For any number $x$ where $x\in\Bbb R$ and where $x\ne0$, what is the mathematical property which states that: $${1-x^2\over x} = {1\over x} - {x^2\over x}$$
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9Distributivity of division over subtraction? Also, you may want $x\neq 0$. – Clement C. Oct 09 '15 at 17:45
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Yes, I forgot to add that. I’m looking for a concrete or published source on this topic for an argument, but I will take what I can get. – m482 Oct 09 '15 at 17:46
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1Why is the tag "discrete mathematics" used here? – mweiss Oct 09 '15 at 17:55
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Just the computation rules for fractions: $$ \frac{a - b}{c} = \frac{a + (-b)}{c} \\ \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c} \\ \frac{(-b)}{c} = -\frac{b}{c} $$ for $a = 1$, $b = -x^2$ and $c = x \ne 0$.
mvw
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In $\Bbb R$ (or any other division ring), note that $\frac{x}{y} = x y^{-1}$, where $y^{-1}$ is defined as the number such that $yy^{-1}=1=y^{-1}y$, and $x-y = x + (-y)$, where $-y$ is defined as the number such that $y+(-y)=0=(-y)+y$, and that $(x+y)z = xz + yz$ (multiplication is left-distributive over addition)
So, in any context in which $b$ has an additive inverse, $c$ has a multiplicative inverse, and multiplication is left-distributive over addition, the expression becomes:
$$\frac{a-b}{c} = (a + (-b))c^{-1} = ac^{-1} + (-b)c^{-1} = \frac{a}{c} - \frac{b}{c}$$
Note that $0$ cannot have a multiplicative inverse, so this is true only when $c\ne0$
AJMansfield
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