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5

They don't. Here is a 9-dimensional associative non-commutative division algebra (over $\Bbb{Q}$): $$ D=\left\{\left(\begin{array}{ccc} x_1&\sigma(x_2)&\sigma^2(x_3)\\ 2x_3&\sigma(x_1)&\sigma^2(x_2)\\ 2x_2&2\sigma(x_3)&\sigma^2(x_1) \end{array}\right)\bigg\vert\ x_1,x_2,x_3\in E\right\}, $$ where $E=\Bbb{Q}(\cos2\pi/7)$ and $\sigma$ ...


4

Perhaps the simplest way to check is to verify that $1010\cdot0.0001100110011\ldots=1$. You can do this quite easily working entirely in binary notation: $$\begin{align*} 1010\cdot0.000\overline{1100}&=(1000+10)\cdot0.000\overline{1100}\\ &=1000\cdot0.000\overline{1100}+10\cdot0.000\overline{1100}\\ &=0.\overline{1100}+0.00\overline{1100}\\ ...


3

The particular family of algebras you are talking about has dimension over $\Bbb R$ a power of $2$ by construction: the Cayley-Dickson construction to be precise.


2

Here is a somewhat more general result. I will use $\oplus$ to denote logical exclusive or. Suppose we have a Boolean system of equations (equivalently, a system of linear equations in several variables $x_1,x_2,\ldots,x_n$ over $\mathbb{Z}/2\mathbb{Z}$). If the equations can be rewritten in a form where all the variables appear only as in XOR'ing pairs ...


2

You can do this with simple arithmetic You have 2000 elements The largest list you'll get after splitting is then 1000 elements Split again and get 500 Split again and get 250 Split again and get 125 Split again and get 63 (worst case) Split again and get 32 Split again and get 16 Split again and get 8 Then 4 Then 2 Then you compare once more And maybe ...


2

All possible bitstrings of length $n$ can be expressed as set via a regular expression: $$ L_n = (0 \mid 1)^n $$ for $n \ge 2$ we can decompose this into four disjoint subsets $$ L_n = (0|1) L_{n-2} (0|1) = (0 L_{n-2} 0) \cup (0 L_{n-2} 1) \cup (1 L_{n-2} 0) \cup (1 L_{n-2} 1) $$ The feasible words are in $$ L = (0 L_{n-2} 0) \cup (1 L_{n-2} 0) \cup (1 ...


2

$2^{n-1}$ bitstrings of length $n$ start with $1$ (fix first position). Similarly, $2^{n-1}$ bitstrings of length $n$ end with $0$. We have double counted the case where $1$ and $0$ are fixed at first and last positions respectively. Number of such cases is $2^{n-2}$. Hence, $$2(2^{n-1}) - 2^{n-2} = 3\cdot 2^{n-2}$$


1

It's because 0.0001100110011001... × 1010 ---------------------- 0.001100110011001... + 0.1100110011001... ---------------------- = 0.1111111111111... ---------------------- = 1.0000000000000...


1

The proof given by @Brian M. Scott is very pedagogical but I do think that understanding the geometric series approach is as well fundamental. It maybe difficult to obtain the adequate series, because of the interference of two "parameters": the (frequent) presence of an part of the development and its period (and, first of all, its length). In this case, ...


1

When you count up to $2^{32}$, you start counting $1$, $2$, ... then the $2^{32}$th number is $2^{32}$. Since the computer has to store the number $0$ in an unsigned int, it is actually starting to count with $0$, then $1$ and so on. That means that the $n$th number for the computer is, in fact, $n-1$. Hence, the biggest number it can store, the $2^{32}$th ...


1

You can think of a binary sequence as a map from the natural numbers, $\mathbb N,$ to the set of values $\{0,1\}$. Whether you write express those maps, informally, as "$f(0),f(1),f(2),\dots$" or "$\dots,f(2),f(1),f(0)$" is essentially irrelevant, unless you want to mix the two notations and give them different meanings. Then you'd have to take some care ...


1

Let's start with some powers of $2$: $2^0 = 1$ $2^1 = 2$ $2^2 = 4$ $2^3 = 8$ $2^4 = 16$ $2^5 = 32$ We need to write out $23$, so this is enough powers of $2$. Start with the highest power of $2$ that is equal to or less than the number. This is $16$. So write a $1$ down and subtract it out: $23-16 = 7$. The next highest power, $8$, is greater than ...



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