Mixing
Show that $F(x,y)=2^x(2y+1)$ is onto from $mathbb{Z}^+ cup {0} times mathbb{Z}^+ cup {0}$ to $mathbb{Z}^+$.
Here is the problem:
Let $C$ be the set of nonnegative integers. Let $A=Ctimes C$ and $B=C$.
Let $F(x,y)=2^x(2y+1)$ for all $x$ and $y$ in C.
How do I show whether $F$ is a function from $A$ onto $B$?
functions
edited Nov 21 '18 at 12:23
amWhy
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
add a comment |
Here is the problem:
Let $C$ be the set of nonnegative integers. Let $A=Ctimes C$ and $B=C$.
Let $F(x,y)=2^x(2y+1)$ for all $x$ and $y$ in C.
How do I show whether $F$ is a function from $A$ onto $B$?
functions
edited Nov 21 '18 at 12:23
amWhy
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
2
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37
add a comment |
Here is the problem:
Let $C$ be the set of nonnegative integers. Let $A=Ctimes C$ and $B=C$.
Let $F(x,y)=2^x(2y+1)$ for all $x$ and $y$ in C.
How do I show whether $F$ is a function from $A$ onto $B$?
functions
edited Nov 21 '18 at 12:23
amWhy
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
Here is the problem:
Let $C$ be the set of nonnegative integers. Let $A=Ctimes C$ and $B=C$.
Let $F(x,y)=2^x(2y+1)$ for all $x$ and $y$ in C.
How do I show whether $F$ is a function from $A$ onto $B$?
functions
functions
edited Nov 21 '18 at 12:23
amWhy
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
edited Nov 21 '18 at 12:23
amWhy
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
edited Nov 21 '18 at 12:23
amWhy
192k28224439
edited Nov 21 '18 at 12:23
amWhy
192k28224439
edited Nov 21 '18 at 12:23
amWhy
192k28224439
192k28224439
asked Nov 21 '18 at 11:32
mrnobody
14818
asked Nov 21 '18 at 11:32
mrnobody
14818
asked Nov 21 '18 at 11:32
mrnobody
14818
14818
By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
2
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37
add a comment |
By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
2
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37
By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
2
2
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37
add a comment |
1 Answer
1
active
oldest
votes
The function is not onto because it can't take value zero. However if you take $B=C- { 0 }$, then it is onto. Simply observe that, by putting $x=0$, you can get all odd positive integers. For any even positive integer, divide it by maximum possible power of 2. Then you'll left with odd integer. Choose your $y$ accordingly. For example, if you want to get 12, $frac{12}{2^2}=3.$ So, take $x= 2$ and $y= 1$.
answered Nov 21 '18 at 12:07
ramanujan
714713
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
The function is not onto because it can't take value zero. However if you take $B=C- { 0 }$, then it is onto. Simply observe that, by putting $x=0$, you can get all odd positive integers. For any even positive integer, divide it by maximum possible power of 2. Then you'll left with odd integer. Choose your $y$ accordingly. For example, if you want to get 12, $frac{12}{2^2}=3.$ So, take $x= 2$ and $y= 1$.
answered Nov 21 '18 at 12:07
ramanujan
714713
add a comment |
The function is not onto because it can't take value zero. However if you take $B=C- { 0 }$, then it is onto. Simply observe that, by putting $x=0$, you can get all odd positive integers. For any even positive integer, divide it by maximum possible power of 2. Then you'll left with odd integer. Choose your $y$ accordingly. For example, if you want to get 12, $frac{12}{2^2}=3.$ So, take $x= 2$ and $y= 1$.
answered Nov 21 '18 at 12:07
ramanujan
714713
add a comment |
The function is not onto because it can't take value zero. However if you take $B=C- { 0 }$, then it is onto. Simply observe that, by putting $x=0$, you can get all odd positive integers. For any even positive integer, divide it by maximum possible power of 2. Then you'll left with odd integer. Choose your $y$ accordingly. For example, if you want to get 12, $frac{12}{2^2}=3.$ So, take $x= 2$ and $y= 1$.
answered Nov 21 '18 at 12:07
ramanujan
714713
The function is not onto because it can't take value zero. However if you take $B=C- { 0 }$, then it is onto. Simply observe that, by putting $x=0$, you can get all odd positive integers. For any even positive integer, divide it by maximum possible power of 2. Then you'll left with odd integer. Choose your $y$ accordingly. For example, if you want to get 12, $frac{12}{2^2}=3.$ So, take $x= 2$ and $y= 1$.
answered Nov 21 '18 at 12:07
ramanujan
714713
edited Nov 21 '18 at 12:31
answered Nov 21 '18 at 12:07
ramanujan
714713
answered Nov 21 '18 at 12:07
ramanujan
714713
answered Nov 21 '18 at 12:07
ramanujan
714713
714713
add a comment |
add a comment |
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By using the definition of "onto" more or less directly.
– Arthur
Nov 21 '18 at 11:34
2
Actually the function is not "onto". $F(x,y)$ cannot take value $0$.
– drhab
Nov 21 '18 at 11:37