Question

Why does $\sqrt{x^2}=|x|$?

score 421 · Answer 1 · 2014-03-18 07:14:05

421

Views

Why does $\sqrt{x^2}=|x|$?

Published on 18 Mar 2014 - 7:14

score 2.9k · Answer 2 · 2026-03-25 21:48:17

2.9k

Views

Is every left maximal ideal the annihilator of a simple left module?

Published on 25 Mar 2026 - 21:48

score 9.8k · Answer 3 · 2026-03-25 12:33:03

9.8k

Views

Converting sums of square-roots to nested square-roots

Published on 25 Mar 2026 - 12:33

score 7.3k · Answer 4 · 2014-03-21 01:37:01

7.3k

Views

What does it mean for a polynomial to be solvable by radicals?

Published on 21 Mar 2014 - 1:37

score 3.7k · Answer 5 · 2026-03-22 22:12:12

3.7k

Views

Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

Published on 22 Mar 2026 - 22:12

score 567 · Answer 6 · 2014-03-22 15:51:55

567

Views

Proving that the limit of $\root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n}$ is $\max(a_1,...a_k)$

Published on 22 Mar 2014 - 15:51

score 223 · Answer 7 · 2014-03-24 07:26:43

223

Views

Why does $\sqrt[\leftroot{-2}\uproot{2}b]{x^a} = x^{a/b}$ and what does it mean?

Published on 24 Mar 2014 - 7:26

score 100 · Answer 8 · 2026-03-27 21:20:29

100

Views

Proof of $\lim_{n \to \infty} a^{1/n} = 1$ and $\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$ where $a>1$

Published on 27 Mar 2026 - 21:20

score 254 · Answer 9 · 2014-03-25 13:39:58

254

Views

-0.73th root not working?

Published on 25 Mar 2014 - 13:39

score 1.6k · Answer 10 · 2026-03-25 07:41:55

1.6k

Views

Proving that $\sqrt{2}+\sqrt{3}$ is irrational

Published on 25 Mar 2026 - 7:41

score 1k · Answer 11 · 2026-03-25 11:15:53

1k

Views

A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$

Published on 25 Mar 2026 - 11:15

score 94 · Answer 12 · 2026-03-27 21:16:46

94

Views

Why expression under root has to be positive?

Published on 27 Mar 2026 - 21:16

score 30 · Answer 13 · 2014-03-30 03:12:42

30

Views

How to find the Domain for this function

Published on 30 Mar 2014 - 3:12

score 180 · Answer 14 · 2014-04-02 10:41:38

180

Views

Show that $\frac{1}{2 \sqrt{n+1}}\le \sqrt{n+1} - \sqrt{n} \le \frac{1}{2 \sqrt{n}} $

Published on 02 Apr 2014 - 10:41

score 422 · Answer 15 · 2026-03-28 10:56:23

422

Views

Limits with cubic root without L’Hopital’s Rule: $\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $

Published on 28 Mar 2026 - 10:56

List Question

Why does $\sqrt{x^2}=|x|$?

Is every left maximal ideal the annihilator of a simple left module?

Converting sums of square-roots to nested square-roots

What does it mean for a polynomial to be solvable by radicals?

Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

Proving that the limit of $\root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n}$ is $\max(a_1,...a_k)$

Why does $\sqrt[\leftroot{-2}\uproot{2}b]{x^a} = x^{a/b}$ and what does it mean?

Proof of $\lim_{n \to \infty} a^{1/n} = 1$ and $\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$ where $a>1$

-0.73th root not working?

Proving that $\sqrt{2}+\sqrt{3}$ is irrational

A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$

Why expression under root has to be positive?

How to find the Domain for this function

Show that $\frac{1}{2 \sqrt{n+1}}\le \sqrt{n+1} - \sqrt{n} \le \frac{1}{2 \sqrt{n}} $

Limits with cubic root without L’Hopital’s Rule: $\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $

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