I need the following answered! Part I: Functions (25 pt.) 1. (20 pt., 4 pt. each

Question

I need the following answered!

Part I: Functions (25 pt.) 1. (20 pt., 4 pt. each) For each of the following functions, determine whether the function is: Injective (one-to-one). .Surjective (onto). » Bijective. Invertible. x5 + 1. f: R+ R+ such that f(x) a. b, f: R R such that f(x) = |x/2j. c. f: R R such that f(x)- d, f: R-(0) R such that f(x)-x+1. e, f: z × z z such that f(x,y) = x + y. 2. (5 pt., 2.5 pt. each) Let g be a function from a set A to a set B and let f be a function from set B to a set C. a. If both f and g are injective (one-to-one), then f g is also injective. Explain why. b. If both f and g are surjective (onto), then f ° g is also surjective. Explain why.

Explanation / Answer

1.

a)

Injective : Yes [No two positive real numbers to the power 5 has same value]

Surjective : No [All the positive real numbers are not covered]

Bijective : No [For a function be be bijective it has to be both injective and surjective]

Invertible : Yes [ f(y) = fifthroot(y-1) is the inverse function]

b)

Injective : No [f(x) gives same value for adjacent even odd numbers. E.g. f(4) & f(5) is same]

Surjective : Yes [All real numbers are generated by dividing their double by half]

Bijective : No [For a function be be bijective it has to be both injective and surjective]

Invertible : No [A function which is not injective is not invertible]

c)

Injective : No [f(x) has same value for a positive and negative number. E.g. 1 & -1, 2 & -2]

Surjective : No [Negative real numbers are not in the range of this functions]

Bijective : No [For a function be be bijective it has to be both injective and surjective]

Invertible :No [A function which is not injective is not invertible]

d)

Injective : Yes [f(x) for no two real numbers would be same, when number divides 1 plus itslef]

Surjective : Yes [f(s) covers all the real numbers]

Bijective : Yes [For a function be be bijective it has to be both injective and surjective]

Invertible : Yes [f(y) = 1/y-1 is the inverse function]

e)

Injective : No [sum of two integers could be same]

Surjective : Yes [f(x,y) convers all the integers]

Bijective : No [For a function be be bijective it has to be both injective and surjective]

Invertible :No [A function which is not injective is not invertible]

2.

a) Let a,bA. If

(fg)(a)=(fg)(b) then

f(g(a))=f(g(b))

Since f is injective, we know that g(a)=g(b). Also g is injective is must be that a=b. Hence fg is injective.

b) Since g is surjective, there is a aA such that g(a)=b. Also f is surjective so that for all cC there is a bB such that f(b)=c. Hence, for all cC there is an aA such that (fg)(a)=f(g(a))=f(b)=c.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.