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It says on the wikipedia page that...

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I am confused as to why a nested Radical of Depth 1 can only be de-nested if $a^2-c$ is the square of a rational number. Here is my below attempt to reason out why...

Suppose you had the quadratic $x^2-ax+\frac{c}{4}$. Then, by the quadratic formula, the roots are $x = \frac{a+\sqrt{a^2-c}}{2}$ and $x = \frac{a-\sqrt{a^2-c}}{2}$. Let the first root written here be $d$ and the latter root be $e$. Hence, by the product and sum relationships of a polynomial with degree 2 (and also just by algebra)

$$d + e = a $$ $$de = \frac{c}{4}$$

but $$\sqrt{d}+\sqrt{e} = d + 2\sqrt{de} + e$$

Hence

$$\sqrt{d}+\sqrt{e} = \sqrt{a + \sqrt{c}}$$

Now, what I observe is that if $a^2 - c$ is a rational square, then the roots d and e will become rational.

What am I missing?

Percival the caterpillar
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1 Answers1

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$$\begin{aligned} &\sqrt{a+\sqrt{b}} = \sqrt{x} + \sqrt{y} \Leftrightarrow a + \sqrt{b} = x + 2\sqrt{xy} + y \Leftrightarrow\\ &a + \sqrt{b} = x + y + \sqrt{4xy} \Rightarrow \begin{cases} a = x + y \\ b = 4xy \end{cases} \Leftrightarrow\\ &\begin{cases} a^2 = x^2 + 2xy + y^2 \\ b = 4xy \end{cases} \Rightarrow \boxed{a^2 - b = (x - y)^2} \Rightarrow \sqrt{a^2 - b} = x - y \end{aligned}$$

Let $c=\sqrt{a^2-b}$, then $c=x-y$, but we know that $a=x+y$. Then

$$x=\frac{a+c}{2},\quad y=\frac{a-c}{2}\Rightarrow \sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$$

Parfig
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  • I can't see how that relates to my question about why $c^2 = a^2 -b$ has to be rational? – Percival the caterpillar Mar 23 '25 at 01:58
  • Indeed, you didn't answer the question. You argued the converse. But look at what you have written. What is $\sqrt{a^2-b}$ equal to? – Ted Shifrin Mar 23 '25 at 02:09
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    Look carefully, I write $\sqrt{a^2 - b} = x - y$ - this is what I got. Now the question to you, how can we get rid of the root and get an expression $x-y$ that does not contain a root? This can be when $a^2-b$ is a perfect square – Parfig Mar 23 '25 at 02:10
  • You don't need to say all that. If $x$ and $y$ are rational, what can you say about $x-y$? – Ted Shifrin Mar 23 '25 at 02:10
  • I thought it was obvious, since we got that the root is equal to some expression, therefore the expression under the radical must be a perfect square – Parfig Mar 23 '25 at 02:11
  • I highlighted this in the solution if you didn't see it – Parfig Mar 23 '25 at 02:34
  • @Parfig, yes, I noticed that in my attempt as I wrote above. What about it, though? – Percival the caterpillar Mar 23 '25 at 02:37
  • @TedShifrin x - y is rational.... – Percival the caterpillar Mar 23 '25 at 02:37
  • @Parfig: It helps to add words (i.e., sentences) to make deductions explicit. It's about concepts and explanations, not just chains of formulas. – Ted Shifrin Mar 23 '25 at 03:09
  • @Parfig How did you prove the first line $\sqrt{a+\sqrt{b}} = \sqrt{x} + \sqrt{y}$ – Percival the caterpillar Mar 23 '25 at 07:51
  • I assumed it was true and then showed it – Parfig Mar 23 '25 at 13:11
  • This is basically right but there are a couple of issues. First, you should justify the step that $a + \sqrt{b} = x + y + \sqrt{4xy}$ leads to $a = x + y$ and $b = 4xy$. Also a minor thing, $\sqrt{a^2 - b}$ could equal either $y - x$ or $x - y$ so you should specify that $x \geq y$ to be clear. – Zarrax Mar 23 '25 at 14:06
  • @Parfig thanks for the effort to produce an answer, by the way. You said "I assumed it was true and then showed it". What do you mean by "it"? How can you assume the first line though? What if there was a nested radical that could equal a non-surdic irrational number? – Percival the caterpillar Mar 24 '25 at 00:22