1i and 1L only Exercises 5.3 1. For each function f, (i) determine whether f is

Question

1i and 1L only Exercises 5.3 1. For each function f, (i) determine whether f is 1-1; (i) determine whether f is onto Prove your answers. (a) f : R R by f(x) = x+1x1. a, b, c, d E R have the propertythat ad-be 0 and c+0 (d) f : (-00,3] [2,oo) by f(x)-(x-3)2 + 2. (e) f : (-00, 1] (-00,4] by f(x) = 4-VI-23 (f) f : R2 R by f(x,y) =x + y. (g) f:R2 R by f(z, y) = (x-y)3. (h) f : R R2,by f(x) =(c,x). (i) f : R2 R2 by f(x,y) (x+ y, x-y). (i) f : R2 R3 by f(x,y) = (x + y, x-y,ry). (k) f : R3 R3 by f(x, y, z) = (x + y, y + z, x + z). (I) f : R3 R2 by f(x, y, z) (x + y, y + z) (m) f : Z+ × Z+ Z+ by f(m, n) = 2m-1(2n-1). (HINT: You will need the Fundamental Theorem of Arithmetic (Theorem 2.3.3).) (n) F : P P by F(p) elements of A. 2. Let f.7 he defined by =p'. (See Exercise 5.1.7.) (o) f : C Z, where C = {A E P(Z) | A is finite) and f(A) is the sum of all

Explanation / Answer

1(i) f(x,y) = (x+y , x-y)

Now f(x1,y1) = f(x2,y2)

=> (x1+y1,x1-y1) = (x2+y2,x2y2)

=> x1+y1 = x2+y2 and

x1-y1 = x2- y2

Add both two , 2x1 = 2x2 => x1 = x2 and hence y1 = y2

Therefore f is one - one

Also f(x,y) = (x+y,x-y)

write x = ((x+y)+(x-y)) /2 and y = ((x+y)-(x-y)) /2

=> f( ((x+y)+(x-y)) /2 ,  ((x+y)-(x-y)) /2 ) = (x+y,x-y)

Therefore clearly f is onto .

1L :

f(x,y,z) = (x+y , y+z)

f(x1,y1,z1) = (x1+y1 , y1+z1)

f(x2,y2,z2) = (x2+y2 , y2+z2)

f(x1,y1,z1)= f(x2,y2,z2)

=> (x1+y1 , y1+z1) = (x2+y2 , y2+z2)  

which does not imply x1 = x2 , y1 = y2 and z1 = z2 therefore , f is not one-one.

f(x,y,z) = (x+y , y+z)

Clealry for any x+y , and y +z , two equations , we can find suitable three variables x , y , z , therefore f is onto

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