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The binary operation defined on a,b∈z such that a*b= min⁡(a,b)then (A,*) is

a)monoid

b)group

c)algebric-structure

d)semi group
in Set Theory & Algebra by Active (2.3k points) | 177 views
0
It is semigroup..
0
monoid too
0
what is identity element ?
0
Semigroup. As identity will go like +∞
0
Can be e
rt?
0
no unique identty element ...

but having closure and assciativity property....so semigroup
0
z could be infinity

now(infinity,x)=x

So, why not identity element infinity
0

can we consider as element ?

it is just a bound right.

even while representing also (-∞,∞) means they are not closed.

0

I don't mean real number here.

i am just asking : can we consider  as element or not?

0
@Anu @hs @gaurav
why infinity couldnot be identity element??
+1
Probably because ∞ is not a number. It is not even a member of set Z.
It is a notation / concept we follow to indicate a very large number which has no bound.

1 Answer

+1 vote
Best answer

For Checking if $(\mathbb{Z},*)$ is a semi-group we need to check 2 conditions

1) $\mathbb{Z}$ is closed under * 
     True
     "
A set is closed under some operation if applying the operation on any elements of the set gives an element which is still in that set"

2) $*$ is an associative operation 
    True
 Let us assume a<b<c | a,b,c $\in \mathbb{Z}$

$(a*b)*c = a*(b*c)$
$a*c = a*b$
$a = a$

Hence it is Semi-group

A monoid is a semi-group with an identity.

There exist no identity for operation (min) in set of integers because there is no minimum element in $\mathbb{Z}$ to satisfy.
Hence it is not a monoid 

by Boss (34.7k points)
selected by
0
@Mk Utkarsh can you explain with the help of example for monoid case?
0
In this same example if the binary operation was performed on $\mathbb{N}$ then identity would have been $1$ and hence monoid.
0

Thanks.

https://gateoverflow.in/20581/firstly-what-is-identity-element-for-this-group

@Mk Utkarsh can you help this..

here the answer should be 8.

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