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말레이어 사전에서 «kerip» 의
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말레이어 사전에서 kerip 의 정의
긁는 것과 같은 방음 소리; 작은 물린 긁기, 쥐어 짜기, 고함 치기. 짐승이나 발굽; 지퍼; 치아의 치아 kerip bunyi spt bunyi mengikis; mengerip menggigit berdikit-dikit, mengerumit, mengunggis; binatang ~ binatang yg mengunggis atau mengerumit; pengerip sesuatu yg mengerip; gigi ~ gigi utk
말레이어 사전에서 «kerip» 의
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«KERIP» 관련 말레이어 책
다음 도서 목록 항목에서
kerip 의 용법을 확인하세요.
kerip 에 관련된 책과 해당 책의 짧은 발췌문을 통해 말레이어 서적에서 단어가 사용되는 맥락을 제공합니다.
1
Lectures on Fuzzy and Fuzzy SUSY Physics - Halaman 18
(3.8) tp descends to a linear map, called ^, from A/Kerip = {a-\- Kerip : a G .A} to C°°(M): ty(a+ Kerip) = ip(a) . (3.9) ip(a) does not depend on the choice of the representative a from a + Kerip because of (3.8). Clearly "J" is an injective map from A/ ...
A. P. Balachandran, Seckin Kürkçüoğlu, Sachindeo Vaidya, 2007
2
Representation Theory and Higher Algebraic K-Theory - Halaman 234
Then Kerip and Coker^s are uniquely ^-divisible. PROOF Follows from theorem 8.2.1 and lemma 2 of [13]. U Corollary 8.2.2 Let R be the ring of integers in a number field F, A any i?-order in a semi-simple .F-algebra E, £ a rational prime.
3
Loops in Group Theory and Lie Theory - Halaman 185
<p(Ker<pk) cannot be contained in the centre of Ker,/. But tor k = 1
Kerip = T\ and <p(Ker <p) = { 1 }. It follows that Mk is a group if and only k = 1 . Let G*+i = Sk+\ x Sk+\, Hk+\ = (ak+\<p(Sk+\), Sk+l). We want to provejhat the set { (y, ;y~'); y e Sk+\}
...
Péter Tibor Nagy, Karl Strambach, 2002
4
Kalipke/Kalipke:Sakai 010754 - Halaman 172
hol ... heraus! mengo- ouk = mengeruk = etw. aus einer Vertierung [Tasche, Mund u. a.] herausholen; -» keko-ouk. ko-ouk-ko-ouk = keko-ouk ko-oup = kerip (bunyi ketika mengunyah kacang) = Knabbergerausch; teko-oup-ko-oup = ber- ...
Hans Kalipke, Mohamad Agar Kalipke, 2001
5
The Jacobson radical of classical rings - Halaman 126
The following conditions are equivalent: (i) There exists ip G A' such that Kerip contains no nonzero right ideals of A. (ii) There exists ip G A' such that Kerip contains no nonzero left ideals of A. (Hi) There exists a bilinear form f : A x A — » F ...
Gregory Karpilovsky, 1991
Since ip is surjective, so M2 = tp(M) C ker /, that is, / = 0. Hence, ip* is injective. To prove the exactness of (M8), it remains to show that ker</>* = Im^*. If / € ker0*, then /</> = 0, so Im0 C ker/. But Inup = ker ip, so kerip Q ker/. Thus we have an ...
Vivek Sahai, Vikas Bist, 2008
7
Handbook of Formal Languages: Beyonds words - Halaman 6
On the other hand, the kernel relation kerip(= ip . ip~i) defined by a = a'(ker^) iff zip — zip (a, a' C A), of any morphism p : A — > B is a congruence of the algebra A, and if p is an epimorphism, then A/kerip = B. These two facts also mean that the ...
Grzegorz Rozenberg, Arto Salomaa, 1997
8
Commutative Ring Theory and Applications - Halaman 167
Now let H be a t- invertible t-ideal of A. Then HAm is principal for each m E A, and hence (HRm)t is principal for each m G A. Thus imp C Kerip. For the reverse inclusion, we adapt the proof of [8, Proposition 3.4]. Let [r] G Kertp. By [4, Corollary ...
Marco Fontana, Salah-Eddine Kabbaj, Sylvia Wiegand, 2002
9
Finite Semigroups and Universal Algebra - Halaman 64
Since, for any homomorphism ip : FnV — > A, there exists ip : FnV — > F such that ip o A = ip, and so kerip C ker^, we have 9n(F) C 9n(A). Hence, we may assume that p is a fully invariant congruence, i.e., p = p^. Let us see that, then, 9n(A) = p, ...
10
Basic Algebra - Halaman 577
The elements e® p generate E'®R P as an abelian group, and hence 1 (X) 1/1 is onto E' (2);; P. To show that ker(l ® (I) g image(1 ® (a), we observe from the exactness of the given sequence at N that E ®R kerip : E (2);; imagego is generated ...
