Question

How many binary operations are defined on set {a,b,c,d,} with "a " being an identity element?

Answer 1

so I'm assuming that a has to be part of the binary operation?

Well there would be 3? AB, AC, AD
I think I"m wrong because I'm not sure what an identify element is

Answer 2

4

Answer 3

i don't this this is right
there are in general \(n^{n^2}\) binary operations on a set of \(n\) elements

Answer 4

in your case, if you dropped the restriction that \(a\) is the identity you would have \(4^{16}\) binary operations, but with the restriction we can compute also, if we accept the previous answer as given

Answer 5

lol....Yeap i thought i saw identity matrix

Answer 6

since \(a\) is the identity element, it its ordered pairs are fixed (no choices) so it is now like asking how many binary operations there are on a set of 3 elements, and you should get
\[3^9\]

Answer 7

at least i think that is right.

Answer 8

@satellite73 , it seems u r right. but r u sure abt ur answer

Answer 9

well no, i am never really sure about anything
i am however sure that there are \(|A|^{|C|}\) functions from from C to A, and since a binary operation is a function from from \(A\times A\) to \(A\) there are \(|A|^{|A|\times |A|}\) such binary operations

Answer 10

in your case \(|A|=4\) so there are \(4^{16}\) binary operations, with no restrictions

Answer 11

but i think all this just comes from the counting principle, and by specifying one element to be the identity gives only 3 remaining elements to count possibilities with

Answer 12

on the other hand it might be \(n^{(n-1)^2}\) hmmm

Answer 13

actually i like the second answer more. i speak to quickly some times
i like \(4^9\) for this one

Answer 14

so , if a is identity element, then we have binary operations are 3 ..

Answer 15

my thinking was muddled, sorry.
if you fix the identity, you still have 4 elements in the target set

Answer 16

but you have no choices for the identity, those ordered pairs are fixed

Answer 17

so there are 9 choices for the "domain" not counting \(a\) there are \(3\times 3=9\) possible ordered pairs getting mapped to 4 choices of "range' elements, making it \(4^9\) combinations

Answer 18

this is the final answer

Answer 19

well, i would not bet my house on it, but i am pretty sure it is right

Answer 20

@sauravshakya , plse check

Answer 21

Sorry, what do u mean by binary operation?

Answer 22

like addition or subtraction
abstractly binary operation on a set \(A\) is a function \((A\times A)\to A\)