Group theory with addition, multiplication and division

[penroseandpaper]

Hi everyone,

I'm working through some group theory questions online. But unfortunately they don't have answers to go with them. So, I'm hoping you can say if I'm on the right track.

If this is a binary operation on ℝ, am I right in thinking it satisfies the closure and associativity axioms but fails the identity axiom because there is no value 'e' which satisfies both a+e=e and ae=a?

a•b=(a+b)/ab

Thanks

 

[member 587159]

Are you sure this is defined on $\Bbb{R}$? Then for example $1 \cdot 0$ gives division by $0$.

Maybe, the domain is $\Bbb{R}\setminus \{0\}$?

Also, I don't believe this is associative ..

 

[PeroK]

but fails the identity axiom because there is no value 'e' which satisfies both a+e=e and ae=a?

If you are learning group theory on your own, then you have got confused over the axioms. That $e$ is a mixture of the additive and multiplicative identities. You'll not find one that does both.

 

[penroseandpaper]

Are you sure this is defined on R? Then for example 1⋅0 gives division by 0.

Maybe, the domain is R∖{0}?

Also, I don't believe this is associative .

My apologies, I thought I'd included a star after ℝ for ℝ*. As for the associativity, you might be right; if I'm correct (big if)

a•(b•c)=c+b+1/a
(a•b)•c=b+a+1/c

 

[penroseandpaper]

If you are learning group theory on your own, then you have got confused over the axioms. That $e$ is a mixture of the additive and multiplicative identities. You'll not find one that does both.

But isn't the fact you won't find one that does both the reason this binary operation doesn't have an identity element? Because a•e=(a+e)/(ae); if e=0 it's undefined and if e=1 it no longer equals a.

Apologies if I'm wrong - I was fine when the operations were just addition, subtraction, multiplication, etc. But can't find much about what to do when they're combined.

 

[PeroK]

My apologies, I thought I'd included a star after ℝ for ℝ*. As for the associativity, you might be right; if I'm correct (big if)

a•(b•c)=c+b+1/a
(a•b)•c=b+a+1/c

Can you show your working? That doesn't look right at all.

 

[PeroK]

But isn't the fact you won't find one that does both the reason this binary operation doesn't have an identity element? Because a•e=(a+e)/(ae); if e=0 it's undefined and if e=1 it no longer equals a.

Apologies if I'm wrong - I was fine when the operations were just addition, subtraction, multiplication, etc. But can't find much about what to do when they're combined.

On the first point you need to look for an identity without preconceptions that it can only be $0$ or $1$.

On the second point, you focus on the binary operation as defined.

 

[penroseandpaper]

Can you show your working? That doesn't look right at all.

I simplified the operation but maybe I shouldn't have.

 

[Office_Shredder]

I think the problem is that 1/(1/ b + 1/c) is not equal to b+c.

 

[penroseandpaper]

I think the problem is that 1/(1/ b + 1/c) is not equal to b+c.

Thanks for pointing that out. Please find corrected solution below (hopefully...). So it's not associative as the two expressions aren't equal.

Before I go in search of something that doesn't exist, is there an identity element for an operation like this?

 

[PeroK]

Thanks for pointing that out. Please find corrected solution below (hopefully...). So it's not associative as the two expressions aren't equal.

Before I go in search of something that doesn't exist, is there an identity element for an operation like this?View attachment 269780

Yes, although technically you should show that these expressions are not equal. It may be fairly obvious, but producing an example $a, b, c$ settles the issue.

What is the algebraic requirement of an identity for this operation? It's important that you are able to write down what being an identity means. If $e$ is an identity for the $\cdot$ operation, then $e$ satisfies ...

 

[Office_Shredder]

For what it's worth, how do you know the things you wrote down aren't equal from some algebraic manipulation you haven't discovered yet? I think a proper proof that this isn't associative is to actually just pick some numbers for a b and c and see if you can get the two things to come out a different numbers (if they aren't equal, almost any choice will work).

For the identity element, the question is if there exists e such that $\frac{a+e}{ae}=a$ for all a. It's pretty simple algebra to pick a value for a and get what e needs to be. Do you get the same e for any choice of a? This is equivalent to asking if you can do algebra to eliminate a from the equation I wrote.

 

[PeroK]

@penroseandpaper this is a very importnat point. Students often struggle with abstract algebra by trying to work generally, when a simple counterexample is all you need. In this case, before you did any algebraic manipulations you could have simply tried a couple of examples for $a, b, c$ to see whether the operation is associative. If this fails, even for a single example, then it is not associative. If it works for your examples, then you could try to prove the general case. So, you could have tried $a, b , c = 1,2, 3$ for example. Just to see what happens.

Also, you made a mistake in trying to simplify $a \cdot (b \cdot c)$. You could have checked your answer by trying a simple set of $a, b, c$ to see if your simplified expression was correct.

Finally, you can do the same with the identity. If $e$ is the identity, then $e \cdot 1 = 1, e \cdot 2 = 2, e \cdot 3 = 3 \dots$. Again, you can work out whether there is such an $e$ by trying these simple examples.

In short, don't forget that you can (and should) try things out with specific numbers to see what happens. For some reason, this is often missing from a student's approach to mathematics.

 

[PeroK]

PS The operation does not satisfy the closure axiom.