Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mathslover

  • 2 years ago

How can we say that in golden ratio : \(\frac{a+b}{a} = \frac{a}{b} \) ??

  • This Question is Closed
  1. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is their any proof for ^ the above equation?

  2. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    that is simply how it is defined.

  3. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    you might as well ask; is there any proof that green is green

  4. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You want to say that : (a+b)/a = a/b --> if a > b ^ that is : 1 + b/a = a/b , is universal ?

  5. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry , I meant that : green = green is OK but a/b + 1 = b/a seems hard to agree.

  6. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    usually when we are to prove x = y then it is agreed that YES IT IS LIKE GREEN = GREEN but not exactly, it is like : color of leaf = green (TO PROVE) . I hope you are getting my point sir.

  7. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    take a line of unit length (1); take some part of it and define it as "x"; which leaves us with the rest of it as (1-x)|dw:1350393541608:dw|the golden ratio is defined as the value such that the ratio\[\frac{1}{x}=\frac{x}{1-x}\]

  8. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    by redefining the parts as x = a x-1 = b 1 = a+b we have your setup

  9. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sir, but what is the proof that : 1-x = x^2 ? Are we estimating this?

  10. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Estimation (in the following sense) : In this special case of golden ratio : (a+b)/b = a/b

  11. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    when 1-x = x^2 the ratio of the parts to the whole IS the golden ratio. this is along the same line of thought as: the ratio of the circumference of a circle to its diameter defines pi. How would you prove the C/d = pi ? its not a matter of proof, but of definition

  12. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So can we regard that as , axiom ? postulate? Well I think it can be defined well as 'definition' , correct? btw, a nice example given by you sir.

  13. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    id say "definition" is a good term to use :) we can prove that the value of the golden ratio is what it is by solving for "x"

  14. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK thanks a lot sir!

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.