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conversely
conversely, adv. (ˈkɒnvɜːslɪ, kənˈvɜːslɪ) [f. converse a.2 + -ly2.] In the converse manner or order; as the converse; by conversion.1806 Hutton Course Math. I. 278 Conversely, if the two angles abc, abd, on both sides of the line ab, make up together two right angles, then cb and bd form one continu...
Oxford English Dictionary
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However vs Conversely - What's the difference? | WikiDiff
Adverb. ( - ) (conjunctive) with a reversed relationship. * 2005 , . Because change itself would absolutely stay-stable, and again, conversely , stability itself would change, if each of them encroached on the other. *. As adverbs the difference between however and conversely is that however is (lb) nevertheless, nonetheless, even so, that said ...
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Conversely vs However: When to Opt for One Term Over Another
However, the forecast predicts rain for the rest of the week. The company's profits have been increasing steadily. However, they have also been facing increased competition. Notice how "however" is used to show a contrast, but also to indicate a shift in focus or thought. It is a useful word for transitioning between two contrasting ideas ...
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what does までに mean in this sentence? = roughly: The cause-association for the case where things conversely went well is written in the brackets. Does the madeni mean' before the reference here?
means "just for reference". * * * According to , the is: > > > **** … * * * Example: > **** > Just for reference [your information], here are statistics for the past five years. > (from _Progressive Japanese-English Dictionary_ )
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Why every small set is a moderate set, but not conversely (as universe itself)? I read here > a subset of the universe, which may be small or large. (every small set is moderate, but not conversely; again, the univer...
The article you link is about foundations in general. In each foundation the precise reason that a universe cannot contain itself may be different. For example, in ZFC, the axiom of regularity forbids that any set contains itself. So in that system, if we have a set that functions as a universe, it ...
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Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive. Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is ne...
1. By definition, if $a$ is positive then $a>0$. 2. By the order axioms, we have that $a+(-a)>0+(-a)$ 3. This simplifies to $0>-a$ (by definition of additive inverse and additive identity) 4. Therefore $-a<0$ and by definition $-a$ is negative.
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Ordering relations : alternatives to the somewhat confusing terminology " a precedes b and conversely b dominates a" If R is an ordering relation, the fact that a _R_ b ( or that (a,b) belongs to R) is often expressed...
The successor is an example for functions. But it really depends on what you are ordering, it could be by inclusion of subsets, meaning b includes a. It could be by something else, dominates is just a general term. Not everything in life means it's greek roots, not all autism is a believe or practic...
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反之 - on the other hand..., conversely... - fǎn zhī | Definition ...
一年之计在于春. yī nián zhī jì zài yú chūn. 1 the whole year must be planned for in the spring (idiom) 2 early planning is the key to success ...
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Continuous linear mapping and bounded subsets 1. > Continuous linear mappings between topological vector spaces preserve boundedness. I was wondering if it means that the inverse image of a bounded subset under a con...
The answer to the first question is clearly _no_ , since the mapping can collapse the domain to the zero vector. A function that simply interchanges two points has an inverse that takes bounded sets to bounded sets, but the function is neither continuous nor linear. The second statement is precisely...
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Are the closed intervals of $\mathbb{R}$ precisely the compact connected sets? Equip $\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\mathbb{R}$ is compact iff it's closed and bounded...
Conversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \in I$, if $x < z<y$, then
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if $A^\times $ is a commutative group, does $A$ necessarily be a commutative ring? Let $A$ be a ring and $A^\times$ be the collection of unit elements of $A$. If $A$ is a commutative ring, then $A^\times$ is a commuta...
Let $K$ be a field. Consider the free $K$-algebra on two generators $x,y$. Its unit group is $K^*$, hence commutative, but $x,y$ don't commute. Here is a more explicit example: Consider the ring of upper-triangular $2 \times 2$-matrices over $\mathbb{F}_2$. It has $8$ elements and it is in fact the ...
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Is every free monoid a pure monoid and vice versa? A pure monoid is a monoid where only the identity has an inverse. Is every free monoid pure, and conversely?
Yes, free monoids are pure. By definition a free monoid on a set is the monoid of words consisting of elements from that set, and there is no cancellation, so no inverses, apart from the empty word. The non-negative rationals under addition is not a free monoid, but it is pure.
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gene expression fold change threshold limit When interpreting qRT-PCR results, which results are acceptable as fold changes, which are acceptable for housekeeping genes? e.j. a values of 0.8 raw fold changes for a "ho...
As I mentioned in your previous post. If your reference gene undergoes changes then replace it with another reference and if even that doesn't work out then use a spike in. You **must** normalize. Interpreting the raw values can be erroneous- there could be a loading bias. Don't think much- repeatin...
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First countable not Lindelöf space It is well known that any second-countable space is a Lindelöf space, but not conversely (see here). Moreover, any second-countable space is, clearly, first-countable. > **Question....
The discrete topology provides an example on any non-countable set $X$. It is clearly first-countable since $\\{x\\}$ is a local basis for any point $x$. The open cover $\\{ \\{x\\} | x\in X\\}$ has no countable subcover.
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Linear equation with form c'Ax Suppose $A$ is a given $n\times n$ matrix, $c$ a given $n \times 1$ vector, and $x$ an unknown $n\times 1$ vector. Given $c'x$, is it possible to find $c'Ax$? Conversely, given $c'Ax$, i...
In general it is impossible to find $x$ given $c'x = y$: you can see it as one equation with $n$ unknowns, which has $\infty^{n-1}$ solutions. All these are parametrized by the $n-1$ values $t_1,\ldots, t_{n-1}$ and are given by (for example) $$\begin{cases} x_1 = t_1,\\\ \vdots \\\ x_{n-1} = t_{n-1...
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