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 Jul 27 comment How do I rotate a given point on $\mathbb{S}^n$ so that I send it to the South Pole But all rotations that send P to S do arise in this way, right? Jul 27 comment Why do real powers need positive bases? (and to handle them) You are absolutely right. It is not true that the power $a^b$ is only well-defined when $a\geq 0$. You can ascribe a well-defined value to $a^b$ for any $a$ when $b$ is a positive integer, for example; in fact, you can extend the function far more than that. But you lose a lot of its nice properties (e.g. $a^{b_1} a^{b_2} = a^{b_1+b_2}$) when you do so. It's fairly safe to say that the power $a^b$ is only defined for all $b$ and in a nice way when $a\geq 0$. Jul 27 comment how to interpret theorem about polynomial factorization over modulo ring? Have you misunderstood the result? You are given a formula - $a_n(1 + \sum a_i)$ - and told that, if it's $0$, then your polynomial has a factor, and if it's not $0$, then it doesn't. There is no need to distinguish between trivial and non-trivial linear factors, because the result is true in both cases. $X+1$ is a factor of $X+1$, even if you want to think of it as somehow "trivial". Jul 27 comment how to interpret theorem about polynomial factorization over modulo ring? (What is a "trivial" linear factor anyway? There are only two linear polynomials - $X$ and $X+1$ - and neither satisfies the condition $a_n(1 + \sum a_i) \neq 0$.) Jul 27 comment how to interpret theorem about polynomial factorization over modulo ring? The claim is that that polynomial has no linear factors if and only if $a_n(1 + \sum a_i) \neq 0$. In the case of $X+1$, $a_n(1 + \sum a_i) = 1(1+1) = 0$. So you should expect a linear factor. This linear factor is $X+1$. Where's the confusion? Jul 27 comment Arithmetic Base Conversion If x is an even number (except 2), write 1 + (x-1). If x is an odd number (except 1, 3, 5, 7), write 1 + 3 + (x-4). Or, if you like, x is already a sum of one odd number (x). Jul 23 comment Prove $\sqrt{k}$ is not a rational number. Sorry, but this is nonsense. The English doesn't even make sense. Can you rewrite your proof carefully? Jul 23 comment The union of a sequence of countable sets is countable. You don't need to pinpoint duplicates. Here's another argument (I'm going to take $\mathbb{N} = \{1, 2, \dots\}$): enumerate all the elements of each $E_n$; send the $m$th element of $E_n$ to $(p_n)^m$, where $p_n$ is the $n$th prime. This is an injection of $\bigcup E_n$ into $\mathbb{N}$. (If the $E_n$ are allowed to overlap, then for each $x\in\bigcup E_n$, just consider $x$ as an element of $E_n$ for the smallest valid $n$, I suppose.) Jul 23 comment Is this infinite series a Fourier series? Noted. You're absolutely right, and it's sloppy of me to talk about "convergence" - I simply wasn't prepared to copy out half a textbook about when a fourier series is useful and when it's not. To the best of my knowledge (though correct me if I'm wrong), an arbitrary choice of coefficients $a_i$ is more often than not meaningless - I was simply trying to get across this idea. Jul 23 comment Why is $1/n^{1/3}$ convergent? @AmireBendjeddou Ah, sorry, I didn't see that you were one of the above (helpful) commenters. But yes, I think that comments of your sort are good (the OP might not have responded because only 2 hours have passed!), and downvotes are kind of unproductive. I tend to find they drive new users away feeling attacked, rather than encouraging them to ask better questions. Jul 23 comment Why is $1/n^{1/3}$ convergent? @AmireBendjeddou So? Help the OP to fix their question instead of just bashing the downvote button. Are you here to help, or not? Jul 22 comment Why is $1/n^{1/3}$ convergent? Why has this got so many downvotes? It looks like a perfectly valid question, albeit one arising from a misunderstanding. Jul 22 comment Showing that if an equation has a unique solution for one variable, then it has unique solutions for all. Your solution looks more or less fine to me (though it needs some patching up). The point is that, regardless of what B is, you solve the equation by row-reducing A, and you can either hit the identity (in which case AX=B has exactly one solution) or you can't (in which case it has no solution or many solutions). Jul 22 comment Searching a function expressing $\sin x$ versus $\cos x$ No, you can't do this. The problem is that, for any one value of $\cos x$, there may be more than one value of $\sin x$ (e.g. $x = \pi/2, 3\pi/2$ have the same $\cos$ but different $\sin$, so $\sin$ cannot be purely a function of $\cos$). If you restrict your attention to $0\leq x < \pi/2$, you can do it as in Amire's answer. (There is also no way of turning $x\ln x$ into $x$ either, unless you work with a restricted range. For example, when $x\ln x = 0.1$, $x$ could be $\approx 0.894$ or $\approx 0.028$.) Jul 22 comment Combinations: A Generalization on a Classic Problem @RGB I don't think there's any inclusion-exclusion going on here! Jul 22 comment Combinations: A Generalization on a Classic Problem Look for the number $u$ such that $(u-1)k < a_m \leq uk$, and the number $v$ such that $vk \leq a_n < (v+1)k$. These are the first and last multiples of k. The multiples of k will be $uk, (u+1)k, (u+2)k, \dots, vk$. (If you're a coder, you may recognise u as ceiling($a_m/k$), and v as floor($a_n/k$).) Then the question is: how many integers are there between u and v inclusive? The answer is $v - u + 1$. Jul 22 comment Rational number to the power of irrational number = irrational number. True? @makerofthings7 It's probably a combination of many things: (a) you speak the way people around you speak, in mathematics or anywhere else - this is just a kind of jargon, (b) there are lots of non-native English speakers reading and writing English papers, so clarity and consistency is very important, (c) "prove that x exists such that..." sounds very sloppy to me, because "such" modifies "x" ("prove that such an x exists that..."??), and in any case it might mislead you into thinking you were given a formula or algorithm for x earlier on, and are being asked to check it makes sense. Jul 21 comment Finding multiplicative inverse modulo n using matrix method This method really just solves the two simultaneous equations $97x = 1$ (true for some specific $x = 97^{-1}$) and $224x = 0$ (true for any $x$, so in particular for $x = 97^{-1}$) without using division. It ends up with $1x = 97$ (and $2x = -30$). Jul 20 comment Combinatorial - Ways to create subcommittees of a certain size out of a committee? That's fine. You should find that it simplifies to $\frac{10!}{6!3!1!}$. (You can interpret this as "arrange the 10 people in any order; the first 6 (in any order) will form one committee, the next three (in any order) will form one committee, and the next one (in any order) will form one committee". This is called a multinomial coefficient.) Jul 20 comment Second Linear Algebra Text for Budding Mathematician That said, I don't know how far Gilbert Strang's course goes. I might be imagining you've gone much further than you have.