Below is a list of claims and attempted \"proofs\", which may have errors and mi

Question

Below is a list of claims and attempted "proofs", which may have errors and missing justifications. Re-write the proof to correct for these issues. If there are no issues, then write "The proof is correct". Claim. The function f: (0, infinity) rightarrow R given by f(x) = ln(x) is surjective. "Proof." Let y elementof R. Then, e^y elementof R and f(e^y) = ln(e^y) = y. We can conclude that f is surjective. Claim. The function f: Z rightarrow Z given by f(n) = 5n - 6 is injective. "Proof." Suppose f(n) = f(m) for some n, m elementof Z. Then, 5n - 6 = 5m - 6. After adding 6 to both sides and then dividing by 5, we get n = m. We can conclude that f is injective. Claim. Let f: X rightarrow Y and g:Y rightarrow Z. If (g f): X rightarrow Z is onto, then g is onto. "Proof." Assume (g f) is onto. If y elementof Y, then g(y) elementof Z. Therefore, g is onto. Claim. Let f: X rightarrow Y and g: Y rightarrow Z. If f and g are one-to-one, then (g compositecuncion f): X rightarrow Z is one-to-one. "Proof.". Let x_1, x_2 elementof X and assume x_1 = x_2. Because f is one-to-one, f(x_1) = f(x_2). Because g is one-to-one, g(f(x_1)) = g(f(x_2)). Thus, (g compositefunciton f)(x_1) = (g f)(x_2). We can conclude that (g f) is one-to-one.

Explanation / Answer

a) A given function is Surjective if for every element y in Y, there exists atleast one element x in X such that f(x) = y.

Now if every element y in Y is a Real number then ey should also be a Real Number.

So for substitute ey as x

f(ey) = ln(ey) = y

And since both y and ey belong to R therefore the given function is Surjective.

b) To show if a given function is injective is show if for every integer x and y, f(x) = f(y)

or f(n) = f(m)

therefore 5n - 6 = 5m - 6

=> n = m

hence the given function is injective.

c) g o f = g(f(x)) is onto if for every element in g(f(x)) there exists an element in f(x).

let the element in f(x) be y

So if g o f is onto then g(y1) = g(y2). This means for every element g(y), there exists an element y. Thus g is onto.

d) To show that g o f is injective (one to one) one needs to show that g o f (x1) = g o f (x2)

Now if f is injective, then

f(x1) = f(x2)

And if g is also injective then

g(f(x1)) = g(f(x2))

Which can be written as g o f (x1) = g o f (x2).

Thus g o f mapping is injective (one to one).

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