# Is division ill-conditioned when divisor is close to zero?

My intuition is that division of real numbers is ill-conditioned when divisor is close to zero. Is this intuition correct?

No. The linear system $$ax=b$$ where $$a \not = 0$$ has condition number $$1$$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $$1$$. However, there is value in applying basic principles. We are interested in the perturbed equation $$(a + \Delta a) (x + \Delta x) = (b + \Delta b).$$ The relevant condition number is $$\kappa =\underset{\epsilon \rightarrow 0_+}{\lim} \, \sup \left \{ \frac{1}{\epsilon} \frac{|\Delta x|}{|x|} \: \Big | \: (a + \Delta a) (x + \Delta x) = (b + \Delta b), \: \frac{|\Delta a|}{|a|} \leq \epsilon, \: \frac{|\Delta b|}{|b|} \leq \epsilon \right\}$$ By direct computation we have $$\Delta x = \frac{b + \Delta b}{a + \Delta a} - \frac{b}{a} = \frac{b}{a} \left(\frac{1 + (\Delta b)/b}{1 + (\Delta a)/a} - 1\right)= \frac{b}{a} \frac{(\Delta b)/b - (\Delta a)/a}{1+(\Delta a)/a}.$$ It follows, that $$\frac{\Delta x}{x}= \frac{(\Delta b)/b - (\Delta a)/a}{1+(\Delta a)/a}.$$ This implies that $$\frac{1}{\epsilon} \left| \frac{\Delta x}{x} \right| \leq \frac{2}{1 - \epsilon}$$ when $$\epsilon < 1$$. Moreover, equality is possible when $$\Delta a = - \epsilon a$$ and $$\Delta b = \epsilon b$$. It follows that $$\kappa = 2$$. In particular, we have $$\left| \frac{\Delta x}{x} \right| \approx 2 \epsilon$$ and the approximation is good for small values of the relative error $$\epsilon$$.
• B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$?? – gaazkam Dec 8 '18 at 22:57
• When error of $a$ is "only" 50%: $\frac{b}{a-\frac12 a}=2\frac ba = 2 x$ – gaazkam Dec 8 '18 at 22:57
• $2x$ is hardly $x\pm 50\% x$ – gaazkam Dec 8 '18 at 22:59