灾难性抵消

形式上，发生灾难性抵消是因为减法运算对邻近数值的输入是病态的：即使近似值${\displaystyle {\tilde {x}}=x(1+\delta _{x})}$ 和${\displaystyle {\tilde {y}}=y(1+\delta _{y})}$ 与真实值${\displaystyle x}$ 和${\displaystyle y}$  相比，相对误差${\displaystyle |\delta _{x}|=|x-{\tilde {x}}|/|x|}$ 和${\displaystyle |\delta _{y}|=|y-{\tilde {y}}|/|y|}$ 不大，近似值差${\displaystyle {\tilde {x}}-{\tilde {y}}}$ 的与真实值差相对误差也会与真实值差${\displaystyle x-y}$ 成反比：

${\displaystyle {\begin{aligned}{\tilde {x}}-{\tilde {y}}&=x(1+\delta _{x})-y(1+\delta _{y})=x-y+x\delta _{x}-y\delta _{y}\\&=x-y+(x-y){\frac {x\delta _{x}-y\delta _{y}}{x-y}}\\&=(x-y){\biggr (}1+{\frac {x\delta _{x}-y\delta _{y}}{x-y}}{\biggr )}\end{aligned}}}$ 

因此，两个近似值的精确差值${\displaystyle {\tilde {x}}-{\tilde {y}}}$ 与真实数字差值${\displaystyle x-y}$ 的相对误差为：

${\displaystyle \left|{\frac {x\delta _{x}-y\delta _{y}}{x-y}}\right|}$ 

如果真实输入${\displaystyle x}$ 和${\displaystyle y}$ 很接近，结果可能会非常大。