39. Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.

f(1)=a,f(2)b,and,f(3)=c If we define g:{a,b,c}→{1,2,3}as,g(a)=1,g(b)=2,g(c)=3, then we have: (fog)(a)=f(g(a))=f(1)=a(fog)(b)=f(g(b))=f(2)=b(fog)(c)=f(g(c))=f(3)=cAnd(gof)(1)=g(f(1))=f(a)=1(gof)(2)=g(f(2))=f(b)=2(gof)(3)=g(f(3))=f(c)=3∴gof=IXand,fog=IYWhere,X={1,2,3},and,Y={a,b,c}. Thus, the inverse of f exists and f−1=g. ∴f−1:{a,b,c}→{1,2,3} is given by, f−1(a)=1,f−1(b)=2,f−1(c)=3 Let us now find the inverse of f−1 i.e., find the inverse of g. If we define h:{1,2,3}→{a,b,c}as h(1)=a,h(2)=b,h(3)=c , then we have (goh)(1)=g(h(1))=g(a)=1(goh)(2)=g(h(2))=g(b)=2(goh)(3)=g(h(3))=g(c)=3And(hog)(a)=h(g(a))=h(1)=a(hog)(b)=h(g(b))=h(2)=b(hog)(c)=h(g(c))=h(3)=c∴goh=IXand,hog=IYWhere,X={1,2,3},and,Y={a,b,c}. Thus, the inverse of g exists and g−1=h⇒(f−1)−1=h. It can be noted that h=f. Hence, (f−1)−1=f.

39. Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f⁻¹ and show that (f⁻¹⁾⁻¹ = f.

9 Views|Posted 9 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

9 months ago

$f (1) = a, f (2) b, a n d, f (3) = c$

If we define $g : {a, b, c} \to {1, 2, 3} a s, g (a) = 1, g (b) = 2, g (c) = 3,$ then we have:

$\begin{array}{l} (f o g) (a) = f (g (a)) = f (1) = a \\ (f o g) (b) = f (g (b)) = f (2) = b \\ (f o g) (c) = f (g (c)) = f (3) = c \\ A n d \\ (g o f) (1) = g (f (1)) = f (a) = 1 \\ (g o f) (2) = g (f (2)) = f (b) = 2 \\ (g o f) (3) = g (f (3)) = f (c) = 3 \\ ∴ g o f = I_{X} a n d, f o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of f exists and $f^{- 1} = g .$

$∴ f^{- 1} : {a, b, c} \to {1, 2, 3}$ is given by,

$\begin{array}{l} f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \end{array}$

Let us now find the inverse of $f^{- 1}$ i.e., find the inverse of g.

If we define $h : {1, 2, 3} \to {a, b, c} a s$

$h (1) = a, h (2) = b, h (3) = c$ , then we have

$\begin{array}{l} (g o h) (1) = g (h (1)) = g (a) = 1 \\ (g o h) (2) = g (h (2)) = g (b) = 2 \\ (g o h) (3) = g (h (3)) = g (c) = 3 \\ A n d \\ (h o g) (a) = h (g (a)) = h (1) = a \\ (h o g) (b) = h (g (b)) = h (2) = b \\ (h o g) (c) = h (g (c)) = h (3) = c \\ ∴ g o h = I_{X} a n d, h o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of g exists and $g^{- 1} = h \Rightarrow {(f^{- 1})}^{- 1} = h .$

It can be noted that h=f.

Hence, ${(f^{- 1})}^{- 1} = f .$

Upvote

Similar Questions for you

Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

R₂ = {(a, b) : a is integral multiple of b} a, b Î s

n(R₁ – R₂) = ?

Alok Kumar Singh

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12)

Alok Kumar Singh

$g o f (x) = {\begin{matrix} f (x), & f (x) < 0 \\ f (x), & f (x) > 0 \end{matrix}$

$= {\begin{array}{l} e^{l n x} = x & (0, 1) & e^{- x} \\ (- \infty, 0) & l n x & (1, \infty) \end{array}$

Alok Kumar Singh

$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Alok Kumar Singh

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$