Whenever I teach a complicated formula, I always demonstrate to the class that it works intuitively in extreme cases. For example, when all the data values are equal, the standard deviation is 0. Not surprisingly, some formulas for the same quantity do a better job than others. Again, for standard deviation, the definition $s = \sqrt{\frac{1}{n-1} \sum (x_i – \overline{x})^2}$ makes the $s=0$ intuitive explanation more obvious than other formulas such as $s = \sqrt{\frac{n\sum{x_i^2 – (\sum{x_i})^2}}{n(n-1)}}$.
I can't seem to do this for when the Pearson product-moment correlation coefficient equals 1.
While looking for an appropriate answer online, I came across this truly amazing(!) intuitive explanation for the covariance. I think$^?$ this can answers my question if we use the formula $\rho_{X,Y}={\mathrm{cov}(X,Y) \over \sigma_X \sigma_Y}$. But covariance is not part of our lesson.
The following are the only formulas we can use in class.
$^?$ I have not come up with an explanation using the covariance either, so an explanation using it will also be much appreciated.
Best Answer
Note that perfect correlation means the deviations of the $y_i$'s about their mean are proportional to the corresponding deviations of the $x_i$'s about their mean; that is, $y_i - \bar y = k(x_i - \bar x)$ for all $i$ and some constant $k$. So using your first expression
$$\begin{align} \rho &= \frac{\sum (x_i - \bar x)(y_i - \bar y)} {\sqrt{\sum (x_i - \bar x)^2}{\sqrt{\sum (y_i - \bar y)^2}}}\\ &= \frac{\sum k(x_i - \bar x)(x_i - \bar x)} {\sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum k^2 (x_i - \bar x)^2}} \\ &= \frac{k \sum (x_i - \bar x)^2} {\pm k \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum (x_i - \bar x)^2}} \\ &= \pm 1 \end{align} $$