By V. Kharchenko

T moi, ... si favait su remark en revenir. One sel'Yice arithmetic has rendered the je n'y serais element aile.' human race. It has positioned logic again Jules Verne the place it belongs, at the topmost shelf subsequent to the dusty canister labelled 'discarded non- The sequence is divergent; consequently we might be sense', capable of do whatever with it. Eric T. Bell O. Heaviside arithmetic is a device for suggestion. A hugely beneficial instrument in an international the place either suggestions and non linearities abound. equally, every kind of elements of arithmetic function instruments for different elements and for different sciences. making use of an easy rewriting rule to the quote at the correct above one reveals such statements as: 'One carrier topology has rendered mathematical physics .. .'; 'One carrier common sense has rendered com puter technology .. .'; 'One carrier class concept has rendered arithmetic .. .'. All arguably precise. And all statements available this manner shape a part of the raison d 'e\re of this sequence.

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**Sample text**

Let us multiply the above equality from the right by an arbitrary element aE R k • Then, due to the flrst property of relation ==, a multiple application of relation (8) enables one to eliminate the wavy lines over the letters. , a m,a ) == O. This fact implies that there is an element tE T, such that the difference s = f( ai' ". , S m of type (9), there exists a quasi-integer polynomial q> of degree m, such that the rnatti. "' S fmm: (: ~ ~J and, hence, qJ is a quasi-integer polynomial for m' of type (9).

We have D. ~nn is is == Q / C. 22, Q / C is a complete nonsingular module. We have already remarked that any module isomorphism is quite continuous. Hence, D inn is also a complete module and, consequently, it is closed in End( R F , + ). The lemma is proved. Let us call the sum of elements a + b of a nonsingular module M orthogonal if the supports of these elements are orthogonal, e(a)' e(b) = O. The subset S of the module M will be referred to as an E-subset (here E is a set of all central idempotents), if it is closed relative to multiplication by the idempotents and, together with any two elements, contains their sum, provided it is orthogonal.

In this case I = V n R is a two-sided ideal of the ring R. Let us prove that its annihilator in the ring R coincides with the annihilator of V in R. Let Ix = O. 9) Iv' v ERn V = I and, hence, Ivvx = O. By property (2), we get vx = O. By property (4) of the ring of quotients, for the identical mapping cp: V ~ V, there exists an element e E R F' such that ve = v for all v E V (in particular, se = s for all s E S) and e annihilates the annihilator L of the set V in the ring R F' This fact implies that for any 1 E L, v E V, X E RF the following equalities are valid: + v)[ x, e] = 0, (1 + v)(e 2 - e) = O.