Dear Uncle Colin,

Does $a > b$ imply $\frac{1}{b} > \frac{1}{a}$?

- Inequality: Mightily Perplexing Logically, Yeah?

Hi, IMPLY, and thanks for your message!

On the face of it, that seems sensible, doesn’t it? It’s tempting to say something like $1 > \frac{b}{a}$, so $\frac{1}{b} > \frac{1}{a}$. With an equation, that would be fine. With an inequality ((I’ve seen them called ‘inequations’, but I really dislike that word.)), you have to be extremely sensitive to sign changes.

There is also the problem that $\frac{1}{a}$ and $\frac{1}{b}$ are not even necessarily defined here: either $a$ or $b$ could be 0.

### How I’d approach it

The first inequality can be rearranged without any funny business to say $a - b > 0$.

The second inequality can be rearranged to say $\frac{1}{b} - \frac{1}{a} > 0$, or $\frac{a - b}{ab} > 0$.

You see the difference? The $ab$ on the bottom is important.

If $ab>0$, the two inequalities give the same result. If $a=0$ or $b=0$, the second is undefined, and if $ab < 0$, the inequalities give opposite results.

In short: no, the implication doesn’t hold in general.

- Uncle Colin