Question

Medals, fandom and general thankfulnes will be given if you can explain why

Answer 1

why?

Answer 2

\[\left( \frac{ 1-\sqrt{5} }{ 2 } \right)^{2}-\frac{ 1-\sqrt{5} }{ 2 }-1=\frac{ 1+5-2\sqrt{5} }{4}-\frac{ 1-\sqrt{5} }{2 }\]

Answer 3

Numerator:(1-√5)²=1²-2*1*√5+(√5)²=1+5-2√5 (using (a-b)²=a²-2ab+b²)
Denominator: 2²=4 ;)

Answer 4

Of course now you have to subract the two fractions to see what happens next...

Answer 5

When you have something ugly like this, sometimes it is a bit more convenient to rewrite it with letters instead of radical signs and so forth. For example, let a = 1, b = sqrt(5), c = 2, and write

\[(frac{a-b}{c})^2 - \frac{a-b}{c} - 1\] do the algebra, and then plug in the values.

Answer 6

@memand: you've lost the "-1"at the end, I think.

Answer 7

Dang it, where did my preview go?

\[\(\frac{a-b}{c})^2-\frac{a-b}{c}-1\]

Answer 8

Happens all the time...

Answer 9

@ZeHanz I'm just trying to get my head around your first reply for starters :)
But yes I have to continue it since I have to prove that \[\phi'^{2} -\phi'-1=0\]

Answer 10

And some other stuff, but first I really have to understand how to reduce stuff like that :)

Answer 11

But I don't see where the 5 comes from in 1+5-2....

Answer 12

Oh, now I see writing\[\left( \frac{ a-b }{ c } \right)^{2}\]is the same as writing \[\frac{ \left( a-b \right)^2 }{ c^2 }\]?

Answer 13

It is...

Answer 14

Cool, then I am ready to prove phi :) (I hope :P )

Answer 15

You can do it!

Answer 16

I did ;)