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I posted a similar but not exact question yesterday, also about a theorem commonly used to de-nest nested radicals - in any case, this one is slightly different. Below is a wikipedia proof https://en.wikipedia.org/wiki/Nested_radical of the following theorem $$\sqrt{a\pm\sqrt{c}} = \sqrt{x}\pm\sqrt{y}$$

where $a,c,x,y$ are rational numbers and c is not a rational square, and where $a^2 - c$ is a rational square. When I read the rest of the proof, it seems to only prove PART of the theorem - that is, if $a^2 - c$ is a rational square, then $x$ and $y$ are rational (which Wiki succeeds in proving lucidly). It seems to me however, that the proof neglects to show $\sqrt{a\pm\sqrt{c}} = \sqrt{x}\pm\sqrt{y}$ which they have blatantly assumed in the first line. In other words, how could the proof-writer just immediately assume right off the bat that a surd inside a surd can be expressed as a sum of seperate surds? Or am I just stupid (likely, yes)?

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Blue
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Percival the caterpillar
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1 Answers1

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It's an assumption that is then used to find cases of $a$ and $c$ for which it's true. The assumption comes from noticing that if $x$ and $y$ are rational, then $$(\sqrt{x} \pm \sqrt{y})^2 = (x + y) \pm \sqrt{4xy}$$ which is in the form of $a \pm \sqrt{c}$.

There is a much more rigorous, non-elementary proof for the general case in this answer by Bill Dubuque using Galois theory of radical (Kummer) extensions along with some useful resources (How to Tangle with Nested Radicals, some Bloemer papers, etc.)

FishDrowned
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  • "The assumption comes from noticing that if and are rational, then$\sqrt{x} \pm \sqrt{y})^2 = (x + y) \pm \sqrt{4xy}$" - but where did the assumption x and y are rational come from? What if there is a case of a nested radical depth 1 that happens to equal the sum of roots of x, y where x and y are non-rational, non-surdic numbers? – Percival the caterpillar Mar 23 '25 at 23:53
  • Another doubt - "$(x + y) \pm \sqrt{4xy} \Rightarrow$ it is in the form $ a\pm \sqrt{c}$" doesn't mean $ a\pm \sqrt{c} \Rightarrow $it is in the form$ (x + y) \pm \sqrt{4xy}$...right? – Percival the caterpillar Mar 24 '25 at 00:00
  • Does the galois theory proof address these points? I am a beginner, unfortunately...do you perhaps know of any elementary proofs that address these points? – Percival the caterpillar Mar 24 '25 at 00:05
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    @Percivalthecaterpillar $x$ and $y$ being rational isn't an 'assumption', it's a given, we're defining $x,y$ to be rational. I suggest you read some proofs to get yourself familiar with them. – FishDrowned Mar 24 '25 at 11:58
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    @Percivalthecaterpillar Correct, $a \pm \sqrt{c}$ isn't always in that form. It's only true if $a^2 - c$ is a perfect square. – FishDrowned Mar 24 '25 at 12:00
  • okay. so because of that observation you made above in your answer, the wiki proof decides its a good idea to assume that $\sqrt{a\pm\sqrt{c}} = \sqrt{x}\pm\sqrt{y}$ where x, y are rational, then it does all that reasoning shown in the screenshot until it finally deduces that, if the assumption is true, then the condition $a^2 - c$ is a rational square must be true. And if we look again, we see that the converse of that must be true. So whenever, we see a nested radical that has $a^2-c$ is rational square, then we know it can be expressed as sum of surds as such. Am I thinking correctly? – Percival the caterpillar Mar 24 '25 at 23:14
  • @Percivalthecaterpillar Yes exactly. – FishDrowned Mar 25 '25 at 00:44