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# Groups/ binary ops watch

1. Hi guys,

Im having some trouble with a practice question, i was wondering if anyone could help me out.
here it is:
Let T = {0,1,2}
i) How many binary operations * can you define on T?

I dont even know where to start
2. What do you think the question is asking? what do you know about binary operations?
3. Well, how many choices are there for what can 0*1 be? What about 2*1? And what about 2*0? And perhaps 1*2? Notice a pattern?
4. (Original post by Unbounded)
Well, how many choices are there for what can 0*1 be? What about 2*1? And what about 2*0? And perhaps 1*2? Notice a pattern?
Your advice is ringing bells but i dont really know where there going

Sorry if im being silly,
5. (Original post by tsmith16)
Your advice is ringing bells but i dont really know where there going

Sorry if im being silly,
Consider all the possibilties basically is what he is saying!
6. Well, 0*1 could be 0. It could also be 1, and it could also be 2. And given that we're just counting arbitrary binary operations, there's no reason why it can't be any of those three options. Hence, there are three choices for 0*1.

Similarly, for 2*1, there will also be three choices. And in general for any pair , a*b will have three different choices.

Further, once we choose the values of a*b for each (a,b), we will have defined uniquely one of the possible binary operations on {0,1,2}. So now we just need to count up how many possible ways we can do all of this.
BIGGER HINT
How many distinct pairs (a,b) are there? How many choices are there for a*b, given a specific pair (a,b) [I have told you this already]? So how many ways are there to assign a value to a*b for all the pairs (a,b)? Thus how many different binary operations are there on the set {0,1,2}?
Remember, a binary operation on a set S is simply a function f : SxS -> S i.e. it is just a function mapping elements from SxS to S.
7. (Original post by Unbounded)
Well, 0*1 could be 0. It could also be 1, and it could also be 2. And given that we're just counting arbitrary binary operations, there's no reason why it can't be any of those three options. Hence, there are three choices for 0*1.

Similarly, for 2*1, there will also be three choices. And in general for any pair , a*b will have three different choices.

Further, once we choose the values of a*b for each (a,b), we will have defined uniquely one of the possible binary operations on {0,1,2}. So now we just need to count up how many possible ways we can do all of this.
BIGGER HINT
How many distinct pairs (a,b) are there? How many choices are there for a*b, given a specific pair (a,b) [I have told you this already]? So how many ways are there to assign a value to a*b for all the pairs (a,b)? Thus how many different binary operations are there on the set {0,1,2}?
This isn't necessarily the best thing to do if {0,1,2} is a group under * (which I assume is the case given the thread title) and not just a set with a binary operation.

My advice would be: you have a choice over which element is the identity; do you have a choice over what the other two elements can do?
8. (Original post by nuodai)
This isn't necessarily the best thing to do if {0,1,2} is a group under * (which I assume is the case given the thread title).
You know what they say about assumptions.

(The question has been posted before over the last few days - it's definitely "how many binary operations" as the 2nd part is "how many groups").
9. (Original post by nuodai)
This isn't necessarily the best thing to do if {0,1,2} is a group under * (which I assume is the case given the thread title) and not just a set with a binary operation.

My advice would be: you have a choice over which element is the identity; do you have a choice over what the other two elements can do?
Well he posted (i) of the question, so I merely assumed he just wanted to know how many (arbitrary) binary operations there were - as opposed to how many binary operations are there that make T into a group [which would presumably be (ii) of the question].
10. (Original post by DFranklin)
You know what they say about assumptions.
Alas, indeed.
11. (Original post by nuodai)
Alas, indeed.
To be clear, it wasn't unreasonable, but given I'd seen the previous thread I was pretty sure it wasn't the right path here.
12. (Original post by Unbounded)
Well, 0*1 could be 0. It could also be 1, and it could also be 2. And given that we're just counting arbitrary binary operations, there's no reason why it can't be any of those three options. Hence, there are three choices for 0*1.

Similarly, for 2*1, there will also be three choices. And in general for any pair , a*b will have three different choices.

Further, once we choose the values of a*b for each (a,b), we will have defined uniquely one of the possible binary operations on {0,1,2}. So now we just need to count up how many possible ways we can do all of this.
BIGGER HINT
How many distinct pairs (a,b) are there? How many choices are there for a*b, given a specific pair (a,b) [I have told you this already]? So how many ways are there to assign a value to a*b for all the pairs (a,b)? Thus how many different binary operations are there on the set {0,1,2}?
Remember, a binary operation on a set S is simply a function f : SxS -> S i.e. it is just a function mapping elements from SxS to S.
hmmm, its making abit more sense, sorry its been a while since i've seen binary ops, can you give me one example to kick start me?
thanks again, i know how frustrating this can be
13. Do you guys want me to post part (ii) of the question?
14. (Original post by nuodai)
This isn't necessarily the best thing to do if {0,1,2} is a group under * (which I assume is the case given the thread title) and not just a set with a binary operation.

My advice would be: you have a choice over which element is the identity; do you have a choice over what the other two elements can do?
Hi, are there 18 binary ops that can be defined on T? I've done a little more research and i've come up with:

(0*1,0*2,1*0,1*2,2*0,2*1), all of which have three options, giving 18 bin ops.

ps, can i have a a pair (0*0) or (1*1) or (2*2)?
EDIT: ive assumed, (i think correctly) that i can use (1*1) and (2*2), but im not sure if i can use (0*0) because you cant do get any other number other than 0 given any binary operation with the pair (0*0).

Thanks
15. (Original post by Unbounded)
Well he posted (i) of the question, so I merely assumed he just wanted to know how many (arbitrary) binary operations there were - as opposed to how many binary operations are there that make T into a group [which would presumably be (ii) of the question].
hey ,
i think youve got the main point of my question, is there 3^(3^2) arbitrary binary ops on any set of 3 elements, giving 19,683 bin ops on T?
thanks

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