# How to prove that Adjusted R^2 is less than R^2

The adjusted R^2 formula is :

$$\overline{R}^{2}=1-\left( \left( 1-R^{2}\right) \cdot > \dfrac{n-1}{n-k}\right)$$

In case of k > 1 , I continue like that;

$$\overline{R}^{2}=1-\left( \dfrac{n-1}{n-k}-\dfrac{n-1}{n-k}R^{2}\right)$$

then

$$\left( n-k\right) \cdot \overline{R}^{2}=n-k-\left( n-1\right) +\left( n-1\right) R^{2}$$

so $$\left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) R^{2}=1-k$$

but after that I don't know how to proceed, is there someone who has an idea?

$$SSRes=\sum_{i=1}^n\left( y_i-\hat y_i \right)^2\\ SSTotal=\sum_{i=1}^n\left( y_i-\bar y \right)^2$$
$$R^2=1-\dfrac{ SSRes/(n-1) }{ SSTotal/(n-1) }$$
$$\bar R^2=1-\dfrac{ SSRes/(n-k) }{ SSTotal/(n-1) }$$
When $$k>1$$, the numerator of the second equation will be larger, meaning that the second equation subtracts a larger number from $$1$$ than does the first equation.
Thus, $$R^2\ge \bar R^2.$$ $$\square$$
A way to think about $$R^2$$ is that it compares the error variance to the total variance, and adjusted $$\bar R^2$$ does the same but with an estimate of the error variance that is slightly larger. Since that estimate of the error variance is slightly larger, adjusted $$\bar R^2$$ must be slightly less favorable (slightly smaller).