RSA and possible range of $\phi(n)$ If $n = p_1 p_2$ for primes $p_1 \ne p_2$, $\phi(n)$ is $\phi(p_1)\cdot\phi(p_2)$. Does that mean that the possible range of $\phi(n)$ is somewhat narrow, that $(p_1-1) \cdot (p_2-1)$ (that is, $\phi(n)$) is close to $p_1\cdot p_2$ (that is, $n$)?
$n - \phi(n) = p + q - 1$.  $p$ and $q$ are secret uniform random 1024-bit prime numbers.  Is a difference of near $2^{1024}$ ‘close’?