If $a+sqrt{a^2+1}= b+sqrt{b^2+1}$, then $a=b$ or not?












1














It might be a silly question but if $$a+sqrt{a^2+1}= b+sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+sqrt{x^2+1}$ is a $1$-$1$ function.






algebra-precalculus functions radicals radical-equations







share|cite|improve this question




















  • 8




    The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
    – Lord Shark the Unknown
    Dec 9 at 18:17
















1














It might be a silly question but if $$a+sqrt{a^2+1}= b+sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+sqrt{x^2+1}$ is a $1$-$1$ function.






algebra-precalculus functions radicals radical-equations







share|cite|improve this question




















  • 8




    The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
    – Lord Shark the Unknown
    Dec 9 at 18:17














1












1








1







It might be a silly question but if $$a+sqrt{a^2+1}= b+sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+sqrt{x^2+1}$ is a $1$-$1$ function.






algebra-precalculus functions radicals radical-equations







share|cite|improve this question















It might be a silly question but if $$a+sqrt{a^2+1}= b+sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+sqrt{x^2+1}$ is a $1$-$1$ function.






algebra-precalculus functions radicals radical-equations




algebra-precalculus functions radicals radical-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 19:02









Zvi

4,670430




4,670430










asked Dec 9 at 18:14









user602338

1607




1607








  • 8




    The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
    – Lord Shark the Unknown
    Dec 9 at 18:17














  • 8




    The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
    – Lord Shark the Unknown
    Dec 9 at 18:17








8




8




The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
– Lord Shark the Unknown
Dec 9 at 18:17




The function $xmapsto x+sqrt{x^2+1}$ is strictly increasing on $Bbb R$.
– Lord Shark the Unknown
Dec 9 at 18:17










4 Answers
4






active

oldest

votes


















4














Hint: Square the equation
$$a-b=sqrt{b^2+1}-sqrt{a^2+1}$$ two times.
The result must be $$(a-b)^2=0$$






share|cite|improve this answer





















  • Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
    – user602338
    Dec 9 at 18:26










  • Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
    – user602338
    Dec 9 at 18:28










  • This depends on the given problem, you can write it here.
    – Dr. Sonnhard Graubner
    Dec 9 at 18:30










  • Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
    – user602338
    Dec 9 at 18:32



















2














Let $f:mathbb R to mathbb R $ such that :



$$f(x) = x + sqrt{x^2+1}$$



Then, it is :



$$f'(x) = 1 + frac{x}{sqrt{x^2+1}}= frac{sqrt{x^2+1}+x}{sqrt{x^2+1}} > 0, ; forall x in mathbb R $$



Thus the function $f(x)$ is strictly increasing for $x in mathbb R$ and thus it is "$1-1$", which then means that :



$$a + sqrt{a^2+1} = b + sqrt{b^2=1} implies a=b$$






share|cite|improve this answer





















  • Just an added question here. The map is not surjective since $0$ does not have a pre-image?
    – Yadati Kiran
    Dec 9 at 18:31





















2














Alternatively, for $a,bin Bbb R$,
begin{align}a+sqrt{a^2+1}&=b+sqrt{b^2+1}\&implies frac{1}{a+sqrt{a^2+1}}=frac{1}{b+sqrt{b^2+1}}wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
&implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
&implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\&implies (a+sqrt{a^2+1})-(sqrt{a^2+1}-a)=(b+sqrt{b^2+1})-(sqrt{b^2+1}-b)
\&implies 2a=2bimplies a=b. end{align}
It is also easy to see that $a=bimplies a+sqrt{a^2+1}=b+sqrt{b^2+1}$. That is,
$$a+sqrt{a^2+1}=b+sqrt{b^2+1}iff a=b.$$






share|cite|improve this answer





























    0














    Hint:



    WLOG $a=cot2A,b=cot2B,0<A,B<dfracpi2$



    we have $cot A=cot Bimpliescot2A=?$






    share|cite|improve this answer





















      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%2f3032720%2fif-a-sqrta21-b-sqrtb21-then-a-b-or-not%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Hint: Square the equation
      $$a-b=sqrt{b^2+1}-sqrt{a^2+1}$$ two times.
      The result must be $$(a-b)^2=0$$






      share|cite|improve this answer





















      • Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
        – user602338
        Dec 9 at 18:26










      • Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
        – user602338
        Dec 9 at 18:28










      • This depends on the given problem, you can write it here.
        – Dr. Sonnhard Graubner
        Dec 9 at 18:30










      • Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
        – user602338
        Dec 9 at 18:32
















      4














      Hint: Square the equation
      $$a-b=sqrt{b^2+1}-sqrt{a^2+1}$$ two times.
      The result must be $$(a-b)^2=0$$






      share|cite|improve this answer





















      • Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
        – user602338
        Dec 9 at 18:26










      • Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
        – user602338
        Dec 9 at 18:28










      • This depends on the given problem, you can write it here.
        – Dr. Sonnhard Graubner
        Dec 9 at 18:30










      • Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
        – user602338
        Dec 9 at 18:32














      4












      4








      4






      Hint: Square the equation
      $$a-b=sqrt{b^2+1}-sqrt{a^2+1}$$ two times.
      The result must be $$(a-b)^2=0$$






      share|cite|improve this answer












      Hint: Square the equation
      $$a-b=sqrt{b^2+1}-sqrt{a^2+1}$$ two times.
      The result must be $$(a-b)^2=0$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 9 at 18:19









      Dr. Sonnhard Graubner

      73.1k42865




      73.1k42865












      • Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
        – user602338
        Dec 9 at 18:26










      • Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
        – user602338
        Dec 9 at 18:28










      • This depends on the given problem, you can write it here.
        – Dr. Sonnhard Graubner
        Dec 9 at 18:30










      • Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
        – user602338
        Dec 9 at 18:32


















      • Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
        – user602338
        Dec 9 at 18:26










      • Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
        – user602338
        Dec 9 at 18:28










      • This depends on the given problem, you can write it here.
        – Dr. Sonnhard Graubner
        Dec 9 at 18:30










      • Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
        – user602338
        Dec 9 at 18:32
















      Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
      – user602338
      Dec 9 at 18:26




      Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems!
      – user602338
      Dec 9 at 18:26












      Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
      – user602338
      Dec 9 at 18:28




      Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that?
      – user602338
      Dec 9 at 18:28












      This depends on the given problem, you can write it here.
      – Dr. Sonnhard Graubner
      Dec 9 at 18:30




      This depends on the given problem, you can write it here.
      – Dr. Sonnhard Graubner
      Dec 9 at 18:30












      Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
      – user602338
      Dec 9 at 18:32




      Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet!
      – user602338
      Dec 9 at 18:32











      2














      Let $f:mathbb R to mathbb R $ such that :



      $$f(x) = x + sqrt{x^2+1}$$



      Then, it is :



      $$f'(x) = 1 + frac{x}{sqrt{x^2+1}}= frac{sqrt{x^2+1}+x}{sqrt{x^2+1}} > 0, ; forall x in mathbb R $$



      Thus the function $f(x)$ is strictly increasing for $x in mathbb R$ and thus it is "$1-1$", which then means that :



      $$a + sqrt{a^2+1} = b + sqrt{b^2=1} implies a=b$$






      share|cite|improve this answer





















      • Just an added question here. The map is not surjective since $0$ does not have a pre-image?
        – Yadati Kiran
        Dec 9 at 18:31


















      2














      Let $f:mathbb R to mathbb R $ such that :



      $$f(x) = x + sqrt{x^2+1}$$



      Then, it is :



      $$f'(x) = 1 + frac{x}{sqrt{x^2+1}}= frac{sqrt{x^2+1}+x}{sqrt{x^2+1}} > 0, ; forall x in mathbb R $$



      Thus the function $f(x)$ is strictly increasing for $x in mathbb R$ and thus it is "$1-1$", which then means that :



      $$a + sqrt{a^2+1} = b + sqrt{b^2=1} implies a=b$$






      share|cite|improve this answer





















      • Just an added question here. The map is not surjective since $0$ does not have a pre-image?
        – Yadati Kiran
        Dec 9 at 18:31
















      2












      2








      2






      Let $f:mathbb R to mathbb R $ such that :



      $$f(x) = x + sqrt{x^2+1}$$



      Then, it is :



      $$f'(x) = 1 + frac{x}{sqrt{x^2+1}}= frac{sqrt{x^2+1}+x}{sqrt{x^2+1}} > 0, ; forall x in mathbb R $$



      Thus the function $f(x)$ is strictly increasing for $x in mathbb R$ and thus it is "$1-1$", which then means that :



      $$a + sqrt{a^2+1} = b + sqrt{b^2=1} implies a=b$$






      share|cite|improve this answer












      Let $f:mathbb R to mathbb R $ such that :



      $$f(x) = x + sqrt{x^2+1}$$



      Then, it is :



      $$f'(x) = 1 + frac{x}{sqrt{x^2+1}}= frac{sqrt{x^2+1}+x}{sqrt{x^2+1}} > 0, ; forall x in mathbb R $$



      Thus the function $f(x)$ is strictly increasing for $x in mathbb R$ and thus it is "$1-1$", which then means that :



      $$a + sqrt{a^2+1} = b + sqrt{b^2=1} implies a=b$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 9 at 18:20









      Rebellos

      14.4k31245




      14.4k31245












      • Just an added question here. The map is not surjective since $0$ does not have a pre-image?
        – Yadati Kiran
        Dec 9 at 18:31




















      • Just an added question here. The map is not surjective since $0$ does not have a pre-image?
        – Yadati Kiran
        Dec 9 at 18:31


















      Just an added question here. The map is not surjective since $0$ does not have a pre-image?
      – Yadati Kiran
      Dec 9 at 18:31






      Just an added question here. The map is not surjective since $0$ does not have a pre-image?
      – Yadati Kiran
      Dec 9 at 18:31













      2














      Alternatively, for $a,bin Bbb R$,
      begin{align}a+sqrt{a^2+1}&=b+sqrt{b^2+1}\&implies frac{1}{a+sqrt{a^2+1}}=frac{1}{b+sqrt{b^2+1}}wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
      &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
      &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\&implies (a+sqrt{a^2+1})-(sqrt{a^2+1}-a)=(b+sqrt{b^2+1})-(sqrt{b^2+1}-b)
      \&implies 2a=2bimplies a=b. end{align}
      It is also easy to see that $a=bimplies a+sqrt{a^2+1}=b+sqrt{b^2+1}$. That is,
      $$a+sqrt{a^2+1}=b+sqrt{b^2+1}iff a=b.$$






      share|cite|improve this answer


























        2














        Alternatively, for $a,bin Bbb R$,
        begin{align}a+sqrt{a^2+1}&=b+sqrt{b^2+1}\&implies frac{1}{a+sqrt{a^2+1}}=frac{1}{b+sqrt{b^2+1}}wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
        &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
        &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\&implies (a+sqrt{a^2+1})-(sqrt{a^2+1}-a)=(b+sqrt{b^2+1})-(sqrt{b^2+1}-b)
        \&implies 2a=2bimplies a=b. end{align}
        It is also easy to see that $a=bimplies a+sqrt{a^2+1}=b+sqrt{b^2+1}$. That is,
        $$a+sqrt{a^2+1}=b+sqrt{b^2+1}iff a=b.$$






        share|cite|improve this answer
























          2












          2








          2






          Alternatively, for $a,bin Bbb R$,
          begin{align}a+sqrt{a^2+1}&=b+sqrt{b^2+1}\&implies frac{1}{a+sqrt{a^2+1}}=frac{1}{b+sqrt{b^2+1}}wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
          &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
          &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\&implies (a+sqrt{a^2+1})-(sqrt{a^2+1}-a)=(b+sqrt{b^2+1})-(sqrt{b^2+1}-b)
          \&implies 2a=2bimplies a=b. end{align}
          It is also easy to see that $a=bimplies a+sqrt{a^2+1}=b+sqrt{b^2+1}$. That is,
          $$a+sqrt{a^2+1}=b+sqrt{b^2+1}iff a=b.$$






          share|cite|improve this answer












          Alternatively, for $a,bin Bbb R$,
          begin{align}a+sqrt{a^2+1}&=b+sqrt{b^2+1}\&implies frac{1}{a+sqrt{a^2+1}}=frac{1}{b+sqrt{b^2+1}}wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
          &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\
          &implies sqrt{a^2+1}-a=sqrt{b^2+1}-b wedge a+sqrt{a^2+1}=b+sqrt{b^2+1}\&implies (a+sqrt{a^2+1})-(sqrt{a^2+1}-a)=(b+sqrt{b^2+1})-(sqrt{b^2+1}-b)
          \&implies 2a=2bimplies a=b. end{align}
          It is also easy to see that $a=bimplies a+sqrt{a^2+1}=b+sqrt{b^2+1}$. That is,
          $$a+sqrt{a^2+1}=b+sqrt{b^2+1}iff a=b.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 at 18:28









          Snookie

          1,30017




          1,30017























              0














              Hint:



              WLOG $a=cot2A,b=cot2B,0<A,B<dfracpi2$



              we have $cot A=cot Bimpliescot2A=?$






              share|cite|improve this answer


























                0














                Hint:



                WLOG $a=cot2A,b=cot2B,0<A,B<dfracpi2$



                we have $cot A=cot Bimpliescot2A=?$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Hint:



                  WLOG $a=cot2A,b=cot2B,0<A,B<dfracpi2$



                  we have $cot A=cot Bimpliescot2A=?$






                  share|cite|improve this answer












                  Hint:



                  WLOG $a=cot2A,b=cot2B,0<A,B<dfracpi2$



                  we have $cot A=cot Bimpliescot2A=?$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 at 18:18









                  lab bhattacharjee

                  223k15156274




                  223k15156274






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3032720%2fif-a-sqrta21-b-sqrtb21-then-a-b-or-not%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

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more