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！

Thus, the assemblage of the \(2^{m}\) classes of discourse is divided into \(2^{n-m}\) series of \(2^{m}\) classes, each series being constituted by the sums of a certain class and of the \(2^{m}\) classes of the first column (sub-classes of \(N\)). Hence we can arrange these \(2^{m}\) sums in the following columns by making them correspond horizontally to the classes of the first column which gave rise to them. Let us take, for instance, the very simple equality \(a=b\), which is equivalent to\[ab^{\prime}+a^{\prime}b=0.\]The logical zero (\(N\)) in this case is \(ab^{\prime}+a^{\prime}b\). It comprises two constituents and consequently four sub-classes: \(0\), \(ab^{\prime}\), \(a^{\prime}b\), and \(ab^{\prime}+a^{\prime}b\). These will compose the first column. The other classes of discourse are \(ab\), \(a^{\prime}b^{\prime}\), \(ab+a^{\prime}b^{\prime}\), and those obtained by adding to each of them the four classes of the first column. In this way, the following table is obtained:\[\begin{array}{cccc}0&ab&a^{\prime}b^{\prime}&ab+a^{\prime}b^{\prime} ab^{\prime}&a&b^{\prime}&a+b^{\prime} a^{\prime}b&b&a^{\prime}&a^{\prime}+b ab^{\prime}+a^{\prime}b&a+b&a^{\prime}+b^{\prime}&1\end{array}\]By construction, each class of this table is the sum of those at the head of its row and of its column, and, by the data of the problem, it is equal to each of those in the same column. Thus we have 64 different consequences for any equality in the universe of discourse of 2 letters. They comprise 16 identities (obtained by equating each class to itself) and 16 forms of the given equality, obtained by equating the classes which correspond in each row to the classes which are known to be equal to them, namely\[\begin{array}{cccc}0=ab^{\prime}+a^{\prime}b,&ab=a+b,&a^{\prime}b^{\prime}=a ^{\prime}+b^{\prime}&ab+a^{\prime}b^{\prime}=1 a=b,&b^{\prime}=a^{\prime},&ab^{\prime}=a^{\prime}b,&a+b^{\prime}=a^{\prime}+b. \end{array}\]Each of these 8 equalities counts for two, according as it is considered as a determination of one or the other of its members. 0.51 Table of CausesThe same table may serve to represent all the causes of the same equality in accordance with the following theorem:When the consequences of an equality \(N=0\) are expressed in the form of determinations of any class \(U\), the causes of this equality are deduced from the consequences of the _opposite_ equality, \(N=1\), put in the same form, by changing \(U\) to \(U^{\prime}\) in one of the two members.For we know that the consequences of the equality \(N=0\) have the form\[U=(N^{\prime}+X)U+NYU^{\prime},\]and that the causes of the same equality have the form\[U=N^{\prime}XU+(N+Y)U^{\prime}.\]Now, if we change \(U\) into \(U^{\prime}\) in one of the members of this last formula, it becomes\[U=(N+X^{\prime})U+N^{\prime}Y^{\prime}U^{\prime},\]and the accents of \(X\) and \(Y\) can be suppressed since these letters represent indeterminate classes. But then we have the formula of the consequences of the equality \(N^{\prime}=0\) or \(N=1\).This theorem being established, let us construct, for instance, the table of causes of the equality \(a=b\). This will be the table of the consequences of the opposite equality \(a=b^{\prime}\), for the first is equivalent to\[ab^{\prime}+a^{\prime}b=0,\]and the second to\[\begin{array}{cccc}(ab+a^{\prime}b^{\prime}=0)=(ab^{\prime}+a^{\prime}b=1). 0&ab^{\prime}&a^{\prime}b&ab^{\prime}+a^{\prime}b ab&a&b&a+b a^{\prime}b^{\prime}&b^{\prime}&a^{\prime}&a^{\prime}+b^{\prime} ab+a^{\prime}b^{\prime}&a+b^{\prime}&a^{\prime}+b&1\end{array}\]To derive the causes of the equality \(a=b\) from this table instead of the consequences of the opposite equality \(a=b^{\prime}\), it is sufficient to equate the negative of each class to each of the classes in the same column. \[\begin{array}{l}a^{\prime}+b^{\prime}=0,\quad a^{\prime}+b^{\prime}=a^{ \prime}b^{\prime},\quad a^{\prime}+b^{\prime}=ab+a^{\prime}b^{\prime}, a^{\prime}+b=a,\quad\ a^{\prime}+b=b^{\prime},\quad\ a^{\prime}+b=a+b^{ \prime};\ldots.\end{array}\]Among the 64 causes of the equality under consideration there are 16 absuridities (consisting in equating each class of the table to its negative); and 16 forms of the equality (the same, of course, as in the table of consequences, for two equivalent equalities are at the same time both cause and consequence of each other).It will be noted that the table of causes differs from the table of consequences only in the fact that it is symmetrical to the other table with respect to the principal diagonal \((0,1)\); hence they can be made identical by substituting the word “row” for the word “column” in the foregoing statement. And, indeed, since the rule of the consequences concerns only classes of the same column, we are at liberty so to arrange the classes in each column on the rows that the rule of the causes will be verified by the classes in the same row

## 发表回复