Q. 134.0( 4 Votes )

Show that f : N <

Answer :

For proving One – One


Case 1 – When x is odd.


Let x1 and x2 be two distinct numbers such that x1,x2 Є N


Now if the function is one – one both distinct elements from the domain can’t give the same image.


So if we suppose f(x1) = f(x2) , on solving we must get x1 = x2 if the function is one-one


Let, f (x1) = f (x2)


x1 + 1 = x2 + 1 {using equation given in question}


x1 = x2


f is an one – one function.


Case 2 – When x is even.


Let x1 and x2 be two distinct numbers such that x1,x2 Є N


Now if the function is one – one both distinct elements from the domain can’t give the same image.


So if we suppose f(x1) = f(x2) , on solving we must get x1 = x2 if the function is one-one


Let, f (x1) = f (x2)


x1 - 1 = x2 - 1 {using equation given in question}


x1 = x2


f is an one – one function


For an Onto Function –


Case 1 – When x is odd.


Let y = f(x) = x + 1


x = y – 1


Thus, y is defined for each and every value of y we have x – defined for it. We can say that each element in the co-domain of f has a pre-image in domain


f is onto.


Case 2 – When x is even.


Let y = f(x) = x - 1


x = y + 1


Thus, y is defined for each and every value of y we have x – defined for it. We can say that each element in the co-domain of f has a pre-image in domain


f is onto.


Hence f is both one-one & onto.


OR


R × R R given by


a * b =|a - b| a,b Є R


For Commutative


a + b = |a - b| a,b Є R


b * a = |b - a| a,b Є R


|a - b|= |b - a|


* is commutative.


For Associative


a * (b *c) = a * |b - c| = |a - |b - c||


(a * b) * c = |a - b| * c =||a - b| - c|


a * (b * c) (a * b) * c


Where a,b,c R


* is not associative.


R × R R given by


a 0 b = a where a,b Є R


For Commutative –


a 0 b = a where a,b Є R


b 0 a = b where a,b Є R


a 0 b b 0 a


0 is not commutative.


For Associative –


a 0 (b 0 c) = a 0 b = a where a,b,c Є R


(a 0 b) 0 c = a 0 c = a where a,b,c Є R


a 0 (b 0 c) = (a 0 b) 0 c


0 is associative.


Hence * is commutative but not associative and 0 is associative but not commutative.


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