Challenge #10 Solution

Have you tried solving Challenge #10 yet? Go try it first if you haven’t. It’s not too hard, I promise!

All of these can be shown by just plugging in the values of $\varphi$ and $\hat{\varphi}$ , but I’d like to show you some easier ways.

The first problem is easy — we know that $\varphi$ is a solution to the equation $x^2 - x - 1 = 0$ ; that is, $\varphi^2 - \varphi - 1 = 0$ . Some simple algebra shows that $\varphi^2 = \varphi + 1$ . This is kind of interesting when you think about it. In English, this says that squaring $\varphi$ is exactly the same as adding 1.

Problems 2 and 3 can be solved simultaneously. Since $\varphi$ and $\hat{\varphi}$ are the roots of the polynomial $x^2 - x - 1$ , we know it can be factored as

$x^2 - x - 1 = (x - \varphi)(x - \hat{\varphi})$ .

Multiplying out the right side, we get

$x^2 - x - 1 = x^2 - (\varphi + \hat{\varphi})x + \varphi \hat{\varphi}$

For this equation to be true, we obviously must have $\varphi + \hat{\varphi} = 1$ and $\varphi \hat{\varphi} = -1$ .

There is not a nice slick way to prove #4 (at least, not that I am aware of). But doing it by plugging in values isn’t so bad:

$\displaystyle \varphi - \hat{\varphi} = \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$ .

Next up: we’re going to use these properties to prove some astonishing facts about the relationship between the golden ratio and Fibonacci numbers! Stay tuned!

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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3 Responses to Challenge #10 Solution

Pseudonym says:

September 9, 2007 at 3:39 am

If you correct the broken LaTeX in my response to the challenge, you can see a cleaner way to prove property 4.
Pseudonym says:

September 9, 2007 at 7:46 am

OK, here’s the proof of property 4 without fixing the LaTeX. I’m going to use φ and φ’ as the two solutions.

First, we need to find out what φ^2+φ’^2 is:

φ^2+φ’^2

And now we can work out:

(φ-φ’)^2
= φ^2 + 2φφ’ + φ’^2
= (φ + 1) + 2φφ’ + (φ’ + 1) (property 1)
= (φ + φ’) + 2 + 2 (regrouping, and property 3)
= 1 + 2 + 2 (property 2)
= 5

And now, because we know that φ>φ’, we take the positive square root.

Oh, by the way. Is there any way that you could add comment previews to this blog? If you’re encouraging commenters to put maths in their responses, it would really help if we could see mistakes before they get published.
Brent says:

September 11, 2007 at 4:27 pm

Pseudonym: nice, I like it! I hadn’t thought of that method of proof. As for comment previews, I don’t see an easy way to do that at the moment. I suppose the real issue is that I need to upgrade to a more recent version of WordPress, perhaps it will support comment previews! I agree that would be a nice feature and I’ll look into it.

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