Thread: Prove rational numbers form a group (associativity - done most of it)

Q: Prove that the set of rational numbers form a group under the operation * where a * b = 2a + 3b + 4ab

I have to prove (a * b) * c = a * (b * c)

After some working and simplifying, I get

(a * b) * c = 12a + 18b + 24ab + 4c

This is the first half of the proof (for associativity)

The second part, a * y = 2a + 6b + 9c + 12bc + 8ab + 12ac + 16abc

I don't think this can be simplified to equal my first result, but it has to; please help me

(a*b)*c=4a+6b+3c+8ab+4ac+12bc+16abc
a*(b*c)=2a+6b+9c+8ab+12ac+12bc+16abc

these 2 equations will not be equal and they cant be simplified.

if you proceed to try and find inverse and identity element you will also run into trouble.

therefore the set as defined is not a group

Originally Posted by jaco

(a*b)*c=4a+6b+3c+8ab+4ac+12bc+16abc
a*(b*c)=2a+6b+9c+8ab+12ac+12bc+16abc

these 2 equations will not be equal and they cant be simplified.

if you proceed to try and find inverse and identity element you will also run into trouble.

therefore the set as defined is not a group

Thanks. I was getting frustrated that I couldn't simplify it...I guess it must have been a trick question.

Thanks again