Problem Statement:
Let I be an ideal in a ring R. Prove that K is an ideal, where $ K $ = { $a\in R$ | $ (\forall r\in R)(ra\in I) $}
What exactly am I supposed to show here? I know I need to show that it's closed under subtraction and that it must absorb products from R for K to be an ideal.
For the last part ( absorption ) I'm assuming I'm supposed to show that if $ c \in R$ then $ cra \in I$ since I is an ideal in R and hence absorbs products, and thus is in K.
The set definition is throwing me off and I'm not even sure what the set K is supposed to be here.
As you said, you have to show that the two defining properties for an ideal (subtraction and absorbtion) hold for $K$:
pick $a,b \in K$ and show that $a-b \in K$.
pick $a \in K$ and $s \in R$ and show that $sa \in K$.
To show this you will need to use the fact that $I$ is an ideal.
Notice that, when $R$ has a $1$, then $K = I$. Indeed, $I \subseteq K$ is clear by the absorbtion property of $I$. You get the other inclusion because the condition $ra \in I$ defining $K$ has to hold for $r=1$.