is indeed a function from $mathbbZ$ to $A$. Now we want to show that it is a bijection.
- Suppose we have $f(x) = f(y)$ for integers $x$ and $y$. This means that $4x+2 = 4y+2$. Can you see how this implies that $x=y$?
- Let's pick any element $a$ in $A$. By the definition of $A$ we know that there is some integer $r$ with $a=4r+2$. So $f(r)=a$.
And now we've shown that $f$ is a bijection between $mathbbZ$ and $A$, therefore the two sets have the same cardinality.
answered Aug 3 at 15:08
Daniel Mroz
851314
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We say that $A$ and $B$ have the same cardinality if there exists a bijection between them, that is, a function $f : A to B$ such that:
- (Injective) Whenever $f(x) = f(y)$ for $x,y$ in $A$ we have $x=y$, and
- (Surjective) For any $y$ in $B$ there is an $x$ in $A$ with $f(x)=y$.
So in this case we'll take $B$ to be $mathbbZ$ (the integers), and $A$ the set $a in mathbbZ : a=4r+2 text for some r in mathbbZ $.
Let's consider the function $f(x) = 4x+2$.
Firstly, if $r$ is in $mathbbZ$, then $f(r) = 4r+2$ for an integer $r$, so $f(r) in A$. This tells us that $f$ is indeed a function from $mathbbZ$ to $A$. Now we want to show that it is a bijection.
- Suppose we have $f(x) = f(y)$ for integers $x$ and $y$. This means that $4x+2 = 4y+2$. Can you see how this implies that $x=y$?
- Let's pick any element $a$ in $A$. By the definition of $A$ we know that there is some integer $r$ with $a=4r+2$. So $f(r)=a$.
And now we've shown that $f$ is a bijection between $mathbbZ$ and $A$, therefore the two sets have the same cardinality.
answered Aug 3 at 15:08
Daniel Mroz
851314
We say that $A$ and $B$ have the same cardinality if there exists a bijection between them, that is, a function $f : A to B$ such that:
- (Injective) Whenever $f(x) = f(y)$ for $x,y$ in $A$ we have $x=y$, and
- (Surjective) For any $y$ in $B$ there is an $x$ in $A$ with $f(x)=y$.
So in this case we'll take $B$ to be $mathbbZ$ (the integers), and $A$ the set $a in mathbbZ : a=4r+2 text for some r in mathbbZ $.
Let's consider the function $f(x) = 4x+2$.
Firstly, if $r$ is in $mathbbZ$, then $f(r) = 4r+2$ for an integer $r$, so $f(r) in A$. This tells us that $f$ is indeed a function from $mathbbZ$ to $A$. Now we want to show that it is a bijection.
- Suppose we have $f(x) = f(y)$ for integers $x$ and $y$. This means that $4x+2 = 4y+2$. Can you see how this implies that $x=y$?
- Let's pick any element $a$ in $A$. By the definition of $A$ we know that there is some integer $r$ with $a=4r+2$. So $f(r)=a$.
And now we've shown that $f$ is a bijection between $mathbbZ$ and $A$, therefore the two sets have the same cardinality.
answered Aug 3 at 15:08
Daniel Mroz
851314
answered Aug 3 at 15:08
Daniel Mroz
851314
answered Aug 3 at 15:08
Daniel Mroz
851314
answered Aug 3 at 15:08
Daniel Mroz
851314
851314
add a comment |Â
add a comment |Â
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2
It's not really clear to me what you mean when you say that "some $a$ can be represented by some even or odd integer". If you want to show that two sets have the same cardinality, the best way is usually to define an explicit function between them and show it is a bijection. In this case, $f(r) = 4r+2$ seems like a good candidate...
– Daniel Mroz
Aug 3 at 12:56
I'm not sure If I follow. What would be the next step to solve for cardinality.
– Jlee
Aug 3 at 14:44