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Show that the set N of natural members is a semi-group under the operation $x * y $= max {x, y}. Is it a monoid?

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  • A non-empty set S with an operation $*$ i.e., $(S, *)$ is said to be a semi-group if it satisfies the following axioms.
  • (1) Closure axiom : $a,b\in S\Rightarrow a*b\in S$
  • (2) Associative axiom : $(a * b) * c = a * (b * c), \forall a, b, c \in S$.
  • A non-empty set M with an operation $* $ i.e., $(M, *)$ is said to be a monoid if it satisfies the following axioms :
  • (1)Closure axiom:$a,b\in M⇒a*b\in M$
  • (2)Associative axiom : $(a*b)*c=a*(b*c)\forall a,b,c\in M$
  • (3)Identity axiom :There exists an element $e \in M$ such that $a * e = e * a = a, \forall a \in M.$
Step 1:
Consider the set $N$ of natural numbers under $x*y$=max$\{x,y\}\;\;x,y\in N$
Closure :$x*y=x$ if $x\geq y$
$\qquad\qquad\quad\;=y\;\;$if $x< y$
In either case $x*y\in S$.So closure property is satisfied.
Step 2:
Let $x,y,z\in N$
Associative :$x*y=max\{x,y\}$
$\therefore (x*y)*z=x*(y*z)$
Associative property is satisfied.$(N,*)$ is a semi group.
Step 3:
Identity element :
For $x,1\in N$
$\therefore 1\in N$ is the identity element.
$(N,*)$ is a monoid.
answered Sep 13, 2013 by sreemathi.v

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