If one always spends the same amount, what's one's price elasticity of demand? Source: p 109, Question 5.9, Principles of Microeconomics, 7 Ed, 2014, by N Gregory Mankiw
= Question 5.7, Principles of Microeconomics, 4 Ed, 2008, by N Gregory Mankiw


  9. $\color{green } { \text { Before looking at the price,}   }$ Jessie spends $d$ dollars on something  (abbreviate this S). What is Jessie’s price
  elasticity of demand?
  
  Given Answer: Jessie's price elasticity of demand is one, because he spends the same amount on S, no
  matter what the price, which means $\color{darkred} { \text { his percentage change in quantity  }   }$ is equal to the percentage
  change in price.
  
  From p 91: Price elasticity of demand $= \dfrac{ \text{
Percentage change in quantity demanded }} {  \text{ Percentage change in price } } $


The green implies Jessie's ignorance of prices, so % change in price = 0. Yet please explain the answer? Especially the red, because the question implies nothing about the % change in quantity demanded?

Footnote: I slightly generalised the question. Mankiw writes 'gas' instead of  'something'.
The intuition is that if Jessie spends the same amount on $S$ no matter the price $p$, then a % change in price of price will reduce the demand by the same % . The thing that might confuse you is that the formula you are given is an approximation for the true price elasticity of demand given by
\begin{align}
     \eta=\frac{dS\Big/ S}{dp\Big/ p}
 \end{align} 
 where $d$ is the derivative operator. The approximation 
 \begin{align}
     \frac{\%\Delta S}{\% \Delta p}
 \end{align}
 only works for very small changes in $p$ and $S$. 
 To see why the elasticity of demand is $\mid 1 \mid$, let $b$ be the fixed amount spend on $S$. Then the quantity demanded $S$ is given by 
\begin{align}
    S=\frac{b}{p}
\end{align}

Assume that there is a small price change from $p$ to $p+\epsilon, \, \epsilon&gt;0$. Then we can define $S_1=\frac{b}{p}$ as the demand before the price change and $S_2=\frac{b}{p+\epsilon}$ as the demand after the price change. Thereby we see that
\begin{align}
    \frac{S_2\Big/ S_1 -1}{\frac{p+\epsilon}{p}-1} = \frac{\frac{b}{p+\epsilon}\Big/\frac{b}{p}-1}{\frac{p+\epsilon}{p}-1}=\frac{\frac{-\epsilon}{p+\epsilon}}{\frac{e}{p}}=-\frac{p}{p+\epsilon}\approx -1
\end{align}
when $\epsilon$ is small enough.