Proving $f(x)=|x|$ is onto












2












$begingroup$


I've been working on proving that this is a onto function:



$f$ : $mathbb R$ $to$ $mathbb R^{geq0}$ is defined by $f(x)=|x|$



My proof so far: Let $yinmathbb R$.



Rough work: $|x|=y Rightarrow sqrt {x^2}=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$



Suppose $f(pm y)=|pm y|=y$.



I know that this function is definitely onto given the co-domain of $mathbb R^{geq0}$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?



Thanks!






proof-verification elementary-set-theory







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$endgroup$








  • 1




    $begingroup$
    Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
    $endgroup$
    – Arrow
    Mar 24 at 15:23






  • 6




    $begingroup$
    You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
    $endgroup$
    – symplectomorphic
    Mar 24 at 15:24
















2












$begingroup$


I've been working on proving that this is a onto function:



$f$ : $mathbb R$ $to$ $mathbb R^{geq0}$ is defined by $f(x)=|x|$



My proof so far: Let $yinmathbb R$.



Rough work: $|x|=y Rightarrow sqrt {x^2}=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$



Suppose $f(pm y)=|pm y|=y$.



I know that this function is definitely onto given the co-domain of $mathbb R^{geq0}$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?



Thanks!






proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
    $endgroup$
    – Arrow
    Mar 24 at 15:23






  • 6




    $begingroup$
    You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
    $endgroup$
    – symplectomorphic
    Mar 24 at 15:24














2












2








2


2



$begingroup$


I've been working on proving that this is a onto function:



$f$ : $mathbb R$ $to$ $mathbb R^{geq0}$ is defined by $f(x)=|x|$



My proof so far: Let $yinmathbb R$.



Rough work: $|x|=y Rightarrow sqrt {x^2}=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$



Suppose $f(pm y)=|pm y|=y$.



I know that this function is definitely onto given the co-domain of $mathbb R^{geq0}$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?



Thanks!






proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$




I've been working on proving that this is a onto function:



$f$ : $mathbb R$ $to$ $mathbb R^{geq0}$ is defined by $f(x)=|x|$



My proof so far: Let $yinmathbb R$.



Rough work: $|x|=y Rightarrow sqrt {x^2}=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$



Suppose $f(pm y)=|pm y|=y$.



I know that this function is definitely onto given the co-domain of $mathbb R^{geq0}$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?



Thanks!






proof-verification elementary-set-theory




proof-verification elementary-set-theory






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 8:00









YuiTo Cheng

2,1512937




2,1512937










asked Mar 24 at 15:21









Nick SabiaNick Sabia

285




285








  • 1




    $begingroup$
    Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
    $endgroup$
    – Arrow
    Mar 24 at 15:23






  • 6




    $begingroup$
    You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
    $endgroup$
    – symplectomorphic
    Mar 24 at 15:24














  • 1




    $begingroup$
    Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
    $endgroup$
    – Arrow
    Mar 24 at 15:23






  • 6




    $begingroup$
    You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
    $endgroup$
    – symplectomorphic
    Mar 24 at 15:24








1




1




$begingroup$
Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23




$begingroup$
Let $ain mathbb R ^{geq 0}$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23




6




6




$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24




$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24










5 Answers
5






active

oldest

votes


















21












$begingroup$

In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.



You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.



You could just say:




Let $yinmathbb R^{ge 0}$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.




(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    (+1) Because this is a vey pedagogical answer.
    $endgroup$
    – José Carlos Santos
    Mar 24 at 15:33



















4












$begingroup$


Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?




No.



You are simply supposed to show that for any general arbitrary $y in mathbb R^{ge 0}$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.



As $|y| = y$ this is very easy. And you are done.



The proof is two lines:



1) Let $y in mathbb R^{ge 0}$.



2) $f(y) = |y| = y$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You should not start with $yinmathbb R$ but rather with $yinmathbb{R}^{geqslant0}$. Then $y=f(y)$. Since this happens for every $yinmathbb{R}^{geqslant 0}$, $f$ is onto.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Well, you want to show that $f$ is onto. So take an arbitrary element $yin{Bbb R}_{geq 0}$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.






        share|cite|improve this answer









        $endgroup$














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          5 Answers
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          5 Answers
          5






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          21












          $begingroup$

          In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.



          You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.



          You could just say:




          Let $yinmathbb R^{ge 0}$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.




          (Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            Mar 24 at 15:33
















          21












          $begingroup$

          In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.



          You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.



          You could just say:




          Let $yinmathbb R^{ge 0}$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.




          (Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            Mar 24 at 15:33














          21












          21








          21





          $begingroup$

          In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.



          You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.



          You could just say:




          Let $yinmathbb R^{ge 0}$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.




          (Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).






          share|cite|improve this answer









          $endgroup$



          In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.



          You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.



          You could just say:




          Let $yinmathbb R^{ge 0}$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.




          (Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 24 at 15:25









          Henning MakholmHenning Makholm

          242k17308552




          242k17308552








          • 2




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            Mar 24 at 15:33














          • 2




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            Mar 24 at 15:33








          2




          2




          $begingroup$
          (+1) Because this is a vey pedagogical answer.
          $endgroup$
          – José Carlos Santos
          Mar 24 at 15:33




          $begingroup$
          (+1) Because this is a vey pedagogical answer.
          $endgroup$
          – José Carlos Santos
          Mar 24 at 15:33











          4












          $begingroup$


          Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?




          No.



          You are simply supposed to show that for any general arbitrary $y in mathbb R^{ge 0}$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.



          As $|y| = y$ this is very easy. And you are done.



          The proof is two lines:



          1) Let $y in mathbb R^{ge 0}$.



          2) $f(y) = |y| = y$.






          share|cite|improve this answer









          $endgroup$


















            4












            $begingroup$


            Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?




            No.



            You are simply supposed to show that for any general arbitrary $y in mathbb R^{ge 0}$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.



            As $|y| = y$ this is very easy. And you are done.



            The proof is two lines:



            1) Let $y in mathbb R^{ge 0}$.



            2) $f(y) = |y| = y$.






            share|cite|improve this answer









            $endgroup$
















              4












              4








              4





              $begingroup$


              Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?




              No.



              You are simply supposed to show that for any general arbitrary $y in mathbb R^{ge 0}$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.



              As $|y| = y$ this is very easy. And you are done.



              The proof is two lines:



              1) Let $y in mathbb R^{ge 0}$.



              2) $f(y) = |y| = y$.






              share|cite|improve this answer









              $endgroup$




              Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?




              No.



              You are simply supposed to show that for any general arbitrary $y in mathbb R^{ge 0}$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.



              As $|y| = y$ this is very easy. And you are done.



              The proof is two lines:



              1) Let $y in mathbb R^{ge 0}$.



              2) $f(y) = |y| = y$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 24 at 16:04









              fleabloodfleablood

              73.6k22891




              73.6k22891























                  2












                  $begingroup$

                  You should not start with $yinmathbb R$ but rather with $yinmathbb{R}^{geqslant0}$. Then $y=f(y)$. Since this happens for every $yinmathbb{R}^{geqslant 0}$, $f$ is onto.






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    You should not start with $yinmathbb R$ but rather with $yinmathbb{R}^{geqslant0}$. Then $y=f(y)$. Since this happens for every $yinmathbb{R}^{geqslant 0}$, $f$ is onto.






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      You should not start with $yinmathbb R$ but rather with $yinmathbb{R}^{geqslant0}$. Then $y=f(y)$. Since this happens for every $yinmathbb{R}^{geqslant 0}$, $f$ is onto.






                      share|cite|improve this answer











                      $endgroup$



                      You should not start with $yinmathbb R$ but rather with $yinmathbb{R}^{geqslant0}$. Then $y=f(y)$. Since this happens for every $yinmathbb{R}^{geqslant 0}$, $f$ is onto.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 24 at 15:41









                      J. W. Tanner

                      4,0171320




                      4,0171320










                      answered Mar 24 at 15:26









                      José Carlos SantosJosé Carlos Santos

                      171k23132240




                      171k23132240























                          0












                          $begingroup$

                          Well, you want to show that $f$ is onto. So take an arbitrary element $yin{Bbb R}_{geq 0}$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Well, you want to show that $f$ is onto. So take an arbitrary element $yin{Bbb R}_{geq 0}$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Well, you want to show that $f$ is onto. So take an arbitrary element $yin{Bbb R}_{geq 0}$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.






                              share|cite|improve this answer









                              $endgroup$



                              Well, you want to show that $f$ is onto. So take an arbitrary element $yin{Bbb R}_{geq 0}$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Mar 24 at 15:24









                              WuestenfuxWuestenfux

                              5,3231513




                              5,3231513























                                  0












                                  $begingroup$

                                  More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.






                                      share|cite|improve this answer









                                      $endgroup$



                                      More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Mar 24 at 20:31









                                      John ColemanJohn Coleman

                                      3,98311224




                                      3,98311224






























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                                          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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                                          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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                                          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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                                          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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