Question

Show that the function
f : N➔N, defined by
f(n) = n² + n + 1 for all n∈N is
one-one but not onto function.
$\rule{190pt}{2pt}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: f : N \: \to \: N \: given \: by$

$\rm \: f(n) = {n}^{2} + n + 1$

Step :- 1 One - One function

Let assume that x, y ∈ N such that

$\rm \: f(x) = f(y)$

$\rm \: {x}^{2} + x + 1 = {y}^{2} + y + 1$

$\rm \: {x}^{2} + x = {y}^{2} + y$

$\rm \: {x}^{2} + x - {y}^{2} - y = 0$

$\rm \: ( {x}^{2} - {y}^{2}) + (x - y) = 0$

$\rm \: (x + y)(x - y) + (x - y) = 0$

$\rm \: (x - y)(x + y + 1) = 0$

$\rm\implies \:\rm \: x - y = 0 \: \: \: \: \{ as \:x + y + 1 \ne \: 0 \: as \: x,y \in \: N \}$

$\rm\implies \:x = y$

$\rm\implies \:f(n) \: is \: one \: - \: one$

Step :- 2 Onto

Let if possible there exist an element y ∈ N such that

$\rm \: f(n) = y$

$\rm \: {n}^{2} + n + 1 = y$

$\rm \: {n}^{2} + n + 1 - y = 0$

Now, its a quadratic in n, so as to get the value of n, we use Quadratic Formula, taking positive square root, we get

$\rm \: n = \dfrac{ - 1 \: + \: \sqrt{ {1}^{2} - 4(1)(1 - y)} }{2(1)}$

$\rm \: n = \dfrac{ - 1 \: + \: \sqrt{ 1 - 4 + 4y} }{2}$

$\rm \: n = \dfrac{ - 1 \: + \: \sqrt{ 4y - 3} }{2}$

Let assume that, y = 1, so we get

$\rm \: n = \dfrac{ - 1 \: + \: \sqrt{ 4 - 3} }{2}$

$\rm \: n = \dfrac{ - 1 \: + \: \sqrt{1} }{2}$

$\rm \: n = \dfrac{ - 1 \: + \: 1 }{2}$

$\rm\implies \:n = 0 \: \cancel \in \: N$

$\rm\implies \:y = 1 \: donot \: have \: any \: pre - image$

$\rm\implies \:f(n) \: is \: not \: onto$

Hence,

$\rm \: f : N \: \to \: N \: given \: by$

$\rm \: f(n) = {n}^{2} + n + 1 \: is \: one - one \: but \: not \: onto.$

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Additional Information :-

1. One - One or injective function :- If f is a function from A to B such that distinct elements in A has distinct images in B.

2. Onto or Surjective function :- If f is a function from A to B such that every element in B has atleast one pre - image in A.

3. Bijective function :- A function f from A to B is bijective iff function is one - one and onto.

Answer 2

Step-by-step explanation:

Given

f(x) = 2x

One-One

$\bf \: f(x_{2}) = 2x_{2} \\ \bf f(x_{1}) = 2x_{1}$

Putting

$\bf \: f(x_{1}) = f(x_{2}) \\ \bf 2x_{1} = 2x_{2} \\ \bf x_{1} = x_{2}$

$\blue{ \bf \: Hence, \: if \: f(x_{1}) = f(x_{2}), x_{1} = x_{2}} \\ \blue{ \bf function \: f \: is \: one-one}$

Onto

f(x) = 2x

$\red{\bf \: Let \: f(x) = y ,such \: that \: y \in \: N 2x = y x = \frac{y}{2} }$$\bf \: If y = 1 \: x = \frac{1}{2} = 0.5 x \: in \: N , \\ \bf which \: is \: not \: possible \: as \: x \: \in N$$\boxed{ \purple{\bf \: Hence, \: f \: is \: not \: onto}}$

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F:N to N defined by f(m)=m^2+m+3 is one one function

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