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 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 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 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 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 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 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 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 at 21:08
Doesbaddel
19910
asked Nov 18 at 21:08
Doesbaddel
19910
edited Nov 18 at 21:11
asked Nov 18 at 21:08
Doesbaddel
19910
asked Nov 18 at 21:08
Doesbaddel
19910
asked Nov 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 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 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 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 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 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 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 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 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 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 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 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 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 18 at 21:12
@EeveeTrainer edited it.
– Doesbaddel
Nov 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 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 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 at 21:14
Eevee Trainer
3,370225
Thanks, everything clear now!
– Doesbaddel
Nov 18 at 21:15
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004121%2fis-the-given-function-injective-surjective%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 at 21:14
Eevee Trainer
3,370225
Thanks, everything clear now!
– Doesbaddel
Nov 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 at 21:14
Eevee Trainer
3,370225
Thanks, everything clear now!
– Doesbaddel
Nov 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 at 21:14
Eevee Trainer
3,370225
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 at 21:14
Eevee Trainer
3,370225
edited Nov 28 at 10:29
answered Nov 18 at 21:14
Eevee Trainer
3,370225
answered Nov 18 at 21:14
Eevee Trainer
3,370225
answered Nov 18 at 21:14
Eevee Trainer
3,370225
3,370225
Thanks, everything clear now!
– Doesbaddel
Nov 18 at 21:15
add a comment |
Thanks, everything clear now!
– Doesbaddel
Nov 18 at 21:15
Thanks, everything clear now!
– Doesbaddel
Nov 18 at 21:15
Thanks, everything clear now!
– Doesbaddel
Nov 18 at 21:15
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004121%2fis-the-given-function-injective-surjective%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 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 at 21:10
Sorry, I assumed that $mathbb{N}={0,1,2,3cdots}$
– Doesbaddel
Nov 18 at 21:10
@EeveeTrainer edited it.
– Doesbaddel
Nov 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 at 21:13