Making a Bijective Function
0
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Q: $N$ and $N$ × {a, b}. So the Cartesian product of $N$ × {a, b} is {(x,y)| $1<=x<=a$, $1<=y<=b$}. I was trying to map the values so I was thinking of doing $1->a_1$, $2->b_1$, $3->a_2$, $4->b_2$ etc. So I was thinking the function can be two pieces, f(n)={n is even then maps to $a_n$ and n is odd then maps to $b_n$
elementary-set-theory
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
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add a comment |
0
$begingroup$
Q: $N$ and $N$ × {a, b}. So the Cartesian product of $N$ × {a, b} is {(x,y)| $1<=x<=a$, $1<=y<=b$}. I was trying to map the values so I was thinking of doing $1->a_1$, $2->b_1$, $3->a_2$, $4->b_2$ etc. So I was thinking the function can be two pieces, f(n)={n is even then maps to $a_n$ and n is odd then maps to $b_n$
elementary-set-theory
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
$endgroup$
$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
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As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31
add a comment |
0
0
0
$begingroup$
Q: $N$ and $N$ × {a, b}. So the Cartesian product of $N$ × {a, b} is {(x,y)| $1<=x<=a$, $1<=y<=b$}. I was trying to map the values so I was thinking of doing $1->a_1$, $2->b_1$, $3->a_2$, $4->b_2$ etc. So I was thinking the function can be two pieces, f(n)={n is even then maps to $a_n$ and n is odd then maps to $b_n$
elementary-set-theory
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
$endgroup$
Q: $N$ and $N$ × {a, b}. So the Cartesian product of $N$ × {a, b} is {(x,y)| $1<=x<=a$, $1<=y<=b$}. I was trying to map the values so I was thinking of doing $1->a_1$, $2->b_1$, $3->a_2$, $4->b_2$ etc. So I was thinking the function can be two pieces, f(n)={n is even then maps to $a_n$ and n is odd then maps to $b_n$
elementary-set-theory
elementary-set-theory
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 1:09
Marcus DillionsMarcus Dillions
63
63
$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
$begingroup$
As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31
add a comment |
$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
$begingroup$
As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31
$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
$begingroup$
As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31
add a comment |
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$begingroup$
Uhm... ${a,b}$ is a two element set which contains the elements $a$ and $b$. $Bbb Ntimes {a,b}$ is the set ${(x,y)~|~xinBbb N,~y=a~text{or}~y=b}$. The set you describe sounds more like $[a]times [b]$ where $[a]$ is intended to mean the $a$-element set beginning from $1$, namely $[a]={1,2,3,dots,a}$
$endgroup$
– JMoravitz
Dec 5 '18 at 1:19
$begingroup$
As for a bijection between these... map the $n$'th odd number to $(n,a)$ and map the $n$'th even number to $(n,b)$.
$endgroup$
– JMoravitz
Dec 5 '18 at 1:20
$begingroup$
The idea is correct. So here I believe you write $(n,a)$ using notation $a_n$. You just need to be careful with the indices. As you can see from the examples that you gave $f(3)=a_2$ so $f(n)neq n$. You should write $f(n)=a_{(n+1)/2)}$ for $n$ odd and $f(n)=b_{n/2}$ for $n$ even.
$endgroup$
– user9077
Dec 5 '18 at 1:24
$begingroup$
Why are you putting the values over 2 and adding one to n. So if you had 2 it would map to b(1/2). I understand the odd since 1 is a(2/2)=a1 but I don't understand the b
$endgroup$
– Marcus Dillions
Dec 5 '18 at 1:31