1. ## ring axioms

Each of the following (with the usual operations of addition and multiplication) fails to be a ring. In each case, you should prove this by showing one of the ring axioms
which is not satisfied.

1) N, the set of natural numbers.

2) 2Z+1, the set of odd integers.

3) The set of invertible 22 real matrices.

4) The set of polynomials in which the coefficient of x3 is zero.

5) The set of vectors in real 3-dimensional space (the multiplication here is cross product).

am not able to start doing this question at all.. It would be helpful if some one could show me how to do part (1), so that I could follow up on doing the other parts... help wpuld be appreciated thanks

2. for the part 1, depending on the convention of your N..
if 0 is not in N, then there is no element in N that will take the place of the additive identity..
but the most obvious is, N is not a group under addition.. all the elements of N have no additive inverse..

3. for part(b) I think multiplicative inverse law doesn't satisfy.. am not sure whether right can some check my answer and prove it...I need some help for the rest of the section aswell thank you

4. Originally Posted by rajr
for part(b) I think multiplicative inverse law doesn't satisfy.. am not sure whether right can some check my answer and prove it...I need some help for the rest of the section aswell thank you
Well we can't check your answer if you don't show it! How do you show that the "multiplicative inverse law" isn't satisfied?

5. hi thx for ur reply... i thought we just have to use the laws that we know from the basics i.e. the multiplicative of law of inverse does not satisfy the integers... that's why I thought that the multiplicative law of inverse doesn't satisfy... but I am a bit confused now as there are no ring axioms for multiplicative inverse law... am reallly lost now..

6. true!! existence of multiplicative inverses is not a requirement for a set to be a ring.. it is a requirement for division ring..

for 2), you may want to check if it is closed under addition..

and yeah, maybe, you want to post your solutions here..