Elasticity of demand equals -1 but income decreases! In my textbook, it's stated that:


  When $\epsilon &lt; -1$, demand is elastic and raising price will result in smaller income, while lowering price will result in bigger income.
  
  When $\epsilon = -1$, demand is neither elastic nor inelastic and change in price won't      result in change in income.
  
  When $\epsilon &gt; -1$, demand is inelastic and raising price will result in bigger income,                 while lowering price will result in smaller income.
  
  $\epsilon = \%\Delta Q / \%\Delta P$.


This is the exercise I found confusing:


  Old price: 5
  
  New price: 6
  
  Old quantity: 25
  
  New quantity: 20
  
  Calculate elasticity


This is my solution:

$\% \Delta P = \frac{\text{new price } - \text{ old price}} {\text{old price}} = \frac{6 - 5} 5 = 0.2$

$\%\Delta Q = \frac{\text{new quantity } - \text{ old quantity}} { \text{old quantity}} = \frac{20 - 25} {25} = -0.2$

$\epsilon = \%\Delta Q / \%\Delta P = -0.2 / 0.2 = -1$

$$$$
This is why I am confused:

$\text{Old income} = \text{old price} \times \text{old quantity} = 5 \times 25 = 125$

$\text{New income} = \text{new price} \times \text{new quantity} = 6 \times 20 = 120$

Old income does not equal new income even though elasticity is -1!

What am I doing wrong? Am I misunderstanding the textbook?

$$$$
Edit: the answer provided is $\epsilon = 1.22$ but I have no idea where it comes from.
We discourage numeric questions as they are unlikely to be useful for future visitors but this is a very good example of why using non-marginal quantities can be misleading.

The exact definition of the price elasticity of demand is
$$
\epsilon(p) = \frac{d D(p)}{d p} \cdot \frac{p}{D(p)}.
$$
(In your notation $D(p) = Q$.) 
By straightforward calculation you can show
\begin{eqnarray*}
\max_p &amp; &amp; p \cdot D(p) \\
\\
p \cdot \frac{d D(p)}{d p} + D(p) &amp; = &amp; 0 \\
\\
p \cdot \frac{d D(p)}{d p} &amp; = &amp; - D(p) \\
\\
\frac{p}{D(p)} \cdot \frac{d D(p)}{d p} &amp; = &amp; - 1 \\
\\
\epsilon(p)  &amp; = &amp; - 1.
\end{eqnarray*}
Whether a marginal price change increases or decreases revenue when $\epsilon(p)  \neq - 1$ can be seen from a similar calculation.

Note that the reasoning used above only stated that $D(p)$ reached its local maximum. It is possible that several local maxima exist and the global maxima is elsewhere.

So $\epsilon(p)  = - 1$ implies that the revenue cannot be increased by small (marginal) changes in $p$. So far I have used the concept of point elasticity. But your textbook uses arc elasticity, which measures non-marginal changes:
$$
\frac{\Delta Q}{Q} / \frac{\Delta p}{p}
$$
Some reasons to do this:


Arc elasticity can be measured empirically, point elasticity can only be approximated.
In case of small changes the two should be reasonably close.
Point elasticity requires some knowledge of calculus.


There are however also some problems with this: 


No guarantuees are made by the $\epsilon(p)  = - 1$ condition for non-marginal price changes.
What base you should use for arc elasticity is not clear at all. You used
$$
\frac{Q_{new} - Q_{old}}{Q_{old}} \cdot \frac{p_{old}}{p_{new} - p_{old}}.
$$
Why is the old quantity and price the reference for the rate of change? Why is it not the new quantity and price
$$
\frac{Q_{new} - Q_{old}}{Q_{new}} \cdot \frac{p_{new}}{p_{new} - p_{old}}
$$
or the average of the two
$$
\frac{Q_{new} - Q_{old}}{\frac{Q_{new} + Q_{old}}{2}} \cdot \frac{\frac{p_{new} + p_{old}}{2}}{p_{new} - p_{old}}?
$$


These bases give you different values for arc elasticity. The second base gives you $1.2\dot{2}$ so it seems to be the one your textbook expects. I guess it is best if you see the difference between point elasticity that is used by the rule and arc elasticity. Other than this you have to ask your professor what definiton of elasticity she/he expects you to use.