Talk:Nth root

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"Radix" as a term for the radical sign?

I cannot anywhere find a reference where "radix" has this meaning. I have accordingly deleted it. — Preceding unsigned comment added by 2603:7000:8100:3600:B9CC:9762:DC72:5A8F (talk) 14:22, 9 December 2024 (UTC)[reply]

It's fine to leave this without synonyms here. The word "radix" just means "root" in Latin. Before the √ symbol some authors used ℞ (as an abbreviation for "radix") to mean square root. You can learn more in e.g. Cajori's History of Mathematical Notations. –jacobolus (t) 15:28, 9 December 2024 (UTC)[reply]

Expla… Explodation!

There says somebody things like "10^0·1·0^0·1^2 + 10^1·2·0^1·1^1 ≤ 1 < 10^0·1·0^0·2^2 + 10^1·2·0^1·2^1".

Let’s have a closer look on this:

"10^0·1·0^0·1^2 + 10^1·2·0^1·1^1 ≤ 1 < 10^0·1·0^0·2^2 + 10^1·2·0^1·2^1"

evaluated powers:

"1·1·1·1 + 10·2·1·1 ≤ 1 < 1·1·1·4 + 10·2·1·2"

evaluated multiplications:

"1 + 20 ≤ 1 < 4 + 40"

evaluated additions:

"21 ≤ 1 < 44"

21 < 44, ok; 1 < 44, ok; but 21 < 1⁇

Nice!

In other words: there is a giant lack of explanation! Even without such strange math-“explanation”. — Preceding unsigned comment added by 84.166.39.233 (talk) 02:50, 11 October 2025 (UTC)[reply]

Another concept of root

There should be some mention of another concept of root in complex numbers, where the nth root function is in a certain way multivalued and it is defined as the set of all n roots (which can also be found in the literature). IMO both concepts have their advantages and disadvantages, the advantage of this concept is that then the equation root(a*b)=root(a)*root(a) holds generally, the disadvantage is that set operations would be needed. (Sorry for my English.) PavelTom (talk) 17:31, 19 October 2025 (UTC)[reply]

The fact that every nonzero number has n complex nth roots is mentioned in various places in the article. D.Lazard (talk) 20:04, 19 October 2025 (UTC)[reply]

Polynomial roots

Hello. I added an {{Importance section}} template to this section but I see it has been removed without really addressing my concerns. Regarding this section, is there any doubt that the nth roots of a number (for any integer n) can all be expressed as radicals? If the answer is no (which I think it should be per Demoivre's theorem), then I'm unclear why the content is needed. Sure, it belongs in a general article on polynomials, but this is a very restricted case. Praemonitus (talk) 05:13, 12 November 2025 (UTC)[reply]

The fact that the same word "root" refers to both ⁠${\displaystyle n}$⁠th root and polynomial roots is a common source of confusion. It is the reason of the weird title of the article, and the reason why the article is not titled root (mathematics)]]. There are several easons for which his section is fundamental: are

• The concept of a root of a polynomial is a direct generalization of that of a ⁠${\displaystyle n}$⁠th roots. It is a good Wikipedia practice to have sections on direct generalizations.
• When confusion is possible, this must be clearly clarified in Wikipedia articles
• The relationship between ⁠${\displaystyle n}$⁠th roots and polynomial roots was a fundamental question of algebra during centuries. It remains essential to explain why, outside very elementary mathematics, polynomial roots are much more importnt than ⁠${\displaystyle n}$⁠th roots.
• Solution in radicals is one of the most important application of ⁠${\displaystyle n}$⁠th roots. The section is mainly about this application. The section should be tagged with {{main|Solution in radicals}}, if the target would not be a stub.

I tried to make these points clearer with my new version of the article. Further work is still needed for this. D.Lazard (talk) 09:31, 12 November 2025 (UTC)[reply]

Hmm, well in that case, a point of confusion that needs to be cleared up for the reader is that the lack of a general solution for a quintic function does not preclude extracting all nth roots of a number. I attempted to address this. Thanks. Praemonitus (talk) 15:05, 12 November 2025 (UTC)[reply]

The lack of a general solution does not preclude calculating the roots of any arbitrary polynomial of 5+ degree numerically. –jacobolus (t) 19:20, 12 November 2025 (UTC)[reply]

Polynomials and n roots

The issue has been resolved

I had this idea for explaining why a number has n roots, rather than just one. It works as follows: It may be unclear why a number has n roots rather than just one. To demonstrate this, for the principal root a of the number x taken to the nth power, the following polynomial relation holds: $${\displaystyle p(x)=x^{n}-a^{n}=0}$$ This polynomial can be factorized as follows:[1] $${\displaystyle {\begin{aligned}p(x)&=x^{n}-a^{n}\\&=(x-a)(x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\cdots +a^{n-1})\\&=(x-a)\left(\sum _{k=0}^{n-1}x^{n-k-1}a^{k}\right)\\\end{aligned}}}$$ Thus, the polynomial ${\displaystyle p(x)}$ is zero for x equal to a, or for any x that solves the equation: $${\displaystyle \sum _{k=0}^{n-1}x^{n-k-1}a^{k}=0}$$ By the fundamental theorem of algebra, this series has ${\displaystyle n-1}$ roots. But I'm not sure where to cite it from so that it isn't WP:OR. Any ideas? Praemonitus (talk) 16:27, 12 November 2025 (UTC)[reply] There are two standard proof, one based on complex numbers, the other based on the fundamental theorem of algebra. The proof based on complex numbers results from the fact that, if ⁠${\displaystyle r}$⁠ is an ⁠${\displaystyle n}$⁠th root of ⁠${\displaystyle a}$⁠ (i.e. ⁠${\displaystyle r^{n}=a}$⁠), the ⁠${\displaystyle n}$⁠th roots of ⁠${\displaystyle a}$⁠ are the numbers ⁠${\displaystyle \textstyle re^{2ik\pi /n}}$⁠ for ⁠${\displaystyle k=0,\ldots ,n-1}$⁠. This is not algebraic and requires knowledge of claculus. A corollary of the fundamental theorem of algebra is that a polynomial of degree ⁠${\displaystyle n}$⁠ witout multiple root (a square-free polynomial) has exactly ⁠${\displaystyle n}$⁠ complex roots. The multiple roots are the common roots of the polynomial and its derivative (a consequene of the derivation rule for ⁠${\displaystyle (x-\alpha )^{2}q(x)}$⁠). Here, we have the polynomial ⁠${\displaystyle x^{n}-a}$⁠ whose derivative ⁠${\displaystyle nx^{n-1}}$⁠ has the unique root 0, which is not a ⁠${\displaystyle n}$⁠th root of ⁠${\displaystyle a}$⁠ if ⁠${\displaystyle a\neq 0}$⁠. For completing your proof, you should prove that the series has exactly ⁠${\displaystyle n-1}$⁠ roots; this is a direct consequence of the above proof, but this is difficult to prove directly. D.Lazard (talk) 17:48, 12 November 2025 (UTC)[reply] I'm not really trying to prove anything: just demonstrating to the casual reader that the polynomial form is not as simple as it first appears, which allows for multiple roots. Praemonitus (talk) 00:10, 13 November 2025 (UTC)[reply] References Pender, William; et al. (2011). Cambridge 3 Unit Mathematics Year 11. Cambridge University Press. ISBN 9781107633322.