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Can someone work out why the following are equal:

  • $3 - \phi = \sqrt5/\phi$

  • $\log_{\phi}{(3 - \phi)} = \log_{\phi}{(\sqrt5)} - 1$

  • $\lfloor\log_{\phi}{(3 - \phi)}\rfloor = \lceil log_{\phi}{\sqrt5} \rceil - 2$

Edit:

The context is on another question which J.W. Tanner linked in comments. I have bad mathematical background and I was reading Knuth's Art of Computer Programming. I can't work this out because I have no mathematical proficiency after elementary school, but I can read when someone works it out. My 'effort' would be relearning math from the ground up or asking a 'stupid' question here where I can see what happens in a matter of minutes.

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closed as off-topic by B. Goddard, cmk, Ak19, The Count, Adrian Keister Jul 3 at 1:41

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  • 2
    $\begingroup$ How much are you willing to pay to have us do your homework? $\endgroup$ – B. Goddard Jul 2 at 14:49
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    $\begingroup$ where are you stuck? the first one at least is simple algebra after you plug in the definition of $\varphi$. The second one follows quickly from the definition of logarithm. $\endgroup$ – graeme Jul 2 at 14:50
  • $\begingroup$ It's not a homework I am just interested because I am not proficient in this. I didn't know that this site is only for experts. $\endgroup$ – Michael Munta Jul 2 at 15:15
  • $\begingroup$ Cf. this question $\endgroup$ – J. W. Tanner Jul 2 at 15:45
  • $\begingroup$ It's not for experts. It's for people who show some effort. $\endgroup$ – The Count Jul 3 at 1:17
1
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It all goes back to

$\displaystyle \phi = \frac{1+\sqrt{5}}{2}$

and

$\displaystyle \phi^2 = \phi + 1$

and

$\displaystyle \frac{1}{\phi} = \phi -1 $

So for the first one,

$3-\phi = 2 - (\phi-1) = 2- 1/\phi = (2\phi-1)/\phi = \sqrt{5}/\phi$

The others will all follow from repeated application of those three equations, plus the definition of the logarithm.

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  • $\begingroup$ How does this step happen? $2 - 1/\phi = (2\phi - 1)/\phi$ $\endgroup$ – Michael Munta Jul 2 at 15:28
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    $\begingroup$ They're both $2\phi/\phi-1/\phi$ $\endgroup$ – J. W. Tanner Jul 2 at 15:47

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