Let R be a ring. An element a ∈ R has a right inverse if there is some x ∈ R such that ax = 1. Suppose that a ∈ R has a right inverse but does not have a left inverse. Show that a has infinitely many distinct right inverses. (Hint: let x be such that ax = 1 but xa 6 = 1. And consider the elements x + (xa − 1)aj for j = 1, 2, . . ..)

Let's prove that if an element a ∈ R has a right inverse but does not have a left inverse, then it has infinitely many distinct right inverses. Given that a has a right inverse, there exists an element x in R such that ax = 1. Now, we want to show that there are infinitely many distinct right inverses for a. Let's construct these right inverses as follows: Let y_j = x + (xa - 1)aj for j = 1, 2, 3, ... We claim that y_j is a right inverse for a. To show this, we need to demonstrate that ay_j = 1 for all j. Let's compute ay_j: ay_j = a(x + (xa - 1)aj) Now, distribute a: ay_j = ax + a(xa - 1)aj Since ax = 1 (given), we have: ay_j = 1 + (xa - 1)aj Now, let's simplify further: ay_j = 1 + (xa - 1)aj ay_j = 1 + (xa)aj - aj Now, notice that aj is a scalar (an element of the ring R), so we can factor it out: ay_j = 1 + aj(xa - 1) Since ax = 1 (given), we have: ay_j = 1 + aj(1 - 1) ay_j = 1 + 0 ay_j = 1 So, we've shown that for each j, y_j is indeed a right inverse for a because ay_j = 1.