How to proceed with this step? Algebra

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I am doing a proof of Fibonacci's sequence. How do I get from



$dfrac1+√5 2 + 1$ to $left(dfrac1+√52right)^2$



Thank you






algebra-precalculus







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    up vote
    1
    down vote

    favorite












    I am doing a proof of Fibonacci's sequence. How do I get from



    $dfrac1+√5 2 + 1$ to $left(dfrac1+√52right)^2$



    Thank you






    algebra-precalculus







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am doing a proof of Fibonacci's sequence. How do I get from



      $dfrac1+√5 2 + 1$ to $left(dfrac1+√52right)^2$



      Thank you






      algebra-precalculus







      share|cite|improve this question















      I am doing a proof of Fibonacci's sequence. How do I get from



      $dfrac1+√5 2 + 1$ to $left(dfrac1+√52right)^2$



      Thank you






      algebra-precalculus




      algebra-precalculus






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      share|cite|improve this question













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      edited 2 hours ago









      lab bhattacharjee

      217k14153267




      217k14153267










      asked 2 hours ago









      Beel

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          4 Answers
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          If $t=dfrac1+sqrt52implies 2t-1=sqrt5$



          Square both sides to find $t^2=t+1$



          Alternatively, $$(1+sqrt5)^2=6+2sqrt5=4+2(1+sqrt5)$$



          Divide both sides by $4$






          share|cite|improve this answer




















          • Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
            – lab bhattacharjee
            2 hours ago

















          up vote
          2
          down vote













          We get $$frac1+sqrt52+1=frac1+2+sqrt52=frac3+sqrt52$$ and
          $$left(frac1+sqrt52right)^2=frac1+5+2sqrt54=…$$






          share|cite|improve this answer



























            up vote
            2
            down vote













            $$frac1+sqrt52+1=frac3+sqrt52=frac6+2sqrt54= frac1+2sqrt5+54 = left(frac1+sqrt52right)^2$$






            share|cite|improve this answer



























              up vote
              0
              down vote













              Let $w$ be a quadratic number, e.g. $w = (1!+!sqrt5)/2,,$ so $,bar w = (1!-!sqrt5)/2.,$ Then



              $$ beginalignoverbrace(w+bar w)^large 1, w - overbracebar w,w^large -1 , = w^2\
              w + 1 = w^2
              endalign$$



              i.e. $w,bar w$ are roots of $ (x !-! bar w)(x! -! bar w), =, x^2 - (w! +! bar w), x + w bar w $ (minimal polynomial)



              When working with algebraic numbers it is usually helpful to know their minimal polynomials, since they imply all other algebraic equations satisfied by the numbers.






              share|cite|improve this answer






















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                4 Answers
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                active

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                4 Answers
                4






                active

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                active

                oldest

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                up vote
                2
                down vote













                If $t=dfrac1+sqrt52implies 2t-1=sqrt5$



                Square both sides to find $t^2=t+1$



                Alternatively, $$(1+sqrt5)^2=6+2sqrt5=4+2(1+sqrt5)$$



                Divide both sides by $4$






                share|cite|improve this answer




















                • Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                  – lab bhattacharjee
                  2 hours ago














                up vote
                2
                down vote













                If $t=dfrac1+sqrt52implies 2t-1=sqrt5$



                Square both sides to find $t^2=t+1$



                Alternatively, $$(1+sqrt5)^2=6+2sqrt5=4+2(1+sqrt5)$$



                Divide both sides by $4$






                share|cite|improve this answer




















                • Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                  – lab bhattacharjee
                  2 hours ago












                up vote
                2
                down vote










                up vote
                2
                down vote









                If $t=dfrac1+sqrt52implies 2t-1=sqrt5$



                Square both sides to find $t^2=t+1$



                Alternatively, $$(1+sqrt5)^2=6+2sqrt5=4+2(1+sqrt5)$$



                Divide both sides by $4$






                share|cite|improve this answer












                If $t=dfrac1+sqrt52implies 2t-1=sqrt5$



                Square both sides to find $t^2=t+1$



                Alternatively, $$(1+sqrt5)^2=6+2sqrt5=4+2(1+sqrt5)$$



                Divide both sides by $4$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                lab bhattacharjee

                217k14153267




                217k14153267











                • Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                  – lab bhattacharjee
                  2 hours ago
















                • Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                  – lab bhattacharjee
                  2 hours ago















                Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                – lab bhattacharjee
                2 hours ago




                Observe that the other root of $$t^2-t-1=0$$ also satisfy the same relation
                – lab bhattacharjee
                2 hours ago










                up vote
                2
                down vote













                We get $$frac1+sqrt52+1=frac1+2+sqrt52=frac3+sqrt52$$ and
                $$left(frac1+sqrt52right)^2=frac1+5+2sqrt54=…$$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  We get $$frac1+sqrt52+1=frac1+2+sqrt52=frac3+sqrt52$$ and
                  $$left(frac1+sqrt52right)^2=frac1+5+2sqrt54=…$$






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    We get $$frac1+sqrt52+1=frac1+2+sqrt52=frac3+sqrt52$$ and
                    $$left(frac1+sqrt52right)^2=frac1+5+2sqrt54=…$$






                    share|cite|improve this answer












                    We get $$frac1+sqrt52+1=frac1+2+sqrt52=frac3+sqrt52$$ and
                    $$left(frac1+sqrt52right)^2=frac1+5+2sqrt54=…$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Dr. Sonnhard Graubner

                    70.6k32863




                    70.6k32863




















                        up vote
                        2
                        down vote













                        $$frac1+sqrt52+1=frac3+sqrt52=frac6+2sqrt54= frac1+2sqrt5+54 = left(frac1+sqrt52right)^2$$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          $$frac1+sqrt52+1=frac3+sqrt52=frac6+2sqrt54= frac1+2sqrt5+54 = left(frac1+sqrt52right)^2$$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            $$frac1+sqrt52+1=frac3+sqrt52=frac6+2sqrt54= frac1+2sqrt5+54 = left(frac1+sqrt52right)^2$$






                            share|cite|improve this answer












                            $$frac1+sqrt52+1=frac3+sqrt52=frac6+2sqrt54= frac1+2sqrt5+54 = left(frac1+sqrt52right)^2$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Henry

                            95.2k473151




                            95.2k473151




















                                up vote
                                0
                                down vote













                                Let $w$ be a quadratic number, e.g. $w = (1!+!sqrt5)/2,,$ so $,bar w = (1!-!sqrt5)/2.,$ Then



                                $$ beginalignoverbrace(w+bar w)^large 1, w - overbracebar w,w^large -1 , = w^2\
                                w + 1 = w^2
                                endalign$$



                                i.e. $w,bar w$ are roots of $ (x !-! bar w)(x! -! bar w), =, x^2 - (w! +! bar w), x + w bar w $ (minimal polynomial)



                                When working with algebraic numbers it is usually helpful to know their minimal polynomials, since they imply all other algebraic equations satisfied by the numbers.






                                share|cite|improve this answer


























                                  up vote
                                  0
                                  down vote













                                  Let $w$ be a quadratic number, e.g. $w = (1!+!sqrt5)/2,,$ so $,bar w = (1!-!sqrt5)/2.,$ Then



                                  $$ beginalignoverbrace(w+bar w)^large 1, w - overbracebar w,w^large -1 , = w^2\
                                  w + 1 = w^2
                                  endalign$$



                                  i.e. $w,bar w$ are roots of $ (x !-! bar w)(x! -! bar w), =, x^2 - (w! +! bar w), x + w bar w $ (minimal polynomial)



                                  When working with algebraic numbers it is usually helpful to know their minimal polynomials, since they imply all other algebraic equations satisfied by the numbers.






                                  share|cite|improve this answer
























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    Let $w$ be a quadratic number, e.g. $w = (1!+!sqrt5)/2,,$ so $,bar w = (1!-!sqrt5)/2.,$ Then



                                    $$ beginalignoverbrace(w+bar w)^large 1, w - overbracebar w,w^large -1 , = w^2\
                                    w + 1 = w^2
                                    endalign$$



                                    i.e. $w,bar w$ are roots of $ (x !-! bar w)(x! -! bar w), =, x^2 - (w! +! bar w), x + w bar w $ (minimal polynomial)



                                    When working with algebraic numbers it is usually helpful to know their minimal polynomials, since they imply all other algebraic equations satisfied by the numbers.






                                    share|cite|improve this answer














                                    Let $w$ be a quadratic number, e.g. $w = (1!+!sqrt5)/2,,$ so $,bar w = (1!-!sqrt5)/2.,$ Then



                                    $$ beginalignoverbrace(w+bar w)^large 1, w - overbracebar w,w^large -1 , = w^2\
                                    w + 1 = w^2
                                    endalign$$



                                    i.e. $w,bar w$ are roots of $ (x !-! bar w)(x! -! bar w), =, x^2 - (w! +! bar w), x + w bar w $ (minimal polynomial)



                                    When working with algebraic numbers it is usually helpful to know their minimal polynomials, since they imply all other algebraic equations satisfied by the numbers.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 1 hour ago

























                                    answered 2 hours ago









                                    Bill Dubuque

                                    204k29189613




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