Recent content by Ragnarok7

Thank you! That is great and very simple. Can I ask how you got the idea to do it this way? I would never have thought of it.

Prove that $$\lfloor nx \rfloor = \sum_{k=0}^{n-1}\lfloor x+k/n \rfloor$$. Note $$\lfloor x\rfloor$$ means the greatest integer less than or equal to $$x$$. I proved the cases where n=2 and n=3 by writing $$x=\lfloor x\rfloor + \{x\}$$, where $$\{x\}$$ is the fractional part of $$x$$, and then...

Thank you! I knew that fact but I didn't think of doing it like that. I can do the other one now.

I'm unsure if this is a calculus or precalculus topic, but it's from a calculus book, so I'm putting it here. (Note $$\lfloor x \rfloor$$ means the floor of $$x$$ or the greatest integer less than or equal to $$x$$.) Prove that $$\lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor...

Let $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$ be the field generated by elements of the form $$a+b\sqrt{2}+c\sqrt{3}$$, where $$a,b,c\in\mathbb{Q}$$. Prove that $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$ is a vector space of dimension 4 over $$\mathbb{Q}$$. Find a basis for $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$. I...

Thank you so much Euge! That is a great proof and I didn't think it could be so simple! I just realized that I forgot to define $$M_p$$. I'll do it now just for completeness. If $$p$$ is a nonzero integer, then the $$p$$th-power mean of $$n$$ positive real numbers $$x_1,x_2,\ldots, x_n$$ is...

This is problem 20b from chapter I 4.10 of Apostol's Calculus I. The geometric mean $$G$$ of $$n$$ positive real numbers $$x_1,\ldots, x_n$$ is defined by the formula $$G=(x_1x_2\ldots x_n)^{1/n}$$. Let $$p$$ and $$q$$ be integers, $$q<0<p$$. From part (a) deduce that $$M_q<G<M_p$$ when...

Thank you mathbalarka, for the explanation! I was surprised at how much I understood of it.

Is $$\mathbb{Q}(\sqrt[3]{3})=\{a+b\sqrt[3]{3}+c\sqrt[3]{9}\mid a,b,c\in\mathbb{Q}\}$$ a ring? If it is a ring, is it a field? I have shown that it is a ring; however, I am not sure that it is a field, since in my calculations it does not seem to be closed under inverses. But I read somewhere...

Thank you so much! That makes a lot of sense.

List every generator of each subgroup of order 8 in $$\mathbb{Z}_{32}$$. I was told to use the following theorem: Let $$G$$ be a cyclic group of order $$n$$ and suppose that $$a\in G$$ is a generator of the group. If $$b=a^k$$, then the order of $$b$$ is $$n/d$$, where $$d=\text{gcd}(k,n)$$...

Thank you both so much! Deveno, I just realized that I asked the wrong question (I shouldn't post when tired). What I meant to ask was how equality implies the given condition. That's the harder part, for me at least. So I got the "if" and "only if" right, I just asked it the other way around...

This is from section I 4.9 of Apostol's Calculus Volume 1. The book states the Cauchy-Schwarz inequality as follows: $$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$$ Then it asks you to show that equality holds in the above if and only if...

Ah! Very nice. Thank you!

This is problem 13 from section I 4.7 of Apostol's Calculus Volume 1: Prove that $$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ if $$n\geq 1$$. Then use this to prove that $$2\sqrt{m}-2<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}-1$$ if $$m\geq 2$$. I have...