How is adjusted coefficient of determination ($R^2$) linked to the F-values of a test against zero when adding a new variable? Someone claims that the  adjusted $R^2$ will increase with the addition of an extra variable. 
I wonder why, as it is called adjusted (in contrast to the normal $R^2$). 
The only condition it has to satisfy (to increase the adjusted $R^2$) is that the F-value (by the way, how is it simple to calculate it?) of the null hypothesis that the new variable is greater than 1.
Can someone give me a hint where the links between the adjusted $R^2$ and the F-Stat of that test are? 
And however, who wants to include a new variable in a multiple OLS regression model anyway, if the beta was tested to be 0? Therefore adjusted $R^2$ always changes. 
 A: The assertion of the question is true. We usually show the inverse situation, i.e. the case of dropping one variable.
In a linear multiple regression model $y_i = X\beta +u_i,\; i=1,...,n$, with $k$ regressors (including the constant term), if the t-ratio $t$ of a variable is less than 1, then dropping this one variable will increase adjusted R_squared, $\bar R^2$. When dealing with dropping one variable, then the corresponding F-statistic (reflecting just one linear restriction) is equal to $t^2$ (see this post). So both should be smaller than unity for $\bar R^2$ to increase.
This result can be proven as follows:
$\bar R^2$ is defined as 
$$ 1- \bar R^2 = \frac {n-1}{n-k} (1-R^2) \qquad [1]$$
Denoting $S_{yy} = \sum_{i=1}^{n}(y_i-\bar y)^2$ and since $R^2 = 1 - \frac{\sum_{i=1}^{n}\hat u_i^2}{S_{yy}}$we can write 
$$ (1- \bar R^2) = \frac {n-1}{n-k} \left(1-1 + \frac{\sum_{i=1}^{n}\hat u_i^2}{S_{yy}}\right) = \frac {n-1}{S_{yy}} \frac{\sum_{i=1}^{n}\hat u_i^2}{n-k}$$
$$\Rightarrow (1- \bar R^2)\frac {S_{yy}}{n-1} = \hat \sigma^2 \qquad [2]$$
By dropping a regressor, $S_{yy}$ and $n$ remain unaffected. So as a matter of mathematical necessity, the term $(1- \bar R^2)$ in the LHS of $[2]$ moves in the same direction as its RHS - meaning that as the OLS estimated variance of the regresion decreases, so is $(1-\bar R^2)$, and hence, $\bar R^2$ increases as $\hat \sigma^2$ decreases.
Consider now dropping one regressor, and index the various quantities related to this restricted regression with $r$. Denote $RSS$ the residuals sum of squares
The F-statistic to test whether the restricted regression with $k-1$ regressors is "better" than the regression with $k$ regressors is
$$F(1,n-k)= \frac{RSS_r -RSS}{RSS/(n-k)} = \frac {(n-k+1)\hat \sigma_r^2 - (n-k)\hat \sigma^2}{\hat \sigma^2}$$
$$=(n-k+1)\frac {\hat \sigma_r^2}{\hat \sigma^2} - (n-k) \Rightarrow \frac {\hat \sigma_r^2}{\hat \sigma^2} = \frac {F+(n-k)}{1+(n-k)} \qquad [2]$$
From $[2]$ it is obvious that if 
$$F<1 \Rightarrow \hat \sigma_r^2 <\hat \sigma^2 \Rightarrow \bar R_r^2 >  \bar R^2$$
And $F(1,n-k)=t^2 <1 \Rightarrow t<1$
Beware that the above results hold only when considering dropping just one regressor. Assume that we run the initial regression with $k$ regressors and we observe that two of them have t-ratios smaller than unity. This does not imply necessarily that if we drop both simultaneously, we will end up with a higher $\bar R^2$.
Now think in reverse - start from the "restricted" model and add one variable.
