# Golden ratio properties [closed]

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.

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

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• How much are you willing to pay to have us do your homework? – B. Goddard Jul 2 at 14:49
• 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. – graeme Jul 2 at 14:50
• 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. – Michael Munta Jul 2 at 15:15
• – J. W. Tanner Jul 2 at 15:45
• It's not for experts. It's for people who show some effort. – The Count Jul 3 at 1:17

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.

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