How is \(\phi - \varphi = \sqrt{5}\) derived in the context of the Golden Ratio?

[morbello]

Im working on a part off my course and it covers this, but its not clear.

$$\phi$$= half (1+$$\sqrt{5}$$)

$$\varphi$$=half (1-$$\sqrt{5}$$)The question asks $$\phi$$-$$\varphi$$ =$$\sqrt{5}$$

It is written 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.

 

[cristo]

What do you get if you try and calculate $\phi-\varphi$ ?

 

[morbello]

the question also say's use the exact forms of each form off the golden ratio to verify the following propertys of$$\phi$$ and $$\varphi$$

 

[morbello]

half (1+$$\sqrt{5}$$) -half(1-$$\sqrt{5}$$)

= half $$\sqrt{5}$$+half $$\sqrt{5}$$= $$\sqrt{5}$$

Is the answer i have in my book but I am lost to how and why its that way.

 

[NoMoreExams]

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)?

 

[morbello]

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.

 

[Dick]

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.

 

[HallsofIvy]

Or would it help to write it as
$$\frac{1+ \sqrt{5}}{2}= \frac{1}{2}+ \frac{\sqrt{5}}{2}$$

 

[morbello]

so the 2's cancels out but does that not leave it as it was.

 

[Mentallic]

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

$$\varphi = \frac{1- \sqrt{5}}{2}$$

Therefore, $$\phi - \varphi = \frac{1+ \sqrt{5}}{2} - \frac{1- \sqrt{5}}{2}$$

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

$$\frac{1+ \sqrt{5}}{2} - \frac{1- \sqrt{5}}{2} = \frac{1}{2}+ \frac{\sqrt{5}}{2} - (\frac{1}{2} - \frac{\sqrt{5}}{2})$$

 

[HallsofIvy]

so the 2's cancels out but does that not leave it as it was.

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] <br /> It's just a matter of "one plus one equals 2"![/tex]