Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Here aare some proof questions, Quizzes of Mathematics

The description is that you can practice these quiz proof questions for infimum/supremum

Typology: Quizzes

2024/2025

Uploaded on 04/10/2025

daniel-farlinger
daniel-farlinger ๐Ÿ‡บ๐Ÿ‡ธ

1 document

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MAT 1362 โ€“ Winter 2025
Quiz 8 โ€“ DGD 2 (Friday) โ€“ Solutions
Professor: Alistair Savage
Date: March 28, 2025
Question (4 points). Consider the function
f:R>0โ†’R>0, f(x) = x+ 1
5.
(a) Is finjective? Justify your answer.
Solution: Yes, fis injective. For x1, x2โˆˆR>0, we have
f(x1) = f(x2) =โ‡’x1+ 1
5=x2+ 1
5
=โ‡’x1+ 1 = x2+ 1
=โ‡’x1=x2.
(b) Is fsurjective? Justify your answer.
Solution: No, fis not surjective. For all xโˆˆR>0, we have
x > 0 =โ‡’x+ 1 >1 =โ‡’f(x) = x+ 1
5>1
5.
Hence, for example, there is no xโˆˆR>0such that f(x) = 1
10 .
(c) Does fhave a left inverse? If it does, give one and show that it is indeed a left inverse.
Otherwise, justify why fdoes not have a left inverse.
Solution: Yes. A function has a left inverse if and only if it is injective. Since fis
injective, it has a left inverse. Define
g:R>0โ†’R>0, g(y) = |5yโˆ’1|.
pf2

Partial preview of the text

Download Here aare some proof questions and more Quizzes Mathematics in PDF only on Docsity!

MAT 1362 โ€“ Winter 2025

Quiz 8 โ€“ DGD 2 (Friday) โ€“ Solutions

Professor: Alistair Savage

Date: March 28, 2025

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 โˆˆ 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 โˆˆ R> 0 , we have

x > 0 =โ‡’ x + 1 > 1 =โ‡’ f (x) = x + 1 5

Hence, for example, there is no x โˆˆ 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 โˆ’ 1 |.

2

Then, for all x โˆˆ R> 0 , we have

(g โ—ฆ f )(x) = g(f (x)) = g

x + 1 5

x + 1 5

โˆ’ 1 = |x| = x,

since x > 0. Thus, g โ—ฆ f = idR> 0 , and so g is a left inverse of f. Important! We cannot define g by g(y) = 5y โˆ’ 1, since this does not always take values in the codomain R> 0. For example 5(0) โˆ’ 1 โˆˆ/ 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.