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The description is that you can practice these quiz proof questions for infimum/supremum
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Question (4 points). Consider the function
f : R> 0 โ R> 0 , f (x) =
x + 1 5
(a) Is f injective? Justify your answer.
Solution: Yes, f is injective. For x 1 , x 2 โ R> 0 , we have
f (x 1 ) = f (x 2 ) =โ x 1 + 1 5
x 2 + 1 5 =โ x 1 + 1 = x 2 + 1 =โ x 1 = x 2.
(b) Is f surjective? Justify your answer.
Solution: No, f is not surjective. For all x โ R> 0 , we have
x > 0 =โ x + 1 > 1 =โ f (x) = x + 1 5
Hence, for example, there is no x โ R> 0 such that f (x) = 101.
(c) Does f have a left inverse? If it does, give one and show that it is indeed a left inverse. Otherwise, justify why f does not have a left inverse.
Solution: Yes. A function has a left inverse if and only if it is injective. Since f is injective, it has a left inverse. Define
g : R> 0 โ R> 0 , g(y) = | 5 y โ 1 |.
2
Then, for all x โ R> 0 , we have
(g โฆ f )(x) = g(f (x)) = g
x + 1 5
x + 1 5
โ 1 = |x| = x,
since x > 0. Thus, g โฆ f = idR> 0 , and so g is a left inverse of f. Important! We cannot define g by g(y) = 5y โ 1, since this does not always take values in the codomain R> 0. For example 5(0) โ 1 โ/ R> 0.
(d) Does f have a right inverse? If it does, give one and show that it is indeed a right inverse. Otherwise, justify why f does not have a right inverse.
Solution: No. A function has a right inverse if and only if it is surjective. Since f is not surjective, it does not have a right inverse.