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Prove function is injective, but not surjective

-1

1

$begingroup$

Prove that the function $f(x) = x^2$ for $xin mathbb N$ is injective, but not surjective.

I know that injective means that $f(x)=f(y)implies x=y$, and
surjective means that for $f(x)=y$. So using this information, how would I prove this problem?

discrete-mathematics logic

share|cite|improve this question

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

asked Mar 26 at 1:47

user643202

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove that $f$ is NOT surjective
  $endgroup$
  – Brian
  Mar 26 at 1:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Injective: if $x^2=y^2$ then $(x+y)(x-y)=0$ so ...; Surjective: is there $xinmathbb N$ such that $x^2=2$?
  $endgroup$
  – J. W. Tanner
  Mar 26 at 1:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is x^2 not subjective?
  $endgroup$
  – user643202
  Mar 26 at 1:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have not specified a codomain.
  $endgroup$
  – N. F. Taussig
  Mar 26 at 1:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  As @N.F.Taussig noted, your definition of $f$ is technically incomplete. It is important to specify both what set $f$ 'takes inputs from' and what set $f$ 'produces elements from. If $f$ where given as a function $f:mathbb N to A$ where $A={ n^2 mid n in mathbb N }$, then $f$ would be surjective.
  $endgroup$
  – Brian
  Mar 26 at 1:56
  
  
  

  
  
  
  

 | 
show 2 more comments

-1

1

$begingroup$

Prove that the function $f(x) = x^2$ for $xin mathbb N$ is injective, but not surjective.

I know that injective means that $f(x)=f(y)implies x=y$, and
surjective means that for $f(x)=y$. So using this information, how would I prove this problem?

discrete-mathematics logic

share|cite|improve this question

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

asked Mar 26 at 1:47

user643202

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove that $f$ is NOT surjective
  $endgroup$
  – Brian
  Mar 26 at 1:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Injective: if $x^2=y^2$ then $(x+y)(x-y)=0$ so ...; Surjective: is there $xinmathbb N$ such that $x^2=2$?
  $endgroup$
  – J. W. Tanner
  Mar 26 at 1:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is x^2 not subjective?
  $endgroup$
  – user643202
  Mar 26 at 1:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have not specified a codomain.
  $endgroup$
  – N. F. Taussig
  Mar 26 at 1:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  As @N.F.Taussig noted, your definition of $f$ is technically incomplete. It is important to specify both what set $f$ 'takes inputs from' and what set $f$ 'produces elements from. If $f$ where given as a function $f:mathbb N to A$ where $A={ n^2 mid n in mathbb N }$, then $f$ would be surjective.
  $endgroup$
  – Brian
  Mar 26 at 1:56
  
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Prove that the function $f(x) = x^2$ for $xin mathbb N$ is injective, but not surjective.

I know that injective means that $f(x)=f(y)implies x=y$, and
surjective means that for $f(x)=y$. So using this information, how would I prove this problem?

discrete-mathematics logic

share|cite|improve this question

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

asked Mar 26 at 1:47

user643202

$endgroup$

share|cite|improve this question

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

asked Mar 26 at 1:47

user643202

share|cite|improve this question

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

asked Mar 26 at 1:47

user643202

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

edited Mar 26 at 16:41

Mauro ALLEGRANZA

68.3k449117

2 Answers
2

active

oldest

votes

3

$begingroup$

It seems that you mean $f: mathbb{N} to mathbb{N}$ given by $x mapsto x^2$ is injective and not surjective.

To show it is not surjective, hint:

Can you find a natural number that isn't a square of some natural number?

To show it is injective, if $x^2 = y^2$ we have $x = y$ or $x = -y$, can you continue from here?

share|cite|improve this answer

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So like numbers like 3, 5, and 7?
  $endgroup$
  – user643202
  Mar 26 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous those are examples yes. Do you understand why that shows the function isn't a surjection?
  $endgroup$
  – Mariah
  Mar 26 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't x = -y not make it injective?
  $endgroup$
  – user643202
  Mar 26 at 2:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous what is our domain?
  $endgroup$
  – Mariah
  Mar 26 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  all natural numbers
  $endgroup$
  – user643202
  Mar 26 at 2:07
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

You just need to give a counter-example here. Consider the number $y=17 in mathbb{N}$. Then $x^2=y Rightarrow x^2 = 17 Rightarrow x = pmsqrt{17} notin mathbb{N}.$ You have found a $y$ such that $x notin mathbb{N}$ for $f(x)=y$. Therefore, $f$ is not surjective.

share|cite|improve this answer

answered Mar 26 at 7:14

YFPYFP

3912

$endgroup$

add a comment | 

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3

$begingroup$

It seems that you mean $f: mathbb{N} to mathbb{N}$ given by $x mapsto x^2$ is injective and not surjective.

To show it is not surjective, hint:

Can you find a natural number that isn't a square of some natural number?

To show it is injective, if $x^2 = y^2$ we have $x = y$ or $x = -y$, can you continue from here?

share|cite|improve this answer

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So like numbers like 3, 5, and 7?
  $endgroup$
  – user643202
  Mar 26 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous those are examples yes. Do you understand why that shows the function isn't a surjection?
  $endgroup$
  – Mariah
  Mar 26 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't x = -y not make it injective?
  $endgroup$
  – user643202
  Mar 26 at 2:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous what is our domain?
  $endgroup$
  – Mariah
  Mar 26 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  all natural numbers
  $endgroup$
  – user643202
  Mar 26 at 2:07
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

It seems that you mean $f: mathbb{N} to mathbb{N}$ given by $x mapsto x^2$ is injective and not surjective.

To show it is not surjective, hint:

Can you find a natural number that isn't a square of some natural number?

To show it is injective, if $x^2 = y^2$ we have $x = y$ or $x = -y$, can you continue from here?

share|cite|improve this answer

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So like numbers like 3, 5, and 7?
  $endgroup$
  – user643202
  Mar 26 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous those are examples yes. Do you understand why that shows the function isn't a surjection?
  $endgroup$
  – Mariah
  Mar 26 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't x = -y not make it injective?
  $endgroup$
  – user643202
  Mar 26 at 2:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anonymous what is our domain?
  $endgroup$
  – Mariah
  Mar 26 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  all natural numbers
  $endgroup$
  – user643202
  Mar 26 at 2:07
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

It seems that you mean $f: mathbb{N} to mathbb{N}$ given by $x mapsto x^2$ is injective and not surjective.

To show it is not surjective, hint:

Can you find a natural number that isn't a square of some natural number?

To show it is injective, if $x^2 = y^2$ we have $x = y$ or $x = -y$, can you continue from here?

share|cite|improve this answer

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

$endgroup$

share|cite|improve this answer

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

edited Mar 26 at 2:13

J. W. Tanner

5,0751520

answered Mar 26 at 2:02

MariahMariah

2,1471718

answered Mar 26 at 2:02

MariahMariah

2,1471718

0

$begingroup$

You just need to give a counter-example here. Consider the number $y=17 in mathbb{N}$. Then $x^2=y Rightarrow x^2 = 17 Rightarrow x = pmsqrt{17} notin mathbb{N}.$ You have found a $y$ such that $x notin mathbb{N}$ for $f(x)=y$. Therefore, $f$ is not surjective.

share|cite|improve this answer

answered Mar 26 at 7:14

YFPYFP

3912

$endgroup$

add a comment | 

0

$begingroup$

You just need to give a counter-example here. Consider the number $y=17 in mathbb{N}$. Then $x^2=y Rightarrow x^2 = 17 Rightarrow x = pmsqrt{17} notin mathbb{N}.$ You have found a $y$ such that $x notin mathbb{N}$ for $f(x)=y$. Therefore, $f$ is not surjective.

share|cite|improve this answer

answered Mar 26 at 7:14

YFPYFP

3912

$endgroup$

add a comment | 

$begingroup$

You just need to give a counter-example here. Consider the number $y=17 in mathbb{N}$. Then $x^2=y Rightarrow x^2 = 17 Rightarrow x = pmsqrt{17} notin mathbb{N}.$ You have found a $y$ such that $x notin mathbb{N}$ for $f(x)=y$. Therefore, $f$ is not surjective.

share|cite|improve this answer

answered Mar 26 at 7:14

YFPYFP

3912

$endgroup$

share|cite|improve this answer

answered Mar 26 at 7:14

YFPYFP

3912

answered Mar 26 at 7:14

YFPYFP

3912

answered Mar 26 at 7:14

YFPYFP

3912