Could someone verify how this simplifies?

[TheGreatEscapegoat]

I don't know why but I'm having this mental block where I can't see how this simplifies.

I have the expression $$ \frac{1}{2}(1+ \sqrt{5}) $$. Now, I substitute for x into the expression $$ \sqrt{x+1}. $$ to make $$ \sqrt{\frac{1}{2}(1+ \sqrt{5})+1}. $$
The result should be $$\frac{1}{2}( \sqrt{5}+1)$$.
However, when I simplify it, I obtain $$ \sqrt{\frac{1}{2}( \sqrt{5}+3)} $$ and for some reason I am not seeing how to simplify that into $$ \frac{1}{2}( \sqrt{5}+1). $$

 

[lurflurf]

notice that

$$x^2=x+1$$

verify this

that is what makes the golden ratio a special number

https://en.wikipedia.org/wiki/Golden_ratioto denest in general observe
https://en.wikipedia.org/wiki/Nested_radical#Denesting_nested_radicals

$$\sqrt{a\pm b \sqrt c}=(\sqrt{d} \pm \sqrt{e})^2=\sqrt{d} \pm \sqrt{e}$$
implies
$$a \pm b \sqrt c=(\sqrt{d}\pm \sqrt{e})^2 =d+e \pm 2 \sqrt{d e}$$
then equate like terms

 

[TheGreatEscapegoat]

notice that

$$x^2=x+1$$

verify this

that is what makes the golden ratio a special number

https://en.wikipedia.org/wiki/Golden_ratio

I don't see how that proves the simplification. If you don't know how to prove it that's fine I have nothing against you for it, and this isn't meant to be insulting, it's just very commonly held that it is rude to interject random comments for your own personal sake that don't actually address the topic.

 

[phinds]

when I simplify it, I obtain $$ \sqrt{\frac{1}{2}( \sqrt{5}+3)} $$ and for some reason I am not seeing how to simplify that into $$ \frac{1}{2}( \sqrt{5}+1). $$

It would be astounding if your COULD see it since it is not true

EDIT: HM ... seems I was mistaken

 

[TheGreatEscapegoat]

http://www.wolframalpha.com/input/?i=(1/2)(1+sqrt(5))

http://www.wolframalpha.com/input/?i=sqrt((1/2)(1+sqrt(5))+1)

 

[berkeman]

It would be astounding if your COULD see it since it is not true

I thought it was wrong too, so I checked it with my calculator. Seems it is true, but I'm not seeing the algebra to make it true yet...

 

[phinds]

I thought it was wrong too, so I checked it with my calculator. Seems it is true, but I'm not seeing the algebra to make it true yet...

Yep. Me too.

 

[phinds]

Ha. Got it. It does simplify. I think lurflurf knew that all along

 

[TheGreatEscapegoat]

Ha. Got it. It does simplify. I think lurflurf knew that all along

That's a fine conjecture, but you're missing proof, so for all I know you are wrong. Remember that the goal is to prove the simplification, not that it is a fixed point of $$ f(x) = \sqrt{x+1}$$

 

[phinds]

That's a fine conjecture, but you're missing proof, so for all I know you are wrong.

You may consider it to be so if you wish. lurflurf gave you the only hint you need

 

[TheGreatEscapegoat]

You may consider it to be so if you wish. lurflurf gave you the only hint you need

Again, that comment doesn't mean anything without proof. Furthermore, it is rude and counter-productive of you to assume you have the right to arbitrarily dictate what others need and don't need, you only determine what you need.

 

[phinds]

Again, that comment does mean anything without proof. Furthermore, you don't determine what others need and don't need, you only determine what you need.

I think you misunderstand the methodology of PF. We don't give answers, we help people figure out how to GET the answer. You've been given a sufficient clue. Ignore it if you so choose.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[jedishrfu]

You can verify that they are equivalent by squaring both values.

The first gives 1/2 ( sqrt 5 + 3 )

And the second gives you 1/4 ( sqrt 5 + 1 ) * ( sqrt 5 + 1 )

Which becomes 1/4 ( 5 + 2*sqrt5 + 1 ) which becomes 1/4 ( 6 + 2*sqrt5)
Which reduces to 2/4 ( sqrt5 + 3 ) or more simply 1/2 ( sqrt 5 + 3 )

So they are indeed equal.

By reversing the verification you could derive the first expression from the second.

 

[Mark44]

That's a fine conjecture, but you're missing proof, so for all I know you are wrong.

You may consider it to be so if you wish.

Again, that comment doesn't mean anything without proof.

I think you misunderstand the methodology of PF. We don't give answers, we help people figure out how to GET the answer.

Since this is your (@TheGreatEscapegoat) problem, the burden of proof is on you, not us.
@phinds is correct in what he says about our principles here at PF.

 

[berkeman]

Thread closed temporarily for Moderation...

After a long PM conversation, the OP has left the building. Thank you to everyone who tried to help him in this thread. Interesting links, BTW.