To show that \textsc {Partition} is in NP, note that we can guess a partition of the set of numbers and then check that they sum to the same value. \par To show that \textsc {Partition} is {\NP }-hard, we will reduce \textsc {Subset-Sum} to this problem. So, suppose we are given a set of numbers $S$ and a goal sum $M$ that comprise an input to the \textsc {Subset-Sum} problem. We begin our transformation of this input into an instance of \textsc {Partition} by including every number in $S$ as input numbers for \textsc {Partition}. Let $N$ denote the total sum of all the numbers in $S$. Then, in addition to the numbers already included, we add the number $|N-2M|$, which is a kind of \emph {enforcer}. If there is a subset $T$ of $S$ that sums to $M$, then $T$ will make a partition such that $T$ is on one side and $S-T$ (the rest of the elements) is on the other, with the enforcer element $|N-2M|$ either on $T$'s side or the other side so as to make the two sides sum to the same value.