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Actually, I've already found a clumsy proof of my statement, but I want to know if there's something deeper going on here. Specifically, Mathematica's "Simplify" function doesn't seem to know how to reduce this, which caused me hours of woe before I realized that Mathematica was giving me the answer I was looking for in a shrouded form. Is there a reasonable way one could "spot" this identity? Does it fit within a broader family of square root identities? I hope my curiosity and bewilderment about this is understandable—I've just never come across this before and I'm surprised Mathematica doesn't know what to do with it.

Here's my proof: $$1+\sqrt{6-2\sqrt{5}}=x$$ $$6-2\sqrt{5}=(x-1)^2$$ $$5-2\sqrt{5}=x(x-2)$$ $$\sqrt{5}(\sqrt{5}-2)=x(x-2)$$ $$x=\sqrt{5}$$

As a more practical question, does anyone know how make Mathematica "catch" things like this? To help understand where I'm coming from, I was expecting Mathematica to return the matrix $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ when instead it returned $$\left( \begin{array}{cc} \frac{\sqrt{5 \left(5+\sqrt{5}\right)}+\sqrt{50-20 \sqrt{5}}}{5 \sqrt{5-\sqrt{5}}} & \frac{1}{10} \left(\sqrt{10 \left(3+\sqrt{5}\right)}-\sqrt{5}+5\right) \\ 0 & \frac{\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}}{\sqrt{10}} \\ \end{array} \right)$$

I arrived at the identity above by playing around with the upper-left entry, and I haven't even begun with the others, but Mathematica doesn't seem to realize they are equal to 1 even though a numerical calculation suggests they are. Insight appreciated!

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  • $\begingroup$ Since you're using Mathematica: RootReduce[1 + Sqrt[6 - 2 Sqrt[5]]] $\endgroup$ Oct 25, 2017 at 3:27

2 Answers 2

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Hint: $6-2\sqrt{5}=(\sqrt{5}-1)^2$

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  • $\begingroup$ More explicit: $6 - 2\sqrt5 = (1+5) - 2\sqrt5 = 1^2 - 2\cdot1\cdot\sqrt5 + (\sqrt5)^2 = (1-\sqrt5)^2$. $\endgroup$
    – md2perpe
    Oct 26, 2017 at 7:33
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$$\sqrt{6-2\sqrt{5}}=\sqrt{5}-1$$ $$6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2$$

$$6-2\sqrt{5}=6-2\sqrt{5}$$

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