mat102 intro to math proofs · mat102 intro to math proofs xinli wang university of toronto...

Course InformationAre they mathematical proofs?

Chapter 1: Numbers, quadratics and inequalitiesPuzzle of the week

MAT102 Intro to Math Proofs

Xinli Wang

University of Toronto Mississauga

[email protected]

January 9, 2019

Xinli Wang MAT102/Week 1



Overview

1 Course InformationAssessment ComponentsHow to reach me?Course DescriptionImportant dates

2 Are they mathematical proofs?

3 Chapter 1: Numbers, quadratics and inequalitiesThe Quadratic FormulaInequalities, and arithmetic/geometric meansThe Triangle InequalityType of Numbers

4 Puzzle of the week




Assessment ComponentsHow to reach me?Course DescriptionImportant dates





How can you find me?

In class.https://wangxinli.youcanbook.me/.Office Hours: Wednesday, 10AM-12PM, DH3060.Email me: [email protected] lunch with me!https://goo.gl/forms/FlMxI3xmdLAjrB2b2


https://wangxinli.youcanbook.me/

https://goo.gl/forms/FlMxI3xmdLAjrB2b2




The goal of MAT102:

Understanding, using and developing precise expressions ofmathematical ideas, including definitions and theorems. Set theory,logical statements and proofs, induction, topics chosen fromcombinatorics, elementary number theory, Euclidean geometry.





Quiz and Term Test Dates

All quizzes and term test will happen on Thursdays, 6-7PM. Markthe following dates on your calendar:Jan 24th, Quiz 1Feb 7th, Quiz 2Feb 28th, Term TestMar 14th, Quiz 3Mar 28th, Quiz 4




https://www.maa.org/press/periodicals/convergence/

proofs-without-words-and-beyond-proofs-without-words-20


https://www.maa.org/press/periodicals/convergence/proofs-without-words-and-beyond-proofs-without-words-20

https://www.maa.org/press/periodicals/convergence/proofs-without-words-and-beyond-proofs-without-words-20



Prove: For any integer a, we have 2a2 > a.

Proof.

This proof is by case analysis. There are two cases:

Case 1: a is positive. Since a is an integer, we must have thata ≥ 1. Hence 2a2 = 2a · a ≥ 2a · 1 > a. This implies the claimholds in Case 1.

Case 2: a is negative. Since a is an integer, we must havethat a ≤ −1. Hence 2a2 ≥ 2 · (−1) · (−1) = 2 > −1 ≥ a.This implies the claim holds in Case 2.

The claim therefore holds in both cases.




The Quadratic FormulaInequalities, and arithmetic/geometric meansThe Triangle InequalityType of Numbers

Our very first theorem!

Thm 1.1.1 The Quadratic Formula

Let a, b, c be three real numbers, with a 6= 0. Then the equationax2 + bx + c = 0 has:

No real solutions if b2 − 4ac < 0.

A unique solution if b2 − 4ac = 0, given by x = − b2a .

Two distinct solutions if b2 − 4ac > 0, given by

x = −b+√b2−4ac2a and x = −b−

√b2−4ac2a .





Basic Facts

Facts

1 if a < b or a ≤ b and c > 0, then ca < cb or ca ≤ cb.

2 a2 ≥ 0.

3 If a ≥ 0, then there is a unique nonnegative number√a

whose square is a.

4 If a < b and b < c, then a < c .





Proposition 1.2.1

Let a and b be two real numbers.

1 If 0 < a < b, then a2 < b2 and√a <√b.

2 Similarly, if 0 < a ≤ b, then a2 ≤ b2 and√a ≤√b.





What’s wrong?

Consider the following incorrect theorem:

Incorrect Theorem

Suppose that x and y are real numbers and x 6= 3. If x2y = 9ythen y = 0.

Proof.

Suppose that x2y = 9y . Then (x2 − 9)y = 0. Since x 6= 3, x2 6= 9,so x2 − 9 6= 0. Therefore we can divide both sides of the equation

(x2 − 9)y = 0

by x2 − 9, which leads to the conclusion that y = 0. Thus ifx2y = 9y , then y = 0.

What’s wrong with this proof? Can you find a counter example?Xinli Wang MAT102/Week 1




Definition

The arithmetic mean of two real numbers, x and y , is x+y2 . If

x , y ≥ 0, then their geometric mean is√xy .

Check the following applet and make a conjecture about how thesetwo quantities are related:https://www.geogebra.org/m/rNv4xR5H


https://www.geogebra.org/m/rNv4xR5H




Proposition 1.2.3(The Arithmetic-Geometric Mean Inequality).

For any two real numbers x and y ,

x · y ≤(x + y

2

)2

, (1)

and equality holds iff x = y . If, in addition, x , y ≥ 0, then

√xy ≤ x + y

2. (2)





Application of Proposition 1.2.3 AGM

Use the AGM inequality to find the maximum of

(5 +√x4 + 1)(9−

√x4 + 1).

Prove that for any two real numbers x , y , with x 6= 0, we have

2y ≤ y2

x2+ x2.





Definition 1.3.1

The absolute value of a real number x , denoted as |x |, is defined as

|x | =

{x if x ≥ 0

−x if x < 0





Proposition 1.3.2

For any two real numbers x , y , we have

√x2 = |x |, |x |2 = x2, x ≤ |x |, and |x · y | = |x | · |y |.





Proposition 1.3.3(The Triangle Inequality).

For any two real numbers x and y , we have

|x + y | ≤ |x |+ |y |.





Applications of the Triangle Inequality

Prove for any real number x and y ,

|x − y | ≥ |x | − |y |





Applications of the Triangle Inequality

If |a| ≥ 2, |b| ≤ 12 . Find an M such that

|a + b| ≥ M.

What’s the maximum value of M?





Example of triangle inequality

Let a, b, c be three real numbers. Prove that

|a− c | ≤ |a− b|+ |b − c |.





Definition 1.4.1

Let a be an integer, and b a nonzero integer. We say that a isdivisible by b (or that b divides a), if there exists an integer m, forwhich a = m · b.





Definition 1.4.2.

1 An integer is even if it is divisible by 2. Otherwise it is odd.

2 A natural number p > 1 is called a prime number,if the onlynatural numbers that divide p are 1 and p.





Exercise

Prove the following statement:There are infinitely many prime numbers.




An elderly shepherd died and left his entire estate to his three sons.To his first son, whom he favored the most, he bequeathed 1

2 hisflock of sheep, to the second son 1

3 , and to the third son, whom heliked the least, 1

9 of his flock.Not wishing to contest their father’s will, the three sons went tothe pasture to begin divvying up the flock. They were alarmed tocount a total of 17 sheep! Is there a means for the three sons tosuccessfully carry out their father’s wishes?




The End


mat102 intro to math proofs · mat102 intro to math proofs xinli wang university of toronto...

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