# Need help with another binary operation.

On the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, I define the operation ⊗ to be like our standard multiplication, except that the result is just the last digit of the product. For example, 4 ⊗ 7 = 8, since 4 × 7 = 28, and 8 is the last digit of 28. Noticing that 1 is the identity element for ⊗, find the inverse of each element of the set with respect to ⊗ (or if an element doesn’t have an inverse,

• This is really just the ring $(\mathbb{Z}/10\mathbb{Z},+,\times)$. Do you know anything about ring? – 3x89g2 Oct 20 '16 at 0:19
• I suggest you make a multiplication table. (or since this is group theory a Cayley table.) You have already pointed out that $1$ is the identity. For every row of your table, if there is an element equal to the identity, then that column is the inverse for that row (and vice versa). – Doug M Oct 20 '16 at 0:21
• I do not... Am I correct in saying that only 1,3 and 7 have inverses for this problem. 1 being 1 3 being 7 and 7 being 3. 1*1=(1) 3*7=2(1) & 7*3=2(1) as well. – Mike Nitti IV Oct 20 '16 at 0:21
• 9 is the inverse of 9. – Doug M Oct 20 '16 at 0:23
• ah, you're right, Doug, forgot about that one.. How would I go about explaining why numbers don't have an inverse? Just by simply stating a*b is not = to x(1)? something in that realm? – Mike Nitti IV Oct 20 '16 at 0:25