Determine if $y = x^2$ is injective











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1
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I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).



However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.



Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.



This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.



However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.



Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?










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  • You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
    – lEm
    Sep 9 '16 at 2:48










  • The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
    – user228113
    Sep 9 '16 at 2:58












  • It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
    – user228113
    Sep 9 '16 at 2:59












  • Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
    – Ham Sandwich
    Sep 9 '16 at 3:11






  • 1




    You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
    – user228113
    Sep 9 '16 at 3:17

















up vote
1
down vote

favorite












I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).



However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.



Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.



This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.



However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.



Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?










share|cite|improve this question
























  • You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
    – lEm
    Sep 9 '16 at 2:48










  • The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
    – user228113
    Sep 9 '16 at 2:58












  • It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
    – user228113
    Sep 9 '16 at 2:59












  • Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
    – Ham Sandwich
    Sep 9 '16 at 3:11






  • 1




    You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
    – user228113
    Sep 9 '16 at 3:17















up vote
1
down vote

favorite









up vote
1
down vote

favorite











I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).



However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.



Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.



This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.



However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.



Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?










share|cite|improve this question















I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).



However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.



Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.



This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.



However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.



Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?







functions






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edited Feb 10 at 22:36









N. F. Taussig

43.2k93254




43.2k93254










asked Sep 9 '16 at 2:45









Ham Sandwich

226316




226316












  • You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
    – lEm
    Sep 9 '16 at 2:48










  • The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
    – user228113
    Sep 9 '16 at 2:58












  • It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
    – user228113
    Sep 9 '16 at 2:59












  • Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
    – Ham Sandwich
    Sep 9 '16 at 3:11






  • 1




    You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
    – user228113
    Sep 9 '16 at 3:17




















  • You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
    – lEm
    Sep 9 '16 at 2:48










  • The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
    – user228113
    Sep 9 '16 at 2:58












  • It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
    – user228113
    Sep 9 '16 at 2:59












  • Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
    – Ham Sandwich
    Sep 9 '16 at 3:11






  • 1




    You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
    – user228113
    Sep 9 '16 at 3:17


















You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48




You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48












The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58






The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58














It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59






It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59














Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11




Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11




1




1




You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17






You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17












3 Answers
3






active

oldest

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up vote
1
down vote



accepted










The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.






share|cite|improve this answer




























    up vote
    1
    down vote













    to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.






    share|cite|improve this answer





















    • First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
      – user228113
      Sep 9 '16 at 3:36




















    up vote
    1
    down vote













    Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.



    If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.



    The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.



    It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.



    Let me take an example. Lets show that $f(x) = x^3$ is injective.



    We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.



    The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.






          share|cite|improve this answer












          The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 '16 at 2:53









          Ross Millikan

          290k23195368




          290k23195368






















              up vote
              1
              down vote













              to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.






              share|cite|improve this answer





















              • First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
                – user228113
                Sep 9 '16 at 3:36

















              up vote
              1
              down vote













              to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.






              share|cite|improve this answer





















              • First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
                – user228113
                Sep 9 '16 at 3:36















              up vote
              1
              down vote










              up vote
              1
              down vote









              to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.






              share|cite|improve this answer












              to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 9 '16 at 3:17









              R.Mor

              728




              728












              • First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
                – user228113
                Sep 9 '16 at 3:36




















              • First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
                – user228113
                Sep 9 '16 at 3:36


















              First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
              – user228113
              Sep 9 '16 at 3:36






              First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
              – user228113
              Sep 9 '16 at 3:36












              up vote
              1
              down vote













              Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.



              If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.



              The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.



              It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.



              Let me take an example. Lets show that $f(x) = x^3$ is injective.



              We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.



              The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.



                If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.



                The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.



                It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.



                Let me take an example. Lets show that $f(x) = x^3$ is injective.



                We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.



                The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.



                  If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.



                  The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.



                  It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.



                  Let me take an example. Lets show that $f(x) = x^3$ is injective.



                  We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.



                  The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.






                  share|cite|improve this answer












                  Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.



                  If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.



                  The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.



                  It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.



                  Let me take an example. Lets show that $f(x) = x^3$ is injective.



                  We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.



                  The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.







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                  answered Nov 23 at 9:59









                  Arjihad

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