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







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  • 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






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share|cite|improve this question













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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





















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      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






























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                      Cabo Verde

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                        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)
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                      Gyllenstierna

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