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







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












$begingroup$


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 21 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









        $endgroup$













          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160658%2fproving-fx-x-is-onto%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          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






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160658%2fproving-fx-x-is-onto%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Masuk log Menu navigasi

                                          Image
                                          JawaSundaBanyumasanBugisAcehBanjarMinangMelayuDeutschEnglishEspañolFrançaisItalianoNederlands (function()var node=document.getElementById("mw-dismissablenotice-anonplace");if(node)node.outerHTML="u003Cdiv class="mw-dismissable-notice"u003Eu003Cdiv class="mw-dismissable-notice-close"u003E[u003Ca tabindex="0" role="button"u003Etutupu003C/au003E]u003C/divu003Eu003Cdiv class="mw-dismissable-notice-body"u003Eu003Cdiv id="localNotice" lang="id" dir="ltr"u003Eu003C/divu003Eu003C/divu003Eu003C/divu003E";()); Masuk log Loncat ke navigasi Loncat ke pencarian Bahasa: Jawa | Sunda | Banyumasan | Bugis | Aceh | Banjar | Minang | Melayu | Deutsch | English | Español | Français | Italiano | Nederlands Nama pengguna Kata sandi   Biarkan saya tetap masuk Masuk log Bantuan masuk log Lupa kata sandi? Belum punya akun?Gabung Wikipedia Diperoleh dari "https:/...
                                          Read more

                                          Identifying “long and narrow” polygons in with PostGISlength and width of polygonWhy postgis st_overlaps reports Qgis' “avoid intersections” generated polygon as overlapping with others?Adjusting polygons to boundary and filling holesDrawing polygons with fixed area?How to remove spikes in Polygons with PostGISDeleting sliver polygons after difference operation in QGIS?Snapping boundaries in PostGISSplit polygon into parts adding attributes based on underlying polygon in QGISSplitting overlap between polygons and assign to nearest polygon using PostGIS?Expanding polygons and clipping at midpoint?Removing Intersection of Buffers in Same Layers

                                          Image
                                          Do I have to worry about players making “bad” choices on level up? Is thermodynamics only applicable to systems in equilibrium? gnu parallel how to use with ffmpeg Has any spacecraft ever had the ability to directly communicate with civilian air traffic control? Reverse the word in a string with the same order in javascript Is creating your own "experiment" considered cheating during a physics exam? How to stop co-workers from teasing me because I know Russian? What is a Recurrent Neural Network? Any examples of headwear for races with animal ears? Confusion about capacitors TikZ how to make supply and demand arrows for nodes? Does a creature that is immune to a condition still make a saving throw? How to verbalise code in Mathematica? Upright [...] in italics quotation Cannot populate data in lightning data table Stark VS Thanos Binary Numbers Magic Trick In gnome-terminal only 2 out of 3 zoom keys work You look catfish vs You look like a...
                                          Read more

                                          Старые Смолеговицы Содержание История | География | Демография | Достопримечательности | Примечания | НавигацияHGЯOLHGЯOL41 206 832 01641 606 406 141Административно-территориальное деление Ленинградской области«Переписная оброчная книга Водской пятины 1500 года», С. 793«Карта Ингерманландии: Ивангорода, Яма, Копорья, Нотеборга», по материалам 1676 г.«Генеральная карта провинции Ингерманландии» Э. Белинга и А. Андерсина, 1704 г., составлена по материалам 1678 г.«Географический чертёж над Ижорскою землей со своими городами» Адриана Шонбека 1705 г.Новая и достоверная всей Ингерманландии ланткарта. Грав. А. Ростовцев. СПб., 1727 г.Топографическая карта Санкт-Петербургской губернии. 5-и верстка. Шуберт. 1834 г.Описание Санкт-Петербургской губернии по уездам и станамСпецкарта западной части России Ф. Ф. Шуберта. 1844 г.Алфавитный список селений по уездам и станам С.-Петербургской губернииСписки населённых мест Российской Империи, составленные и издаваемые центральным статистическим комитетом министерства внутренних дел. XXXVII. Санкт-Петербургская губерния. По состоянию на 1862 год. СПб. 1864. С. 203Материалы по статистике народного хозяйства в С.-Петербургской губернии. Вып. IX. Частновладельческое хозяйство в Ямбургском уезде. СПб, 1888, С. 146, С. 2, 7, 54Положение о гербе муниципального образования Курское сельское поселениеСправочник истории административно-территориального деления Ленинградской области.Топографическая карта Ленинградской области, квадрат О-35-23-В (Хотыницы), 1930 г.АрхивированоАдминистративно-территориальное деление Ленинградской области. — Л., 1933, С. 27, 198АрхивированоАдминистративно-экономический справочник по Ленинградской области. — Л., 1936, с. 219АрхивированоАдминистративно-территориальное деление Ленинградской области. — Л., 1966, с. 175АрхивированоАдминистративно-территориальное деление Ленинградской области. — Лениздат, 1973, С. 180АрхивированоАдминистративно-территориальное деление Ленинградской области. — Лениздат, 1990, ISBN 5-289-00612-5, С. 38АрхивированоАдминистративно-территориальное деление Ленинградской области. — СПб., 2007, с. 60АрхивированоКоряков Юрий База данных «Этно-языковой состав населённых пунктов России». Ленинградская область.Административно-территориальное деление Ленинградской области. — СПб, 1997, ISBN 5-86153-055-6, С. 41АрхивированоКультовый комплекс Старые Смолеговицы // Электронная энциклопедия ЭрмитажаПроблемы выявления, изучения и сохранения культовых комплексов с каменными крестами: по материалам работ 2016-2017 гг. в Ленинградской области

                                          Image
                                          Населённые пункты по алфавитуНаселённые пункты Волосовского районаМызы Ингрии Беседском сельском поселенииВолосовского районаЛенинградской областиПисцово...
                                          Read more