# Extracted Mathematical Notes: – The equation $ab=0$ is used to simplify the expression. – The parameter $u$ in the solution is indeterminate and can take all possible values, including 0 and 1. – The formula $x=a^{\prime}u+bu^{\prime}$ replaces the limited variable $x$ with the unlimited variable $u$. – The solution $x=a^{\prime}u+bu^{\prime}$ is not equivalent to the given equation, but $x=a^{\prime}x+bx^{\prime}$ is. – Poretsky concluded that the solution of an equation is neither a consequence nor a cause of the equation. Title: Indeterminate Parameters in Equation Solutions

\]This last form is needlessly complicated, since, by hypothesis,$ab=0.$Therefore there remains$x=a^{\prime}b+ua^{\prime}b^{\prime}$which again is equivalent to$x=b+ua^{\prime},$since$a^{\prime}b=b\quad\text{and}\quad a^{\prime}=a^{\prime}b+a^{\prime}b^{\prime}.$Whatever form we give to the solution, the parameter $u$ in it is absolutely indeterminate, _i.e._, it can receive all possible values, including $0$ and $1$; for when $u=0$ we have$x=b,$and when $u=1$ we have$x=a^{\prime},$and these are the two extreme values of $x$.Now we understand that $x$ is determinate in the particular case in which $a^{\prime}=b$, and that, on the other hand, it is absolutely indeterminate when$b=0,\quad a^{\prime}=1,\quad(\text{or}\ a=0).$Summing up, the formula$x=a^{\prime}u+bu^{\prime}$replaces the “limited” variable $x$ (lying between the limits $a^{\prime}$ and $b$) by the “unlimited” variable $u$ which can receive all possible values, including $0$ and $1$._Remark.42_–The formula of solutionPorétsky. _Sept lois_, Chaps. XXXIII and XXXIV.$x=a^{\prime}x+bx^{\prime}$is indeed equivalent to the given equation, but not so the formula of solution$x=a^{\prime}u+bu^{\prime}$as a function of the indeterminate $u$. For if we develop the latter we find$ab^{\prime}x+a^{\prime}bx^{\prime}+ab(xu+x^{\prime}u^{\prime})+a^{\prime}b^{ \prime}(xu^{\prime}+x^{\prime}u)=0,$and if we compare it with the developed equation$ab+ab^{\prime}x+a^{\prime}bx^{\prime}=0,$we ascertain that it contains, besides the solution, the equality$ab(xu^{\prime}+x^{\prime}u)=0,$and lacks of the same solution the equality$a^{\prime}b^{\prime}(xu^{\prime}+x^{\prime}u)=0.$Moreover these two terms disappear if we make$u=x$and this reduces the formula to$x=a^{\prime}x+bx^{\prime}.$From this remark, Poretsky concluded that, in general, the solution of an equation is neither a consequence nor a cause of the equation. It is a cause of it in the particular case in which$ab=0,$and it is a consequence of it in the particular case in which$(a^{\prime}b^{\prime}=0)=(a+b=i).$But if $ab$ is not equal to 0, the equation is unsolvable and the formula of solution absurd, which fact explains the preceding paradox. If we have at the same time$ab=0\quad\text{and}\quad a+b=1,$the solution is both consequence and cause at the same time, that is to say, it is equivalent to the equation. For when $a=b$ the equation is determinate and has only the one solution$x=a^{\prime}=b.$Thus, whenever an equation is solvable, its solution is one of its causes; and, in fact, the problem consists in finding a value of $x$ which will verify it, _i.e._, which is a cause of it.To sum up, we have the following equivalence:$(ax+bx^{\prime}=0)=(ab=0)\sum_{u}(x=a^{\prime}u+bu^{\prime})$which includes the following implications:\[(ax+bx^{\prime}=0)