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Introduction to Proofs - Triangle Inequality Prof Mike Pawliuk UTM June 11, 2020 Slides available at: mikepawliuk.ca This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.5 Canada License. Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 1 / 11
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Page 1: Introduction to Proofs - Triangle Inequality

Introduction to Proofs - Triangle Inequality

Prof Mike Pawliuk

UTM

June 11, 2020

Slides available at: mikepawliuk.ca

This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.5 Canada License.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 1 / 11

Page 2: Introduction to Proofs - Triangle Inequality

Learning Objectives (for this video)

By the end of this video, participants should be able to:

1 Apply the absolute value function in various contexts (algebraicallyand geometrically).

2 Relate/distinguish problem solving and rough work to a proof.

3 Break up a sum using the triangle inequality.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 2 / 11

Page 3: Introduction to Proofs - Triangle Inequality

Motivation

“The sum of any two sides of a triangle is always greater or equal to thethird side.”

Motivation

The triangle inequality is one of the two important inequalities we see inthis course.We will understand it geometrically and algebraically.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 3 / 11

Page 4: Introduction to Proofs - Triangle Inequality

Absolute value

Definition (Absolute value)

If x is a rel number, then

|x | ={

x x ≥ 0−x x < 0

Special case. |a− b| is the distance between a and b.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 4 / 11

Page 5: Introduction to Proofs - Triangle Inequality

Absolute value

Definition (Absolute value)

If x is a rel number, then

|x | ={

x x ≥ 0−x x < 0

Special case. |a− b| is the distance between a and b.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 4 / 11

Page 6: Introduction to Proofs - Triangle Inequality

More visualizations of absolute value

This picture shows us the inequality: for all real x , x ≤ |x |.

What is the max value of |x2 − 1| when |x | ≤ 1? At x = 0, |x2 − 1| = 1.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 5 / 11

Page 7: Introduction to Proofs - Triangle Inequality

More visualizations of absolute value

This picture shows us the inequality: for all real x , x ≤ |x |.

What is the max value of |x2 − 1| when |x | ≤ 1?

At x = 0, |x2 − 1| = 1.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 5 / 11

Page 8: Introduction to Proofs - Triangle Inequality

More visualizations of absolute value

This picture shows us the inequality: for all real x , x ≤ |x |.

What is the max value of |x2 − 1| when |x | ≤ 1? At x = 0, |x2 − 1| = 1.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 5 / 11

Page 9: Introduction to Proofs - Triangle Inequality

Basic Facts about absolute value

Facts

Let x , y ∈ R.1 x2 = |x |2

2

√x2 = |x |

3 x ≤ |x |4 |xy | = |x ||y |

Special case of (4). If y = −1, get | − x | = |x || − 1| = |x |.Special case of this. |b − a| = | − (a− b)| = |a− b|.The distance from a to b is the same as the distance from b to a.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 6 / 11

Page 10: Introduction to Proofs - Triangle Inequality

Basic Facts about absolute value

Facts

Let x , y ∈ R.1 x2 = |x |2

2

√x2 = |x |

3 x ≤ |x |4 |xy | = |x ||y |

Special case of (4). If y = −1, get | − x | = |x || − 1| = |x |.

Special case of this. |b − a| = | − (a− b)| = |a− b|.The distance from a to b is the same as the distance from b to a.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 6 / 11

Page 11: Introduction to Proofs - Triangle Inequality

Basic Facts about absolute value

Facts

Let x , y ∈ R.1 x2 = |x |2

2

√x2 = |x |

3 x ≤ |x |4 |xy | = |x ||y |

Special case of (4). If y = −1, get | − x | = |x || − 1| = |x |.Special case of this. |b − a| = | − (a− b)| = |a− b|.

The distance from a to b is the same as the distance from b to a.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 6 / 11

Page 12: Introduction to Proofs - Triangle Inequality

Basic Facts about absolute value

Facts

Let x , y ∈ R.1 x2 = |x |2

2

√x2 = |x |

3 x ≤ |x |4 |xy | = |x ||y |

Special case of (4). If y = −1, get | − x | = |x || − 1| = |x |.Special case of this. |b − a| = | − (a− b)| = |a− b|.The distance from a to b is the same as the distance from b to a.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 6 / 11

Page 13: Introduction to Proofs - Triangle Inequality

Motivation for Triangle Inequality

|x − y | = |(x − z) + (z − y)|≤ |x − z |+ |z − y |

Intuition. Direct routes are shorter than detours (to z).

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 7 / 11

Page 14: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |

=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 15: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |

=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 16: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 17: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 18: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 19: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |

=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 20: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy |

←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 21: Introduction to Proofs - Triangle Inequality

Triangle Inequality - Rough work

Theorem (Triangle Inequality)

For all a, b ∈ R, |a+ b| ≤ |a|+ |b|.

Rough work / Ugly garbage

|x + y | ≤ |x |+ |y |=⇒ |x + y |2 ≤ (|x |+ |y |)2

=⇒ (x + y)2 ≤ |x |2 + 2|x ||y |+ |y |2

=⇒ x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2

=⇒ 2xy ≤ 2|xy |=⇒ xy ≤ |xy | ←− This is the basic fact we will start from.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 8 / 11

Page 22: Introduction to Proofs - Triangle Inequality

Proof of Triangle inequality

Proof.

Let x , y ∈ R.Note that xy ≤ |xy | by Fact 3.So 2xy ≤ 2|xy | = 2|x ||y |.Since, by Fact 1, x2 = |x |2 and y2 = |y |2 we get

x2 + 2xy + y2 ≤ |x |2 + 2|xy |+ |y |2.

Factoring gives |x + y |2 ≤ (|x |+ |y |)2. By Fact 2 of inequalities,|x + y | ≤ ||x |+ |y ||.Note that ||x |+ |y || = |x |+ |y |, since |x | ≥ 0, |y | ≥ 0 and |x |+ |y | ≥ 0.So |x + y | ≤ |x |+ |y |.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 9 / 11

Page 23: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|

≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 24: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|

≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 25: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 26: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |

≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 27: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 28: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 29: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 30: Introduction to Proofs - Triangle Inequality

Bounding argument

Example

Find an M so that |x5 − 2x − 5| ≤ M for all |x | ≤ 2.

Proof.

Solution Use the Triangle inequality.

|x5 − 2x − 5| = |x5 + (−2x − 5)|≤ |x5|+ | − 2x − 5| By Triangle ineq

= |x5|+ |2x + 5| |y | = | − y |≤ |x5|+ |2x |+ |5| By Triangle ineq

= |x |2 + 2|x |+ 5

≤ 25 + 2 · 2 + 5 = 41 since |x | ≤ 2

So take M = 41.

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 10 / 11

Page 31: Introduction to Proofs - Triangle Inequality

Reflection

What’s the value in understanding the absolute value functiongeometrically as well as algebraically?

How is the triangle inequality related to triangles?

How come Desmos tells us that |x5 − 2x − 5| ≤ 33 on the interval−2 ≤ x ≤ 2, but we got an answer of 41?

Prof Mike Pawliuk (UTM) Intro to Proofs June 11, 2020 11 / 11


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