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similary
† ˈsimilary, a. Obs. Also 7 -iary. [See similar a. and -ary2. Common in the 17th cent.] 1. = similar a. 1.(a) 1564 P. Moore Hope Health i. iv. 7 Soche members are compounded and doe consiste of the saied similarie and like partes. 1594 T. B. La Primaud. Fr. Acad. ii. 29 The partes then of the body a...
Oxford English Dictionary
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consimilary
† conˈsimilary, a. Obs. rare. [f. as consimilar: cf. similary (obs.).] = consimilar.1736 H. Brooke Univ. Beauty iii. 236 The flood consimilary ducts receive, And glands refine the separated wave.
Oxford English Dictionary
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verisimilary
† veriˈsimilary, a. Obs.—1 [Cf. prec. and similary a.] Verisimilar.1653 Urquhart Rabelais ii. vi. 31 Like verisimilarie [F. verisimiles] amorabons, we captat the benevolence of the..fæminine sexe.
Oxford English Dictionary
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similariness
† ˈsimilariness Obs.—1 [f. similary a.] Similarity, homogeneity.1669 W. Simpson Hydrol. Chym. 44 It makes no alteration in the water, because of similariness of parts.
Oxford English Dictionary
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Entries of Unitarily Similary Matrics If two matrices $A$ and $B$ are unitarily equivalent ($QAQ^{*} = B$ for some unitary matrix $Q$), what can we say about the entries of $B$? Is there a concise way to express its e...
Suppose that $Q$ has columns $q_1,\dots,q_n$. We can then write the matrix product as $$ Q^*AQ = \pmatrix{q_1^*\\\q_2^* \\\ \vdots \\\ q_n^*}\ A \ \pmatrix{q_1 & q_2 & \cdots & q_n} $$ With block-matrix multiplication, we can write this product as $$ Q^*AQ = \pmatrix{q_1^*Aq_1 & \cdots & q_1^*A q_n\...
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Class of Ordinals $On$ not a Set I have a elementary question about the Ordinals $On$. I following thread the Class of Ordinals is a set? is proved that the class $On$ can't be a set. One argument is: If we assume th...
> One argument is: If we assume that $On$ is a set, hen it is well-ordered and transitive, hence an ordinal and an element of itself, there $\\{On\\} \in On$. Why does it contradicts to the wellorder? This doesn't contradict well-orderedness (as Hagen von Eitzen's answer seems to mistakenly claim); ...
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Classification of connected subsets of the real line (up to homeomorphism) I want to describe, up to homeomorphism, all the proper connected subsets of $\mathbb{R}$. I know the theorem that $A \subset \mathbb{R}$ is c...
Not quite. It’s actually five: 1. the type of $(0,1)$ (and all open intervals and rays, including $\mathbb{R}$ itself); 2. the type of $[0,1)$ (and all half-open intervals and closed rays); 3. the type of $[0,1]$ (and all non-trivial closed intervals); 4. the type of $\\{0\\}$ (and all singletons); ...
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Galois group of order 2^4 Find the galois group the polynomial $f(X)=(X^2-2)(X^2-3)(X^2-5)(X^2-7)$ over $\mathbb{Q}$. A splitting field for $f(X)$ is $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7})$. We must ha...
You've explicitly identified the generators of the Galois group! You don't need to invoke the classification of finite(ly-generated) abelian groups here. You know that the Galois group must have order $16$, and it's easy to show that the group generated by the maps $$\sqrt{2} \mapsto -\sqrt{2}, \qua...
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Difference between sd0/sda, hd0/hda From what I understand, `hdx` is naming convention for IDE (PATA) drive, `sdx` is for SCSI (SATA) drive. Sometimes I see sd0 and sda, similary hd0/hda. What are their differences ?...
hd0 - is a Linux grub2.gfg/grub.conf designation of hard drive 0, even it is sata `/dev/sd0s1` \- usualy a name of first slice in FreeBSD OS on sata0 `/dev/sda1` \- a name of Linux partition on sata0
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If $A_n$ are fields (of sets) satisfying $A_n \subset A_{n+1}$ show $\cup_n A_{n}$ is also a field. If $A_n$ are fields satisfying $A_n \subset A_{n+1}$ show $\cup_n A_{n}$ is also afield. Reminder: In order for som...
Define $\mathcal{A}:=\bigcup_{n}\mathcal{A}_{n}$ (1) You allready proved that yourself. (2) If $A\in\mathcal{A}$ then $A\in\mathcal{A}_{n}$ for some $n$ and consequently $A^{c}\in\mathcal{A}_{n}\subseteq\mathcal{A}$. (3) If $A,B\in\mathcal{A}$ then $A\in\mathcal{A}_{n}$ for some $n$ and $B\in\mathca...
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Compositum $FL/F$ Galois if $L/K$ Galois This question arises from following thread of mine: Solvable Field Extension Let $K$ a field, $F /K$ an algebraic and $L/K$ a Galois extension of $K$. Why is the compositum $F...
Take an automorphism $\alpha$ of $FL$ that fixes $F$. Then it fixes $K$ in particular, so as $L/K$ is normal, $\alpha(L)=L$. From this you conclude that $\alpha(FL)=FL$ and so $FL/F$ is normal. Regarding separability, you sketched the argument yourself.
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Implication value (for x which isn't in a domain of antecedent) We define two functions: $f(x): \dfrac{1}{x} \ge 0$ $g(x): x \ge 0$ and we want to find out for which $x$ implication $f(x) \Rightarrow g(x)$ is true....
Terminologically, we'd normally call $f$ and $g$ predicates and not functions. From a logical perspective, there's no way of talking about a partially defined predicate typically. So you do need to say whether $f(0)$ is true or not. This comes down to what we mean by $\frac{1}{x}$. Taking a relation...
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Limits of complex numbers > "We say $z_n \rightarrow \infty$ if, for each positive number $M$ (no matter how large), there is an integer $N$ such that $|z_n |>M$ whenever $n > N$; similary $\lim_{z\rightarrow z_0}f(z)...
The precise definitions are given in what you wrote. If you want a more intuitive explanation, the first clause states that a sequence $\\{z_n\\}$ approaches $\infty$ if given any $M$, there is some index $N$ in the sequence such that every term after that point in the sequence is at least distance ...
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Unitary orbit of the Jordan matrices Let $$ \mathcal{J}=\\{A\in M_n(\mathbb{C}):\ A \text{ is a Jordan matrix}\\} $$ Then it is well-known that the similary orbit of $\mathcal{J}$ is all of $M_n(\mathbb{C})$. > What ...
It cannot be dense except in the trivial case of $n=1$. The (real) dimension of $U(n)$ is $n^2$ while the dimension of $Gl_n(\mathbb{C})$ is $2n^2$ Since $\mathcal{J}$ has dimension $n$ (in the sense that it's a union of the diagonal matrices together with unions of various Jordan blocks things of s...
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$n$ points and $n-1$ dimensions > 2 points make a line -1D > > 3 points at most make a plane -2D similary can we find $n$ points making up a $n-1$-dimensional object in $\mathbb{R}^n$?
Yes, under certain conditions. Let $p,q$ be points. Then denote by $v=p-q$ the vector from $q$ to $p$. Let $p_0,...,p_{n-1}$ be $n$ point in $n$-space, make the $n-1$ vectors $v_1=p_1-p_0,...,v_{n-1}=p_{n-1}-p_0$. If this set of vectors is linearly independent, then the points uniquely define an $n-...
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