Mathematical Notes: Propositions, Variables, and Equations

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Thus we have a means of expressing universal and particular propositions when they are applied to variables, especially those in the form: “For every value of \(x\) such and such a proposition is true”, and “For some value of \(x\), such and such a proposition is true”, etc.For instance, the equivalence\[(a=b)=(ac=bc)(a+c=b+c)\]is somewhat paradoxical because the second member contains a term (\(c\)) which does not appear in the first. This equivalence is independent of \(c\), so that we can write it as follows, considering \(c\) as a variable \(x\)\[\prod_{x}[(a=b)=(ax=bx)(a+x=b+x)],\]or, the first member being independent of \(x\),\[(a=b)=\prod_{x}[(ax=bx)(a+x=b+x)].\]In general, when a proposition contains a variable term, great care is necessary to distinguish the case in which it is true for _every_ value of the variable, from the case in which it is true only for some value of the variable.37 This is the same as the case in which it is true only for some value of \(x\); that is to keep purpose that the symbols \(\prod\) and \(\sum\) serve.This is the same as the distinction made in mathematics between _identities_ and _equations_, except that an equation may not be verified by any value of the variable.Thus when we say for instance that the equation\[ax+bx^{\prime}=0\]is possible, we are stating that it can be verified by some value of \(x\); that is to say,\[\sum_{x}(ax+bx^{\prime}=0),\]and, since the necessary and sufficient condition for this is that the resultant \((ab=0)\) is true, we must write\[\sum_{x}(ax+bx=0)=(ab=0),\]although we have only the implication\[(ax+bx=0) indeterminates occur in one member of an equivalence, which are not present in the other. For instance, certain authors predicate the two following equivalences\[(a indeterminates “. Now, each of the two equalities has the inclusion \((a indeterminates , that it is true for _all_; in particular, it is not true for the values\[u=1,\quad v=0,\]for then \((a=bu)\) and \((a+v=b)\) become \((a=b)\), which obviously asserts more than the given inclusion \((a


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