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Homework Help: Golden ratio

  1. Feb 4, 2009 #1
    Im working on a part off my course and it covers this, but its not clear.

    [tex]\phi[/tex]= half (1+[tex]\sqrt{5}[/tex])

    [tex]\varphi[/tex]=half (1-[tex]\sqrt{5}[/tex])


    The question asks [tex]\phi[/tex]-[tex]\varphi[/tex] =[tex]\sqrt{5}[/tex]

    It is writen in my book, the answer but it does not explain how the maths cancels and manipilates.

    Could you show me a way that the answer is derived.
     
  2. jcsd
  3. Feb 4, 2009 #2

    cristo

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    What do you get if you try and calculate [itex]\phi-\varphi[/itex] ?
     
  4. Feb 4, 2009 #3
    the question also say's use the exact forms of each form off the golden ratio to verify the following propertys of[tex]\phi[/tex] and [tex]\varphi[/tex]
     
  5. Feb 4, 2009 #4
    half (1+[tex]\sqrt{5}[/tex]) -half(1-[tex]\sqrt{5}[/tex])

    = half [tex]\sqrt{5}[/tex]+half [tex]\sqrt{5}[/tex]= [tex]\sqrt{5}[/tex]

    Is the answer i have in my book but im lost to how and why its that way.
     
  6. Feb 4, 2009 #5
    Which part confuses you? The fact that the 1/2 - 1/2 = 0 or the fact that 1/2*sqrt(5) + 1/2*sqrt(5) = sqrt(5)?
     
  7. Feb 4, 2009 #6
    its the part that makes the 1/2 -1/2 =0 why is the 1+sqrt(5) and the 1-sqrt (5) taken out off the equation what dicided this.
     
  8. Feb 4, 2009 #7

    Dick

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    half(1+sqrt(5))=(1+sqrt(5))/2=1/2+sqrt(5)/2.
    half(1-sqrt(5))=(1-sqrt(5))/2=1/2-sqrt(5)/2. Subtract them.
     
  9. Feb 5, 2009 #8

    HallsofIvy

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    Or would it help to write it as
    [tex]\frac{1+ \sqrt{5}}{2}= \frac{1}{2}+ \frac{\sqrt{5}}{2}[/tex]
     
  10. Feb 5, 2009 #9
    so the 2's cancels out but does that not leave it as it was.
     
  11. Feb 6, 2009 #10

    Mentallic

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    [tex]\phi = \frac{1+ \sqrt{5}}{2}[/tex]

    [tex]\varphi = \frac{1- \sqrt{5}}{2}[/tex]

    Therefore, [tex]\phi - \varphi = \frac{1+ \sqrt{5}}{2} - \frac{1- \sqrt{5}}{2}[/tex]

    If you cannot understand how to simplify this to get your answer of [tex]\sqrt{5}[/tex] then maybe manipulating the fractions in the same way hallsofivy has done will help you out.

    [tex]\frac{1+ \sqrt{5}}{2} - \frac{1- \sqrt{5}}{2} = \frac{1}{2}+ \frac{\sqrt{5}}{2} - (\frac{1}{2} - \frac{\sqrt{5}}{2})[/tex]
     
  12. Feb 6, 2009 #11

    HallsofIvy

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    I wouldn't use the word "cancel": for any number a,
    [tex]\frac{a}{2}+ \frac{a}{2}= a(\frac{1}{2}+ \frac{1}{2})= a(\frac{2}{2}= a(1)= a[/itex]
    It's just a matter of "one plus one equals 2"!
     
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