Commutative and Associative relations of binary operations

**leshields** · May 15th 2011, 04:35 AM

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 $z^+$ say whether or not it is associative, commutative
i) $a*b=a+b^2$

This is neither commutative nor associative

ii) $a*b=a+1$

Is again neither commutative nor associative

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

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

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

I think this fails because $(a*b)*c=1+c$ , but $a*(b*c)= 2+c$ 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 $a+b=b+a=0$ .

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

Then $(a+b)*c=0$ , but $(b+c)*a=a*b+a*c$ 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!

**emakarov** · May 15th 2011, 05:33 AM

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 $a+b=b+a=0$ .

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

Then $(a+b)*c=0$ , but $(b+c)*a=a*b+a*c$ 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?

**leshields** · May 15th 2011, 05:55 AM

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

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

e) Not true!!

Reals obviously a subring of Complex with unity

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

Hoping thats a bit better??

**emakarov** · May 15th 2011, 06:10 AM

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.

**leshields** · May 15th 2011, 06:39 AM

Brill, thanks for your help!!

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