for a,b∈N , a*b=max{ a,b} . then what is not true
1= * is binary operation on N
2=* IS COMMUTATIVE
3=identity element of N UNDER the operation * is 1
4= every element of N has inverse element in N under the binary operation
Answers
Answer:
4 is not true plzzzzzzz make it a brainlist answer plzzzzzzz
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Check the binary operation * is commutative :
We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
L.H.S =(a, b) * (c, d)
=(a + c, b + d)
R. H. S = (c, d) * (a, b)
=(a + c, b + d)
Hence, L.H.S = R. H. S
Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
* is commutative (a, b) * (c, d) = (a + c, b + d)
Check the binary operation * is associative :
We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
L.H.S = (a, b) * ( (c, d) * (x, y) ) = (a+c+x, b+d+y)
R.H.S = ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)
Thus, L.H.S = R.H.S
Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
Thus, the binary operation * is associative
Checking for Identity Element:
e is identity of * if (a, b) * e = e * (a, b) = (a, b)
where e = (x, y)
Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)
= (x + a , b + y) = (a, b)
Now, (a + x, b + y) = (a, b)
Now comparing these, we get:
a+x = a
x = a -a = 0
Next compare: b +y = b
y = b-b = 0
Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.