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
    yesterday






  • 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
    yesterday
















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
    yesterday






  • 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
    yesterday














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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 16 hours ago









YuiTo Cheng

2,1362837




2,1362837










asked yesterday









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
    yesterday






  • 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
    yesterday














  • 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
    yesterday






  • 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
    yesterday








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
yesterday




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




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
yesterday




$begingroup$
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
yesterday










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









$endgroup$









  • 1




    $begingroup$
    (+1) Because this is a vey pedagogical answer.
    $endgroup$
    – José Carlos Santos
    yesterday



















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









          $endgroup$









          • 1




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            yesterday
















          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









          $endgroup$









          • 1




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            yesterday














          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 yesterday









          Henning MakholmHenning Makholm

          242k17308551




          242k17308551








          • 1




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            yesterday














          • 1




            $begingroup$
            (+1) Because this is a vey pedagogical answer.
            $endgroup$
            – José Carlos Santos
            yesterday








          1




          1




          $begingroup$
          (+1) Because this is a vey pedagogical answer.
          $endgroup$
          – José Carlos Santos
          yesterday




          $begingroup$
          (+1) Because this is a vey pedagogical answer.
          $endgroup$
          – José Carlos Santos
          yesterday











          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 yesterday









              fleabloodfleablood

              73.4k22791




              73.4k22791























                  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 yesterday









                      J. W. Tanner

                      3,8001320




                      3,8001320










                      answered yesterday









                      José Carlos SantosJosé Carlos Santos

                      170k23132238




                      170k23132238























                          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 yesterday









                              WuestenfuxWuestenfux

                              5,2981513




                              5,2981513























                                  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 yesterday









                                      John ColemanJohn Coleman

                                      3,98311224




                                      3,98311224






























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                                          Engineer refusing to file/disclose patents Extending the spectral theorem for bounded self adjoint operators to bounded normal operators Do varchar(max), nvarchar(max) and varbinary(max) columns affect select queries? Why is it that I can sometimes guess the next note? What linear sensor for a keyboard? Why don’t women say שלא עשני גויה and שפחה, according to Ashkenazim? Did arcade monitors have same pixel aspect ratio as TV sets? anything or something to eat Query about absorption line spectra Can I use my Chinese passport to enter China after I acquired another citizenship? Fuse symbol on toroidal transformer Create all possible words using a set or letters Does the Mind Blank spell prevent the target from being frightened? Can I sign legal documents with a smiley face? When quoting, must I also copy hyphens used to divide words that continue on the next line? ...
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                                          Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! 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                                          SlayerMusikgrupper bildade 1981Amerikanska thrash metal-grupper amerikansktthrash metalHuntington Park1981gitarristernaJeff HannemanKerry KingbasistenTom ArayatrummisenDave LombardoGary HoltPaul Bostaphcertifierats med guldMetallicaAnthraxMegadeththrash metal1986extrema metalgenrerdeath metal22 januari2018USAKanadaEuropaHuntington ParkKalifornien1981gitarristernaJeff HannemanKerry KingbassångDave Lombardoheavy metalbrittiskaIron MaidenJudas PriestthrashMetallicaRattSteeler1982demoundergroundMetal Blade1983high schoolGene HoglanDeathStrapping Young Ladkassettvinyl1984LondonMarquee Club1984Eldorado StudiosTrack Record StudiosLos Angelesmastrade1985CDCombat Recordsextreme metalVenomExodusStudio 54 club1985VHSCombat RecordsBelgienUFORick RubinDef Jam Recordship hoprockakterThe Black CrowesRun–D.M.C.LL Cool JBeastie BoysthrashbpmAngel of DeathRaining BloodAndy WallaceHowie Weinberg1986bildskivaBillboard 200RIAAMetal HammerKerrang!Angela Gossowdeath metalArch Enemythrash metalOverk...
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