I worked through an a RSA Encryption example, where I am given $p,q,e$ and I have to work out $n,\phi(n),d$. I don't have any difficulty determining all the of other items, however I get a negative value for $d$. Namely, $d=-37$. I've heard suggestions from my classmates that I should do $\phi(n)+d=113-37=76$ but some say I should do $\phi(n) - d = 150$, and I'm not sure which one is correct.
PS - The values for $d$ and $\phi(n)$ are not the ones I worked out in my example. The ones above are just random numbers I picked to understand what to do in my case.
Assuming $\ p,q\ $ are odd primes, all you need for $\ d\ $ to do is satisfy the relation $$ ed\equiv 1\pmod{\lambda(n)}\ $$ where $\ \lambda\ $ is Carmichael's lambda function (i.e. $\ \lambda(n)=$$\text{lcm}(p-1,$$q-1)\ $). If $\ d=-37\ $ satisfies that congruence, and $\ r=m^e \pmod{pq}\ $ is a received RSA-encrypted message, then the congruence $\ m\equiv r^{-37} \pmod{pq}\ $ will hold, so you don't need to do anything at all to $\ d\ $. Nevertheless, if you really want to get a positive decryption exponent that will work, you can do so by adding an appropriate integer multiple of $\ \lambda(n)\ $ (or of $\ \phi(n)\ $, since $\ \phi(n)\ $ is an integer multiple of $\ \lambda(n)\ $) to it. So you could use $\ \phi(n)+d\ $ (or $\ \lambda(n)+d\ $) if you wish, but definitely not $\ \phi(n)-d\ $, which will not work.
Note that it is impossible for $\ \phi(n)\ $ to have a value of $113$, since its only odd values are $\ \phi(1)=\phi(2)=1\ $.