- Author:
- Amanda Hager
- Subject:
- Mathematics
- Material Type:
- Lesson
- Level:
- Academic Lower Division
- Provider:
- University of Texas at Austin
- Tags:

- License:
- Creative Commons Attribution
- Language:
- English
- Media Formats:
- Text/HTML, Video

# Education Standards

# Adapting Proofs - PDF

# 2.2 Rational numbers

## Overview

TCCNS Course | MATH 1332: Contemporary Mathematics |

UT Austin Course | M 302: Introduction to Mathematics |

# Introduction

**Suggested Resources and Preparation**

**Materials and Technology**

For the instructor: Handouts that allow students to edit complete proofs may be useful.

For the student: a calculator that can take square roots and a reference sheet for algebraic properties may be useful.

**Prerequisite Assumptions**

- Students should be able to manipulate algebraic equations involving exponents, square roots, or variables using appropriate techniques and properties.
- Students should be able to use numeric or variable substitution while working with expressions.

**Overview and Student Objectives**

**Lesson Length**

50 minutes

**Lesson Objectives**

Students will understand that:

- Infinitely many real numbers are rational, and infinitely many real numbers are irrational.
- To prove that a number is rational, you express it in the form of an integer divided by a non-zero integer.
- To prove that a number is irraitonal, you must prove that there is no fraction of integers that could equal the number.

Students will be able to:

- Use proof by construction to prove that terminating decimals are rational numbers.
- Adapt a demonstrated indirect proof that \(\sqrt{2}\) is irrational to create other indirect proofs that square roots of prime numbers are irrational.
- Create simple direct proofs that sums, products, and most quotients of rational numbers are rational.

# Rational numbers

# Algebra review

- An
is a whole number that can be positive, negative, or zero. One way to write them: ..., -3, -2, -1, 0, 1, 2, 3, ...*integer* - A
is a number that can be written as a fraction of integers**rational number***a/b*, where*a*and*b*are integers and*b*is not 0. - An
is a number that**irrational number***cannot*be written as a fraction of integers*a/b*, where*a*and*b*are integers and*b*is not 0.

# Proving numbers are rational

If you want the prove that a number is rational, all you have to do is literally display the fraction that the number is equal to. Just remember that it has to be integer over integer, and the denominator can't be zero. *You do not have to reduce fractions!* Here are some short and easy proofs; notice how I make a claim "NUMBER is rational" and then I supply the reason "because it's equal to this legal fraction:"

- .5 is rational because .5 = 5/10.
- .123 is rational because .123 = 123/1000.
- -5.01 is rational because -5.01 = -501/100.

You might be tempted to say ".5 is rational because .5 = .5/1" but remember that the fraction has to be integer over integer, so this argument doesn't count!

# Is the square root of 2 a rational number?

So are all the numbers rational? There was this cult, the Pythagoreans, that thought that all numbers could be written as fractions of integers, and they really loved fractions. But this belief didn't play well with their more famous Pythagorean Theorem. If you have a square of side length 1, the diagonal has to be length \(\sqrt{2}\), right? So what fraction is that?

The next four videos teach you the proof, and the text version is below:

Video: can sqrt(2) be written as integer over integer?

## Algebra review

- If you have a fraction of integers, then that fraction is
if the numerator and denominator have no factors in common. For example, 32/6 is not fully reduced because the top and bottom have a common factor of 2.**fully reduced** - To reduce a fraction fully, factor out the numerator and denominator all the way down to primes and then cross out all the factors that they have in common. For example:

\(\frac{60}{42}=\frac{2\cdot2\cdot3\cdot5}{2\cdot3\cdot7}=\frac{2\cdot5}{7}=\frac{10}{7}\)

Video: Algebra skills for rational numbers

Video: Turns out that a and b are both even

Video: a/b is fulled reduced and not fully reduced! This is bad!

# Summary/text version of proof

Claim: The square root of 2 is irrational.

Proof: Pretend that \(\sqrt{2}\) is rational. Then \(\sqrt{2}=\frac{a}{b}\), where *a* and *b* are some mysterious integers and *b* is not zero. Let's make sure before we go any further that *a/b* is fully reduced, so that means that *a* and *b* have no factors in common. Then we do algebra, squaring both sides and multiplying both sides by *\(b^2\)*:

\(2=\frac{a^2}{b^2}\)

\(2b^2=a^2\)

The left side is even because it's multiplied by 2. That means the right side has to be even because we have an equation. That means that* \(a\) *has to be even because if *\(a\)* were odd, then *\(a^2\)* would be odd too.

Let's write \(a=2n\), where *n* is exactly half of *a*. This is okay to do because *\(a\) *is even. Then:

\(2b^2=(2n)^2\)

\(2b^2=4n^2 \)

\(b^2=2n^2\)

The right side is even because it's multiplied by 2. That means the left side has to be even because we have an equation. That means that * b* has to be even because if *\(b\) *were odd, then *\(b^2\) *would be odd too.

WAIT! If *\(a\) *and *\(b\) *are both even, then they share a common factor. This contradicts what we said above, that they don't share a common factor. Both of these sentences can't be true at the same time. This means that *\(a\)* and *\(b\) *don't actually exist at all.

Conclusion: the square root of 2 is irrational.

# Extending theorems

Whenever math people do this much work to prove something, we look around for other things we can use the proof for, or other "bonus" facts we can prove. Like we try to get the most of our hard work, right?

For starters, we can copy the proof line for line, changing the 2's for 3's, and prove that the square root of 3 is irrational:

Video: What else can we use this proof for?

## Terms

- The word
is synonymous with*theorem*in this class. It is a statement that we prove step-by-step using logic.**claim** - A
**corollary**

So it's a fact now that the square root of 2 is irrational. Now we can use that fact without reproving it every time. Watch how I use it to prove a corollary that the square root of is 18 irrational. It's a much shorter proof!

Video: Corollaries or using this proof to get more results

# Quiz Questions

Question | Answer |

1 | 2 |

2 | 1 |

3 | 2 |

4 | 2 |

5 | 2 |

6 | 3 |

# Question 1

When we proved that the square root of 2 was irrational, we pretended that the square root of 2 was rational, so there was some fraction \(\sqrt{2}=\frac{a}{b}\). What did we have to make sure was true about \(a\) and \(b\)?

- We needed both a and b to be positive.
- We needed a/b to be a fully reduced fraction.
- We needed both a and b to be prime.
- We needed a and b to be even.

# Question 2

True/false: The square root of 5 is irrational.

- True
- False

# Question 3

True/false: The square root of every number is irrational.

- True
- False

# Question 4

True/false: The square root of 121 is irrational.

- True
- False

# Question 5

What is the definition of a rational number?

- Any fraction.
- Any number that can be written as a fraction of integer divided by integer.
- Any positive number that could be written as a fraction of integers.
- Any number that can be written as a fraction.
- Any number that can be written as a fraction of natural number divided by natural number.

# Question 6

Is the number \(\frac{\frac{5}{2}\times\frac{6}{5}}{\frac{2}{3}}\) rational?

- No because we can't tell what that number is.
- Yes because there are fractions in it.
- Yes because the number could be simplified to 92, which is a fraction of integers.
- No because it isn't in the right form of a single fraction of integers.

# Homework Questions

# Question 2.2.1

A *rational number* is any number that can be expressed as a fraction of two integers, with the denominator being non-zero.

# Question 2.2.2

\(1.25=\frac{5}{4}, 3\frac{2}{3}=\frac{11}{3}, -4.2=\frac{-42}{10}, \frac{\sqrt{25}}{3}=\frac{5}{3}, 3.1415926=\frac{31415926}{10000000}\)

# Question 2.2.3

Claim: The square root of 7 is irrational.

Proof: Pretend that \(\sqrt{7}\) is rational. Then \(\sqrt{7}=\frac{a}{b}\), where *a* and *b* are some mysterious integers and *b* is not zero. Let's make sure before we go any further that *a/b* is fully reduced, so that means that *a* and *b* have no factors in common. Then we do algebra, squaring both sides and multiplying both sides by *\(b^2\)*:

\(7=\frac{a^2}{b^2}\)

\(7b^2=a^2\)

The left side is **divisible by 7** because it's multiplied by 7. That means the right side has to be divisible by 7 because we have an equation. That means that* \(a\) *has to be divisible by 7 because if *\(a\)* were not divisible by 7, then *\(a^2\)* would not be divisible by 7 either.

Let's write \(a=7n\), where *n* is exactly one seventh of *a*. This is okay to do because *\(a\) *is divisible by 7. Then:

\(7b^2=(7n)^2\)

\(7b^2=49n^2 \)

\(b^2=7n^2\)

The right side is divisible by 7 because it's multiplied by 2. That means the left side has to be divisible by 7 because we have an equation. That means that * b* has to be divisible by 7 because if *\(b\) *were not divisible by 7, then *\(b^2\) *would be not divisible by 7 either.

WAIT! If *\(a\) *and *\(b\) *are both divisible by 7, then they share a common factor. This contradicts what we said above, that they don't share a common factor. Both of these sentences can't be true at the same time. This means that *\(a\)* and *\(b\) *don't actually exist at all.

Conclusion: the square root of 7 is irrational.

# Question 2.2.4

Proof: Pretend that \(5\sqrt{2}\) is rational. Then \(5\sqrt{2}=\frac{a}{b}\), where a and b are some integers and b is nonzero. Then \(\sqrt{2}=\frac{a}{5b}\), which implies that \(\sqrt{2}\) is rational. This is impossible, since we already know that the square root of 2 is irrational. Therefore, \(5\sqrt{2}\) must be irrational also.

# Question 2.2.5

Let a/b and c/d represent two rational numbers. That means a, b, c, and d are integers, and b and d are nonzero. Then \(\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\). Both numerator and denominator are integers, and we can tell that bd is nonzero. This is a rational number.

# Question 2.2.6

Let a/b and c/d represent two rational numbers. That means a, b, c, and d are integers, and b and d are nonzero. Then \(\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\). Both numerator and denominator are integers, and we can tell that bd is nonzero. This is a rational number.

# Question 2.2.7

Sometimes. Let a/b and c/d represent two rational numbers. That means a, b, c, and d are integers, and b and d are nonzero. Then \(\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}\). Both numerator and denominator are integers, but we aren't sure if *bc *is zero or not. This is a rational number if \(c\neq0\), and it's not a number at all if \(c=0\).

# Question 2.2.8

Sometimes. For example, \(\sqrt{2}\cdot\sqrt{2}=2\). But \(\sqrt{2}\cdot\sqrt{5}\) is irrational.

# Question 2.2.9

Mindset question. Responses vary.

# Question 2.2.1

Write the definition of a rational number.

# Question 2.2.2

Prove that these numbers are rational: \(1.25, 3 \frac{2}{3}, -4.2, \frac{\sqrt{25}}{3}, 3.1415926\).

# Question 2.2.3

By copying and adapting the proof that \(\sqrt{2}\) is irrational, prove that \(\sqrt{7}\) is irrational.

# Question 2.2.4

Prove that \(5\sqrt{2}\) is irrational. You have to use a proof by contradiction. You get to use that \(\sqrt{2}\) is irrational without reproving it.

# Question 2.2.5

Prove that the sum of two rational numbers has to be a rational number.

# Question 2.2.6

Prove that the product of two rational numbers has to be a rational number.

# Question 2.2.7

Is the quotient of two rational numbers always, sometimes, or never rational?

# Question 2.2.8

Is the product of two irrational numbers always, sometimes, or never irrational?

# Question 2.2.9

Reflection time! What are some things you did really well on this assignment, and why?