Question

Show that the function f : N ---> N defined by
f(x) = { x + 1 if x is odd
{ x - 1 if x is even

is one - one and onto

CBSE
CLASS 12TH mathematics

Need step by step explanation

Answer 1

Answer:

Step-by-step explanation:

Answer 2

solution:

one-one:

Here we discuss the following possible case:

$(1).when \: x_{1} \: is \: odd \: and \: x_{2} \: is \: even. \\ \\ here \: f(x_{1}) = f(x_{2}) = > x_{1} + 1 = x_{2} - 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x_{2} - x_{1} = 2 </p><p>$

which is impossible.

$(2).when \: x_{1} \: is \: even \: and \: x_{2} \: is \: odd. \\ \\ here \: f(x_{1}) = f(x_{2}) = > x_{1} - 1 = x_{2} + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x_{1} - x_{2} = 2 \\ \\ </p><p>$

Which is impossible.

$(3).when \: x_{1} \: and \: x_{2} \: are \: both \: odd.\\ \\ here \: f(x_{1}) = f(x_{2}) = > x_{1} + 1 = x_{2} + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x_{1} = x_{2} \\ \\ </p><p>$

∴ 'f' is one- one.

$(4).when \: x_{1} \: and \: x_{2} \: are \: both \:even.\\ \\ here \: f(x_{1}) = f(x_{2}) = > x_{1} - 1 = x_{2} - 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x_{1} = x_{2} </p><p>$

∴'f' is one-one.

Onto:

Let 'x' be an arbitrary natural number.

when x is an odd natural number , then there exists an even natural number (x+1) such that:

$f(x + 1) = (x + 1) - 1 = x$

when x is an even natural number , then there exists an odd natural number (x-1) such that:

$f(x - 1) = (x - 1) + 1 = x$

∴ Each x ∈ N has its pre-image in N.

Thus 'f' is Onto.

Hence , 'f' is both one- one and onto.