Artificial intelligent assistant

attemptation

answer Answers

ProphetesAI is thinking...

MindMap

Sources

1
attemptation
† attempˈtation Obs. rare—1. [ad. L. attemptātiōn-em, n. of action f. attemptāre to attempt. See also attentation.] An attempting.1425 Paston Lett. 5 I. 21 The attemptacion of diverses matieres a geyn summe frendes of the seyd John. Oxford English Dictionary
prophetes.ai 0.0 3.0 0.0
2
attend, v. meanings, etymology and more - Oxford English Dictionary
The earliest known use of the verb attend is in the Middle English period (1150—1500). OED's earliest evidence for attend is from around 1315, in the writing of Shoreham. attend is a borrowing from French. Etymons: French atendre. See etymology. Nearby entries. attemptation, n. 1425;
www.oed.com 0.0 1.5 0.0
3
attentation
† attenˈtation Obs. [ad. L. attentātiōn-em, n. of action f. attentāre: see attent v.] = attemptation; ‘a trying or essaying.’ Bullokar 1676.a 1670 Hacket Abp. Williams i. 99 (D.) The Devil that spies the first spark of attentation, and blows it into a flame. Oxford English Dictionary
prophetes.ai 0.0 0.90000004 0.0
4
attemption
† aˈttemption Obs. rare—1. [irreg. for attemptation.] An attempt.1565 R. Lindsay Hist. Scot. (1728) 33 The English Attemptions were punished in the last Battle. Oxford English Dictionary
prophetes.ai 0.0 0.6 0.0
5
Prove that $ AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime. If $A$ and $B$ are finite subgroups, of orders $m$ and $n$, respectively, of the abelian group $G$, Prove that $AB$ is a subgroup of or...
$AB $ is a subgroup because $G$ is abelian. We know that $$|AB| = \frac{|A| \cdot |B|}{|A \cap B|}$$ But if $\gcd(m,n) = 1$ then $|A \cap B| = 1 $ because $|A \cap B| \mid |A| = m$ and $|A \cap B| \mid |B| = n$ by Lagrange theorem. So $$|AB| = |A| \cdot |B| = mn$$
prophetes.ai 0.0 0.6 0.0
6
If A, B are subgroups of G such that $b^{-1}Ab\subset A$ for all $b\in B$, show that AB is a subgroup of G My attemptation: First I wanted to prove that the set AB is closed under multiplication. That means if $a,a'\...
Note that $aba'b'= a(ba'b^{-1})bb'$ and $(ab)^{-1}=b^{-1}a^{-1}= (b^{-1}a^{-1}b)b^{-1}$.
prophetes.ai 0.0 0.3 0.0
7
a Problem about Special Sequence I am doing homework. And this question may related to #Proving statement about sequences. However, I want to consider the limit of it. Let we assume $a_1=a,b_1=b,b>a>0$. Then we consi...
Yes, $(2)$ is easy once you notice that $a_{n+1}b_{n+1}=a_nb_n$, so the product (and hence also the geometric mean) is invariant. As it is well-know that it is between the harmonic and arithmetic mena, the limit of our nested intervals must equal $\sqrt{ab}$. A similar approach works for $(1)$: Show...
prophetes.ai 0.0 0.3 0.0