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 Sep 21 comment Is zero a scalar? To the down-voter: please state the reason for the downvote. Sep 21 awarded Good Answer Sep 21 comment What's special about measurable functions? @StefanHansen: That's a restriction on $f$, not on $g$. That $f$ needs do be simple doesn't translate to $g$ needing to be simple (otherwise the definition would be pointless). Note that non-measurable sets $S$ have measurable subsets (like the empty set, and single-element sets for each element of $S$). Given the completeness of real numbers, the only possibility I see for the supremum not to exist is if the set is unbounded. Sep 21 comment How to use convergence definitions? There are no perfect dice in the real world (not that you could actually prove it, as you can only throw each one finitely many times). Sep 21 comment Qualitatively, what is the difference between a matrix and a tensor? Did you see this question? Sep 19 comment Weird question about inverses @coldnumber: No, the axiom is not that the parentheses can be moved around, but that $\mathrm{times_{Scalar}}(\alpha,\mathrm{times_{Scalar}}(\beta,v)) = \mathrm{times_{Scalar}}(\mathrm{times_{Field}}(\alpha,\beta),v)$. It's just that we've decided to denote both operations by juxtaposition, and to denote the order of operation by parentheses, that this translates into the ability to move parentheses around. Note that there are not even the same operations on both sides of the equal sign! Sep 19 answered Weird question about inverses Sep 19 comment Can $s_n=1+x+x^2+x^4+x^8+\ldots+x^{2^n}$ be simplified? It is more clear if you write $s_n=1+\sum_{k=0}^n x^{2^k}$. I wonder, however, why you've got the extra $1$ term. Sep 18 comment Possible bisection method? I would start with the observation that neither $R$ not the roots of the polynomial change if you multiply the polynomial with a non-zero number. Thus you can just assume $a_n=1$ and get rid of the division. The case $\left|\alpha\right|\le1$ is easily handled by the condition $R>1$. This leaves the case $\left|\alpha\right|>1$. I'd use the fact that for $a_n=1$, $p(x) = (x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$ and use that to express the $a_n$ in terms of $\alpha_n$, and look whether I can find some inequalities that can be applied. Sep 18 answered Is zero a scalar? Sep 16 comment What are properties, operators, and where do we get numbers from? … when you decided to give the successor of $3$ the name $4$. However you are right that operations like addition and multiplication are ultimately defined through the successor function. For example $a+b$ is defined by the two formulas $a+0=a$ and $a+S(b) = S(a+b)$. Saying essentially "If you add nothing, you arrive at the same number, and if you add the number following $b$, you get the number following $a+b$. Similarly, multiplication is defined by $a\cdot 0=0$ and $a\cdot S(b) = a\cdot b + a$. Sep 16 comment What are properties, operators, and where do we get numbers from? @user271090: In mathematics, the basic assumption is that all things are already there, and we just describe them. So basically, the axiom says "there is a successor $S(n)$ for every $n$; we just might create a new name for it, since it's not nice to always have to say "the successor of the successor of the successor of the successor of zero". But of course if you have four marbles, they didn't turn into four marbles at the point when you counted them and arrived at $4$. So it makes sense to assume the number $4$ existed from the beginning, and you just gave it a name … Sep 15 answered What are properties, operators, and where do we get numbers from? Sep 13 comment Show that 2n “1” digits subtract n “2” digits is a perfect square. @MCKapur: The original expression quite obviously gives an integer, and since everything done was just equivalence transformations, it is obvious that it still describes an integer. The number that is squared is clearly a rational number (quotient of two integers). Any rational number whose square is integer is itself an integer. Sep 13 comment Show that 2n “1” digits subtract n “2” digits is a perfect square. While you've gotten an answer, here's how you could have seen it for yourself: First, $10^{2n} = (10^n)^2$. Then just sort the numerator for powers, to get $(10^n)^2 + 2\cdot 10^n + 1$. At that point, you should be able to see the pattern of the binomial formula $a^2 + 2 a b + b^2 = (a+b)^2$ with $a=10^n$ and $b=1$. Sep 12 comment I roll a die repeatedly until I get 6, and then count the number of 3s I got. What's my expected number of 3s? @zhoraster: No. Why? Sep 12 comment I roll a die repeatedly until I get 6, and then count the number of 3s I got. What's my expected number of 3s? Well, that's probably the difference. I don't like the excessively sparse arguments. I think an answer is there to inform the reader, not to challenge the reader. Sep 12 comment I roll a die repeatedly until I get 6, and then count the number of 3s I got. What's my expected number of 3s? @zhoraster: The difference is that you don't need to intensely study it to recognize what it says. ;-) When I wrote my answer, I hadn't recognized yours to say the same. Indeed, on first glance it just looks like a big mess of symbols. Already a few explicit multiplication signs (or even just \, for some extra spacing) would have greatly increased the readability of your answer. Sep 11 answered Geometric meaning of a vector space Sep 10 comment Lemma, theorem, corollary… which one is a suitable term for an observation? @user153465: I cannot compare my answer with any other answer because I didn't write an answer to your question. I only wrote a comment concerning your comment claiming that you know that it is not new, contradicting the claim in your question that you don't know if anyone had made that observation.