I'm trying to show how to reduce the Partition problem to the 3-Partition problem.I'll first describe the problem using the definitions and notations I'm familiar with (I hope they're legit), and will then ask my question.So, the Partition problem is as follow:Given a multiset of integers $X$, we have to decide whether or not $X$ can be partitioned into two sub-sets $A,B$ such that $\sum_{x_i\in A}x_i=\sum_{x_i\in B}x_i$.In the 3-Partition problem, given a multiset of integers $Y$, we have to decide whether or not $Y$ can be partitioned into three sub-sets $A,B,C$ such that $\sum_{x_i\in A}x_i=\sum_{x_i\in B}x_i=\sum_{x_i\in C}x_i$.  I want to show a reduction $Partition\leq _p 3Partition$, That is, to show that given $X$ as input for the Partition problem, I can build $Y$ as input for the 3-Partition problem, s.t. $X$ can be partitioned to two iff $Y$ can be partitioned to three.So given $X$, let $\sum_{x_i\in X}x_i=m$, and define $Y$ as:$Y=\left \{ x_1, x_2,...,x_n,\frac{m}{2}|x_i\in X \right \}$.   Proof:The first direction is easy: suppose that $X$ can be partitioned to two sub-sets $A,B$ s.t. the above conditions hold, then $Y$ can be partitioned to three: $A,B,C$, while $C=\left \{\frac{m}{2} \right \}$.I'm stuck with the second direction: suppose that $Y$ can be partitioned to three, why does that guarantee that $X$ can be partitioned to two?b.t.w, how can I be sure that $\frac{m}{2}$ is an integer? do I even have to pay attention to it?
Reduction from Partition problem to 3-Partition problem
reductions
To answer your bolded question: The sum of the numbers in $Y$ is $\frac{3m}2$. Thus $A$, $B$, and $C$ each need to be of size $\frac m2$. This can only be achieved by partioning as in the first direction.Regarding your second question: You can not be sure that $\frac m2$ is an integer and you have to pay attention to it, in order to get a formally correct reduction. The easiest way to address this is by having your reduction function distinguish two cases: If $\frac m2$ is an integer, it proceeds as you describe. If $\frac m2$ is not an integer, it exploits the fact that in this case the answer to the original question (Can $X$ be partitioned?) is easy. I'll leave it as an exercise to figure out the details of this case.