By V. Kharchenko

T moi, ... si favait su remark en revenir. One sel'Yice arithmetic has rendered the je n'y serais aspect aile.' human race. It has positioned good judgment again Jules Verne the place it belongs, at the topmost shelf subsequent to the dusty canister labelled 'discarded non- The sequence is divergent; for this reason we should be sense', capable of do anything with it. Eric T. Bell O. Heaviside arithmetic is a device for idea. A hugely beneficial device in an international the place either suggestions and non linearities abound. equally, all types of components of arithmetic function instruments for different components and for different sciences. utilising 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 good judgment has rendered com puter technological know-how .. .'; 'One provider classification conception has rendered arithmetic .. .'. All arguably real. And all statements available this fashion shape a part of the raison d 'e\re of this sequence.

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**Example text**

Indeed, if I 2 r = 0, then 0 = II(II aA)g, where A is such a 4} CHAPTER } set that A g= I} r. This set exists since I} is an ideal and I} ~ I g . , r = 0, which fact proves that I2 Let us now define the mapping a g: I2 ~ i E I 2 , then i =j g for a certain j E :? I a I, R E F . in the following way. If and we set ia g = ( ja) g , in which the right part of the equality is determined because ja ~ I. 9). We can now easily check if a ~ a g is the sought extension of g. In this case the formula other extension, j gag = (ja ) g g I, we j gag - j g a g'= ( ja)g - (ja)g = 0, g = g.

Any ideal I of a semiprime ring R has a zero intersection with its annihilator ann I. The direct sum I+ annI belongs to IF. Proof. The intersection III annI has a zero multiplication and, hence, equals zero. If (I + ann I)x =0, then Ix = 0 and, hence, x belongs to the ideal A= ann I. , xE All annA= 0, which is the required proof. 4. Definition. An ideal of the ring R is called essential if it has a zero intersection with any nonzero ideal of the ring R. 5. Lemma. The ideal I of a semiprime ring R belongs to IF iff it is essential.

By induction over quite integer over T of a certain degree m. Let k = 1 and a 1= diag( rl'0, •. O)E ~. ')' Then for k n - k rows and n - k we will show R to be Let us find an element any ~E ~ we have (a1 - t) ~ = 0, and, hence, at a 2 = ta 2 which is the required proof. Let k> 1. Let us present an arbitrary matrix aE Rk as a= [ 0] * * ° °°° a' r" (7) where a' is the (k - 1) x (k - 1) matrix, k - 1. Let us set (a'O) iI= 00 at = ~ E R k _ t' Then for all the elements at, r'i iff the columns and r'i ~E Rk we shall introduce the relation corresponding to the matrices at and r;.