Thread: Commutative and Associative relations of binary operations

1. Commutative and Associative relations of binary operations

Hi I've got some problems and just need a bit of reassurance that I'm doing them right!

a) For each of the following binary operations on $\displaystyle z^+$ say whether or not it is associative, commutative
i) $\displaystyle a*b=a+b^2$

This is neither commutative nor associative

ii)$\displaystyle a*b=a+1$

Is again neither commutative nor associative

b)Give an example of a binary operation on Z which is commutative but not associative

I think $\displaystyle a*b =$ $\displaystyle a^2+b^2$ or $\displaystyle a*b=ab+1$

c) On the set of integers Z define a multiplication $\displaystyle *$ by $\displaystyle a*b=1+b$ for any $\displaystyle a,b \in Z$ . Show that <Z,+,*> is not a ring, where + is the usual addition.

I think this fails because $\displaystyle (a*b)*c=1+c$, but $\displaystyle a*(b*c)= 2+c$ so not associative under multiplication. However this is for 6 marks so feel like I must be missing something!

d) Prove the assertion using only axioms of a ring. Let R be a ring with a an element in R. Show there is exactly one element in which $\displaystyle a+b=b+a=0$.

For this I assume two distinct elements b and c satisfy this i.e.

Then $\displaystyle (a+b)*c=0$, but $\displaystyle (b+c)*a=a*b+a*c$ not zero so assumption fails. Not sure if this is precise enough??

e) Is it true that every subring in a commutative domain R with unity is an ideal in R? Prove your claim.

I think this is true but no idea how to prove it

Sorry I know this is quite a few questions just want a bit of verification that what I'm doing is right and a bit of help with the last question!

2. You could have added counterexamples to false statements to make it easier for people to check.

I agree for a) -- c).

Originally Posted by leshields
d) Prove the assertion using only axioms of a ring. Let R be a ring with a an element in R. Show there is exactly one element in which $\displaystyle a+b=b+a=0$.

For this I assume two distinct elements b and c satisfy this i.e. ???

Then $\displaystyle (a+b)*c=0$, but $\displaystyle (b+c)*a=a*b+a*c$ not zero so assumption fails.
This I don't understand at all. Yes, (a + b) * c = 0, but why a * b + a * c != 0? What is the contradiction?

Originally Posted by leshields
e) Is it true that every subring in a commutative domain R with unity is an ideal in R? Prove your claim.
What about reals as a subring of complex numbers?

3. Oh yeah that doesn't make sense! and i missed i.e. $\displaystyle a+b=0$and$\displaystyle a+c=0$

how about if I say $\displaystyle (b+a)+c=b+(a+c)$ implies$\displaystyle 0+c=b+0$ hence $\displaystyle b=c$

e) Not true!!

Reals obviously a subring of Complex with unity

If $\displaystyle a\in R$ and $\displaystyle b\in C$ then R is an ideal of C iff $\displaystyle b.a=a.b \in R$this is clearly not the case for example taking a=1 and b=i

Hoping thats a bit better??

4. Yes, I believe this is correct. I would only replace "iff" in the second last sentence by "only if" because the fact that ba = ab is in R for these particular a and b is not a sufficient condition for R to be an ideal.

5. Brill, thanks for your help!!