2. Check the injectivity and surjectivity of the following functions: (i) f: N → N given by f(x) = x2 (ii) f: Z → Z given by f(x) = x2 (iii) f: R → R given by f(x) = x2 (iv) f: N → N given by f(x) = x3 (v) f: Z → Z given by f(x) = x3

(i) f:N→N given by f(x)=x2 For, x1,x2∈N , f(x1)=f(x2) ⇒x12=x22 ⇒x1=x2∉N So, f is one-one/ injective For x∈N , i.e., x=1,2,3.... Range of f(x)={12,22,32...}={1,4,9...}≠N i.e., co-domain of N So, f is not onto/ subjective (ii) f:Z→Z given by f(x)=x2 For, x1,x2∈Z , f(x1)=f(x2) ⇒x12=x22 ⇒x1=±x2∉Z i.e., x1=x2 and x1=−x2 So, f is not one-one/ injective For x∈Z , x=0,±1,±2,±3.... Range of f(x)={02,(±1)2,(±2)2,(±3)2...} {0,1,4,9....}≠ co-domain Z So, f is not onto/ subjective (iii) f:R→R given by f(x)=x2 For, x1,x2∈R , f(x1)=f(x2) ⇒x12=x22 ⇒x1=±x2 So, f is not injective For x∈R Range of f(x)={x2,x∈R} gives a set of all positive real numbers Hence, range of f(x≠) co-domain of R So, f is not subjective (iv) f:N→N given by f(x)=x3 For, x1,x2∈N , f(x1)=f(x2) ⇒x13=x23 ⇒x1=x2∈N So, f is injective For x∈N, Range of f(x)={13,23,33....} {1,8,27....}≠ co-domain of N So, f is not subjective (v) f:R→R given by f(x)=x3 For, x1,x2∈N , f(x1)=f(x2) ⇒x13=x23 ⇒x1=x2∈N So, f is injective For x∈R, Range of f(x)={13,23,33....} {1,8,27....}≠ co-domain of R So, f is not subjective

2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

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Vishal Baghel | Contributor-Level 10

9 months ago

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives

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