Fdtxjr

validation vs test vs training accuracy, which one to compare for claiming overfit?

My story is written in English, but is set in my home country. What language should I use for the dialogue?

Is it true that real estate prices mainly go up?

Who is our nearest neighbor

Why do Australian milk farmers need to protest supermarkets' milk price?

Unreachable code, but reachable with exception

Welcoming 2019 Pi day: How to draw the letter π?

Is a lawful good "antagonist" effective?

Can "semicircle" be used to refer to a part-circle that is not a exact half-circle?

When were linguistics departments first established

Want to switch to tankless, but can I use my existing wiring?

Touchscreen-controlled dentist office snowman collector game

Why don't MCU characters ever seem to have language issues?

What has been your most complicated TikZ drawing?

Is it ok to include an epilogue dedicated to colleagues who passed away in the end of the manuscript?

What wound would be of little consequence to a biped but terrible for a quadruped?

What is the difference between "shut" and "close"?

If Invisibility ends because the original caster casts a non-concentration spell, does Invisibility also end on other targets of the original casting?

Do Bugbears' arms literally get longer when it's their turn?

Best approach to update all entries in a list that is paginated?

The three point beverage

What does it mean when multiple 々 marks follow a 、?

Plywood subfloor won't screw down in a trailer home

Identifying the interval from A♭ to D♯

Discrete Mathematics Surjective injective bijective proof

Surjective, Injective, Bijective Functions from $mathbb Z$ to itselfSufficient / necessary conditions for $g circ f$ being injective, surjective or bijectiveFunctions Type (Injective and surjective)Injective? Surjective? Bijective? None?Is this function bijective, surjective and injective?Intuitive definition of injective, surjective and bijectiveStability of injective, surjective and bijective functions under compositionWhich values of $(a,b)$ make a function bijective/surjective/injective.Why is the composition of a surjective and injective function neither surjective nor injective?Proving a multi variable function bijective

0

$begingroup$

1. Prove that the following function f : R+ → R>2 is both injective and
   surjective and therefore, bijective.

$$f(x) = 2 + sqrtx$$

2. Use the method of disproof by counterexample to show that function $g:
   Z to Z$ with the following general formula is neither injective nor surjective: $g(n) = n^2 − 8$
   

functions

share|cite|improve this question

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement is true taking f from R+ to R+ but I don't understand what "R> 2" means.
  $endgroup$
  – user247327
  Mar 10 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think OP means $mathbbR_>2 = (2,infty)$ or $mathbbR_geq 2 = [2,infty)$.
  $endgroup$
  – rolandcyp
  Mar 10 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations please use MathJax.
  $endgroup$
  – dantopa
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

1. Prove that the following function f : R+ → R>2 is both injective and
   surjective and therefore, bijective.

$$f(x) = 2 + sqrtx$$

2. Use the method of disproof by counterexample to show that function $g:
   Z to Z$ with the following general formula is neither injective nor surjective: $g(n) = n^2 − 8$
   

functions

share|cite|improve this question

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement is true taking f from R+ to R+ but I don't understand what "R> 2" means.
  $endgroup$
  – user247327
  Mar 10 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think OP means $mathbbR_>2 = (2,infty)$ or $mathbbR_geq 2 = [2,infty)$.
  $endgroup$
  – rolandcyp
  Mar 10 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations please use MathJax.
  $endgroup$
  – dantopa
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

1. Prove that the following function f : R+ → R>2 is both injective and
   surjective and therefore, bijective.

$$f(x) = 2 + sqrtx$$

2. Use the method of disproof by counterexample to show that function $g:
   Z to Z$ with the following general formula is neither injective nor surjective: $g(n) = n^2 − 8$
   

functions

share|cite|improve this question

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

dantopa

6,62442245

edited 2 days ago

dantopa

6,62442245

asked Mar 10 at 20:31

RoxyRoxy

1

New contributor

Roxy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 20:31

RoxyRoxy

1

1 Answer
1

active

oldest

votes

1

$begingroup$

I will try to sketch out the argument/steps required to solve this problem fully. I leave it to you to write out the details.

To prove that $f$ is injective, start by assuming that $f(x) = f(y)$ for two $x,y geq 0$. Written explicitly, this means that
$$
sqrtx + 2 = sqrty + 2.
$$
Through simple manipulations, you can deduce that $x = y$. This means that $f$ is injective.

As for the surjectivity of $f$ fix $y in [2,infty)$. Our job is then to prove that there exists $x geq 0$ such that $f(x) = y$. That is, we want $x geq 0$ such that
$$
y = f(x) = sqrtx + 2.
$$
Again, you can check that this $x$ exists. This would prove the surjectivity of $f$.

Now we look at 2. Consider the function
$$
g : mathbbZ to mathbbZ, quad g(n) := n^2 - 8.
$$
To prove that $g$ is not surjective, we must find $m in mathbbZ$ such that $g(n) neq m$ for all $n in mathbbZ$. (Hint: consider $m=0$). As for injectivity, it should be clear that $g(-1) = g(1)$.

share|cite|improve this answer

edited Mar 10 at 20:54

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

$endgroup$

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
);



);

Roxy is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142854%2fdiscrete-mathematics-surjective-injective-bijective-proof%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

I will try to sketch out the argument/steps required to solve this problem fully. I leave it to you to write out the details.

To prove that $f$ is injective, start by assuming that $f(x) = f(y)$ for two $x,y geq 0$. Written explicitly, this means that
$$
sqrtx + 2 = sqrty + 2.
$$
Through simple manipulations, you can deduce that $x = y$. This means that $f$ is injective.

As for the surjectivity of $f$ fix $y in [2,infty)$. Our job is then to prove that there exists $x geq 0$ such that $f(x) = y$. That is, we want $x geq 0$ such that
$$
y = f(x) = sqrtx + 2.
$$
Again, you can check that this $x$ exists. This would prove the surjectivity of $f$.

Now we look at 2. Consider the function
$$
g : mathbbZ to mathbbZ, quad g(n) := n^2 - 8.
$$
To prove that $g$ is not surjective, we must find $m in mathbbZ$ such that $g(n) neq m$ for all $n in mathbbZ$. (Hint: consider $m=0$). As for injectivity, it should be clear that $g(-1) = g(1)$.

share|cite|improve this answer

edited Mar 10 at 20:54

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

$endgroup$

add a comment | 

1

$begingroup$

I will try to sketch out the argument/steps required to solve this problem fully. I leave it to you to write out the details.

To prove that $f$ is injective, start by assuming that $f(x) = f(y)$ for two $x,y geq 0$. Written explicitly, this means that
$$
sqrtx + 2 = sqrty + 2.
$$
Through simple manipulations, you can deduce that $x = y$. This means that $f$ is injective.

As for the surjectivity of $f$ fix $y in [2,infty)$. Our job is then to prove that there exists $x geq 0$ such that $f(x) = y$. That is, we want $x geq 0$ such that
$$
y = f(x) = sqrtx + 2.
$$
Again, you can check that this $x$ exists. This would prove the surjectivity of $f$.

Now we look at 2. Consider the function
$$
g : mathbbZ to mathbbZ, quad g(n) := n^2 - 8.
$$
To prove that $g$ is not surjective, we must find $m in mathbbZ$ such that $g(n) neq m$ for all $n in mathbbZ$. (Hint: consider $m=0$). As for injectivity, it should be clear that $g(-1) = g(1)$.

share|cite|improve this answer

edited Mar 10 at 20:54

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

$endgroup$

add a comment | 

$begingroup$

I will try to sketch out the argument/steps required to solve this problem fully. I leave it to you to write out the details.

To prove that $f$ is injective, start by assuming that $f(x) = f(y)$ for two $x,y geq 0$. Written explicitly, this means that
$$
sqrtx + 2 = sqrty + 2.
$$
Through simple manipulations, you can deduce that $x = y$. This means that $f$ is injective.

As for the surjectivity of $f$ fix $y in [2,infty)$. Our job is then to prove that there exists $x geq 0$ such that $f(x) = y$. That is, we want $x geq 0$ such that
$$
y = f(x) = sqrtx + 2.
$$
Again, you can check that this $x$ exists. This would prove the surjectivity of $f$.

Now we look at 2. Consider the function
$$
g : mathbbZ to mathbbZ, quad g(n) := n^2 - 8.
$$
To prove that $g$ is not surjective, we must find $m in mathbbZ$ such that $g(n) neq m$ for all $n in mathbbZ$. (Hint: consider $m=0$). As for injectivity, it should be clear that $g(-1) = g(1)$.

share|cite|improve this answer

edited Mar 10 at 20:54

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

$endgroup$

share|cite|improve this answer

edited Mar 10 at 20:54

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

answered Mar 10 at 20:42

rolandcyprolandcyp

714311

answered Mar 10 at 20:42

rolandcyprolandcyp

714311