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如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! 0.20 Postulate of ExistenceOne final axiom may be formulated here, which we will call the _postulate of existence_:Ax. 9whence may be also deduced \(1\neq 0\).In the conceptual interpretation (C. I.) this axiom means that the universe of discourse is not null, that is to say, that it contains some elements, at least one. If it contains but one, there are only two classes possible, \(1\) and \(0\). But even then they would be distinct, and the preceding axiom would be verified.In the propositional interpretation (P. I.) this axiom signifies that the true and false are distinct; in this case, it bears the mark of evidence and necessity. The contrary proposition, \(1=0\), is, consequently, the type of _absurdity_ (of the formally false proposition) while the propositions \(0=0\), and \(1=1\) are types of _identity_ (of the formally true proposition). Accordingly we put\[(1=0)=0,\quad(0=0)=(l=1)=1.\]More generally, every equality of the form\[x=x\]is equivalent to one of the identity-types; for, if we reduce this equality so that its second member will be \(0\) or \(1\), we find\[(xx^{\prime}+xx^{\prime}=0)=(0=0),\quad(xx+x^{\prime}x^{\prime}=1)=(1=1).\]On the other hand, every equality of the form\[x=x^{\prime}\]is equivalent to the absurdity-type, for we find by the same process,\[(xx+x^{\prime}x^{\prime}=0)=(1=0),\quad(xx^{\prime}+xx^{\prime}=1)=(0=1).\]property is evident from the preceding formulas. The other results from the fact that any two constituents differ at least in the “sign” of one of the terms which serve as factors, _i.e._, one contains this term as a factor and the other the negative of this term. This is enough, as we know, to ensure that their product be null.The maxima of discourse possess analogous and correlative properties; their combined product is equal to 0, as we have seen; and the sum of any two of them is equal to 1, inasmuch as they differ in the sign of at least one of the terms which enter into them as summands.For the sake of simplicity, we shall confine ourselves, with Boole and Schroder, to the study of the constituents or minima of discourse, _i.e._, the developments of 1. We shall leave to the reader the task of finding and demonstrating the corresponding theorems which concern the maxima of discourse or the developments of 0. 0.23 Logical FunctionsWe shall call a _logical function_ any term whose expression is complex, that is, formed of letters which denote simple terms together with the signs of the three logical operations.30In this algebra the logical function is analogous to the _integral function_ of ordinary algebra, except that it has no powers beyond the first.A logical function may be considered as a function of all the terms of discourse, or only of some of them which may be regarded as unknown or variable and which in this case are denoted by the letters \(x,y,z\). We shall represent a function of the variables or unknown quantities, \(x,y,z\), by the symbol \(f(x,y,z)\) or by other analogous symbols, as in ordinary algebra. Once for all, a logical function may be considered as a function of any term of the universe of discourse, whether or not the term appears in the explicit expression of the function. 0.24 The Law of DevelopmentThis being established, we shall proceed to develop a function \(f(x)\) with respect to \(x\). Suppose the problem solved, and let\[ax+bx^{\prime}\]be the development sought. By hypothesis we have the equality\[f(x)=ax+bx^{\prime}\]for all possible values of \(x\). Make \(x=1\) and consequently \(x^{\prime}=0\). We have\[f(1)=a.\]and so on. Moreover these formulas may be directly obtained by multiplying by \(a\) both members of each development of 1._Cor_. 1. We have the equivalence\[(a+x^{\prime})(b+x)=ax+bx+ab=ax+bx^{\prime}.\]For, if we develop with respect to \(x\), we have\[ax+bx^{\prime}+abx+abx^{\prime}=(a+ab)x+(b+ab)x^{\prime}=ax+bx^{\prime}.\]_Cor_. 2. We have the equivalence\[ax+bx^{\prime}+c=(a+c)x+(b+c)x^{\prime}.\]For if we develop the term \(c\) with respect to \(x\), we find\[ax+bx^{\prime}+cx+cx^{\prime}=(a+c)x+(b+c)x^{\prime}.\]Thus, when a function contains terms (whose sum is represented by \(c\)) independent of \(x\), we can always reduce it to the developed form \(ax+bx^{\prime}\) by adding \(c\) to the coefficients of both \(x\) and \(x^{\prime}\). Therefore we can always consider a function to be reduced to this form.In practice, we perform the development by multiplying each term which does not contain a certain letter (\(x\) for instance) by \((x+x^{\prime})\) and by developing the product according to the distributive law. Then, when desired, like terms may be reduced to a single term. 0.25 The Formulas of De Morgan_In any development of 1, the sum of a certain number of constituents is the negative of the sum of all the others._For, by hypothesis, the sum of these two sums is equal to 1, and their product is equal to 0, since the product of two different constituents is zero.From this proposition may be deduced the formulas of De Morgan:\[(a+b)^{\prime}=a^{\prime}b^{\prime},\quad(ab)^{\prime}=a^{\prime}+b^{\prime}.\]_Demonstration_.–Let us develop the sum \((a+b)\):\[a+b=ab+ab^{\prime}+ab+a^{\prime}b=ab+ab^{\prime}+a^{\prime}b.\]Now the development of 1 with respect to \(a\) and \(b\) contains the three terms of this development plus a fourth term \(a^{\prime}b^{\prime}\). This fourth term, therefore, is the negative of the sum of the other three.We can demonstrate the second formula either by a correlative argument (_i.e._, considering the development of 0 by factors) or by observing that the development of \((a^{\prime}+b^{\prime})\),\[a^{\prime}b+ab^{\prime}+a^{\prime}b^{\prime},\]differs from the development of 1 only by the summand \(ab\).How De Morgan’s formulas may be generalized is now clear; for instance we have for a sum of three terms,\[a+b+c=abc+abc^{\prime}+ab^{\prime}c+ab^{\prime}c^{\prime}+a^{\prime}bc+a^{\prime }bc^{\prime}+a^{\prime}b^{\prime}c.\]This development differs from the development of 1 only by the term \(a^{\prime}b^{\prime}c^{\prime}\). Thus we can demonstrate the formulas\[(a+b+c)^{\prime}=a^{\prime}b^{\prime}c^{\prime},\quad(abc)^{\prime}=a^{\prime }+b^{\prime}+c^{\prime},\]which are generalizations of De Morgan’s formulas.The formulas of De Morgan are in very frequent use in calculation, for they make it possible to perform the negation of a sum or a product by transferring the negation to the simple terms: the negative of a sum is the product of the negatives of its summands; the negative of a product is the sum of the negatives of its factors.These formulas, again, make it possible to pass from a primary proposition to its correlative proposition by duality, and to demonstrate their equivalence. For this purpose it is only necessary to apply the law of contraposition to the given proposition, and then to perform the negation of both members._ \[ab+ac+bc=(a+b)(a+c)(b+c).\]_Demonstration:_\[(ab+ac+bc)^{\prime}=[(a+b)(a+c)(b+c)],\]\[(ab)^{\prime}(ac)^{\prime}(bc)^{\prime}=(a+b)^{\prime}+(a+c)^{\prime}+(b+c)^{ \prime},\]\[(a^{\prime}+b^{\prime})(a^{\prime}+c^{\prime})(b^{\prime}+c^{\prime})=a^{ \prime}b^{\prime}+a^{\prime}c^{\prime}+b^{\prime}c^{\prime}.\]Since the simple terms, \(a,b,c\), may be any terms, we may suppress the sign of negation by which they are affected, and obtain the given formula图片描述

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