# Product of increasing functions need not be increasing

## Contents

## Statement

It is possible to have two real-valued functions , both defined on all real numbers, such that both and are increasing functions but the product (defined as ) is *not* an increasing function on all of .

We can also adapt this to functions defined on any open interval or closed interval instead of all real numbers.

## Proof

### Counterexample

Since we need to disprove a general statement, it is enough to exhibit one counterexample.

The example is with both the identity function:

These are both increasing on all real numbers.

The product is:

This is the square function which is decreasing on and increasing on .

### Generic idea behind counterexample

A product of *negative* increasing functions is decreasing.

### Compatibility with product rule for differentiation

The product rule for differentiation states that:

In particular, the sign of depends not merely on the signs of and but also on the signs of and . In particular, if is negative, then multiplying by a positive gives a negative product. This is why can be negative even though are both positive.