$$\text{RSE}=\sqrt{\frac{1}{n-2}\text{RSS}}=\sqrt{\frac{1}{n-2}\sum_{i=1}^n (y_i-\hat{y_i})^2}$$
Why should we put square root to get RSE ? what for?
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Suppose that $y$ is in units of meters. Then the squared errors $\sum (y_i - \hat y_i)^2$ are in units of meters squared, whereas RSE is in units of meters again. This puts it on a comparable scale to the data points, making the resulting numbers much more interpretable.
Geometrically, you can think of the square root as corresponding to taking the length of the error vector (the distance between the true vector of all the labels and the estimated vector of all labels), so the square-then-root structure shouldn't be too surprising.
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$\begingroup$ +1. The OP may find a reminder of Pythagoras' theorem to be helpful. $\endgroup$– Nick CoxCommented Mar 29, 2016 at 8:17