# Show that f : N <

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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