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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edited 2 hours ago

lab bhattacharjee

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asked 2 hours ago

Beel

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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

edited 2 hours ago

lab bhattacharjee

217k14153267

asked 2 hours ago

Beel

333

add a comment | 

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

edited 2 hours ago

lab bhattacharjee

217k14153267

asked 2 hours ago

Beel

333

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

share|cite|improve this question

edited 2 hours ago

lab bhattacharjee

217k14153267

asked 2 hours ago

Beel

333

share|cite|improve this question

edited 2 hours ago

lab bhattacharjee

217k14153267

asked 2 hours ago

Beel

333

share|cite|improve this question

share|cite|improve this question

edited 2 hours ago

lab bhattacharjee

217k14153267

edited 2 hours ago

lab bhattacharjee

217k14153267

edited 2 hours ago

lab bhattacharjee

217k14153267

217k14153267

asked 2 hours ago

Beel

333

asked 2 hours ago

Beel

333

asked 2 hours ago

Beel

333

333

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add a comment | 

4 Answers
4

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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

answered 2 hours ago

lab bhattacharjee

217k14153267

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
2
down vote

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

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answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

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up vote
2
down vote

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

share|cite|improve this answer

answered 2 hours ago

Henry

95.2k473151

add a comment | 

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

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

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

answered 2 hours ago

lab bhattacharjee

217k14153267

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 2 hours ago

lab bhattacharjee

217k14153267

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 2 hours ago

lab bhattacharjee

217k14153267

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

answered 2 hours ago

lab bhattacharjee

217k14153267

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

lab bhattacharjee

217k14153267

answered 2 hours ago

lab bhattacharjee

217k14153267

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

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

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

add a comment | 

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

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

add a comment | 

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

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

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

share|cite|improve this answer

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

answered 2 hours ago

Dr. Sonnhard Graubner

70.6k32863

70.6k32863

add a comment | 

add a comment | 

up vote
2
down vote

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

share|cite|improve this answer

answered 2 hours ago

Henry

95.2k473151

add a comment | 

up vote
2
down vote

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

share|cite|improve this answer

answered 2 hours ago

Henry

95.2k473151

add a comment | 

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

answered 2 hours ago

Henry

95.2k473151

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

share|cite|improve this answer

answered 2 hours ago

Henry

95.2k473151

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

Henry

95.2k473151

answered 2 hours ago

Henry

95.2k473151

answered 2 hours ago

Henry

95.2k473151

95.2k473151

add a comment | 

add a comment | 

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

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

add a comment | 

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

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

add a comment | 

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

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

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

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 2 hours ago

Bill Dubuque

204k29189613

answered 2 hours ago

Bill Dubuque

204k29189613

answered 2 hours ago

Bill Dubuque

204k29189613

204k29189613

add a comment | 

add a comment | 

 

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