binary operation on a set

Oct 2009
255
20
St. Louis Area
Given a set S, a function f: S X S \(\displaystyle \longrightarrow\) S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?

I have no clue on this one? Could someone point me in the right direction?

I realize that if S has n elements then SxS would have \(\displaystyle n^2\) possible ordered pairs. However, I'm not sure what is meant by how many different binary operations. Would we have addition, subtraction, multiplication, division, exponentiation, etc. times \(\displaystyle n^2\) binary operations?
 

Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
Given a set S, a function f: S X S \(\displaystyle \longrightarrow\) S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?

I have no clue on this one? Could someone point me in the right direction?

I realize that if S has n elements then SxS would have \(\displaystyle n^2\) possible ordered pairs. However, I'm not sure what is meant by how many different binary operations. Would we have addition, subtraction, multiplication, division, exponentiation, etc. times \(\displaystyle n^2\) binary operations?
It's asking how many functions are there from a set with \(\displaystyle n^2\) elements to a set with \(\displaystyle n\) elements.
 
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Oct 2009
255
20
St. Louis Area
Given a set S, a function f: S X S \(\displaystyle \longrightarrow\) S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?
With the help of Drexel28, we have a set S with n elements which gives \(\displaystyle n^2\) mapped to n. This should give us \(\displaystyle n^3\) binary operations, correct?
 
Apr 2009
678
140
With the help of Drexel28, we have a set S with n elements which gives \(\displaystyle n^2\) mapped to n. This should give us \(\displaystyle n^3\) binary operations, correct?

Shouldn't it be (n)^(n^2)
 
Jun 2012
1
0
Missouri
Yes I believe that (n)^(n^2) is correct
 

Deveno

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Mar 2011
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Tejas
in general, the number of functions f:A→B is:

|B||A|.

it is easiest to see this when |B| = 2, such as when B = {0,1}, so that the functions f:A→{0,1} can be put in a 1-1 correspondence with the subset of A:

given a subset S of A, we define:

f(a) = 1, if a is in S
f(a) = 0, if a is not in S.

thus the number of functions f:A→{0,1} is the same number as 2|A|, the cardinality of the power set of A.

so, yes, the correct answer is n(n2).