Is the given function injective, surjective?
The following function is obviously injective, surjective and
bijective, but I'm not really sure how to prove this. N = {0,1,2,...}
$f: mathbb{N} times mathbb{N} to mathbb{N}$, $f(x,y)=x+y$
functions discrete-mathematics
asked Nov 18 '18 at 21:08
Doesbaddel
19910
|
show 1 more comment
The following function is obviously injective, surjective and
bijective, but I'm not really sure how to prove this. N = {0,1,2,...}
$f: mathbb{N} times mathbb{N} to mathbb{N}$, $f(x,y)=x+y$
functions discrete-mathematics
asked Nov 18 '18 at 21:08
Doesbaddel
19910
1
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
1
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
1
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13
|
show 1 more comment
The following function is obviously injective, surjective and
bijective, but I'm not really sure how to prove this. N = {0,1,2,...}
$f: mathbb{N} times mathbb{N} to mathbb{N}$, $f(x,y)=x+y$
functions discrete-mathematics
asked Nov 18 '18 at 21:08
Doesbaddel
19910
The following function is obviously injective, surjective and
bijective, but I'm not really sure how to prove this. N = {0,1,2,...}
$f: mathbb{N} times mathbb{N} to mathbb{N}$, $f(x,y)=x+y$
functions discrete-mathematics
functions discrete-mathematics
asked Nov 18 '18 at 21:08
Doesbaddel
19910
asked Nov 18 '18 at 21:08
Doesbaddel
19910
edited Nov 18 '18 at 21:11
asked Nov 18 '18 at 21:08
Doesbaddel
19910
asked Nov 18 '18 at 21:08
Doesbaddel
19910
asked Nov 18 '18 at 21:08
Doesbaddel
19910
19910
1
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
1
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
1
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13
|
show 1 more comment
1
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
1
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
1
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13
1
1
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
1
1
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
1
1
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13
|
show 1 more comment
1 Answer
1
active
oldest
votes
In this case, the function is clearly surjective, yes. Note that for any $n in mathbb{N}$ you have $(0,n) in mathbb{N} times mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)neq(0,3)$, thus not injective (and in turn, not bijective).
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
add a comment |
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1 Answer
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In this case, the function is clearly surjective, yes. Note that for any $n in mathbb{N}$ you have $(0,n) in mathbb{N} times mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)neq(0,3)$, thus not injective (and in turn, not bijective).
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
add a comment |
In this case, the function is clearly surjective, yes. Note that for any $n in mathbb{N}$ you have $(0,n) in mathbb{N} times mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)neq(0,3)$, thus not injective (and in turn, not bijective).
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
add a comment |
In this case, the function is clearly surjective, yes. Note that for any $n in mathbb{N}$ you have $(0,n) in mathbb{N} times mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)neq(0,3)$, thus not injective (and in turn, not bijective).
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
In this case, the function is clearly surjective, yes. Note that for any $n in mathbb{N}$ you have $(0,n) in mathbb{N} times mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.
However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)neq(0,3)$, thus not injective (and in turn, not bijective).
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
edited Nov 28 '18 at 10:29
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
answered Nov 18 '18 at 21:14
Eevee Trainer
4,9971734
4,9971734
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
add a comment |
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
Thanks, everything clear now!
– Doesbaddel
Nov 18 '18 at 21:15
add a comment |
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1
Are you sure that "obvious" is the right word to use? It's not injective, nor is it surjective if $mathbb{N} = {1, 2, 3, …}$.
– T. Bongers
Nov 18 '18 at 21:09
1
Specifying the codomain of the function would be important to determine if the function is truly surjective as well.
– Eevee Trainer
Nov 18 '18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 '18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 '18 at 21:12
1
Ok, so surjectivity is now ok (and you should be able to write a very specific $f(**, **) = n$ for any given $n in mathbb{N}$. But injectivity still isn't ok.
– T. Bongers
Nov 18 '18 at 21:13