Real Numbers · Lesson 4 of 5
Revisiting Irrational Numbers
“An introductory look into the mathematically unhinged numbers that talk forever without ever repeating themselves.”
• Recall what irrational numbers are. • Understand why proving irrationality is important. • Learn and prove the theorem: If p is a prime number and p divides a², then p divides a. • Use the theorem to prove √2 is irrational. • Extend the same idea to prove √3 is irrational. • Identify other irrational numbers.
Introduction
We already know that numbers like √2 and √3 are irrational.
But how do mathematicians know this?
Can we actually prove that these numbers can never be written as fractions?
Mathematics accepts statements only after they are logically proved. In this lesson, we will learn an important theorem and then use it to prove that √2 and √3 are irrational.
Revisiting Irrational Numbers
A number that cannot be expressed in the form p/q, where p and q are integers and q ≠ 0, is called an irrational number.
| Rational Numbers | Irrational Numbers |
|---|---|
| 1/2 | √2 |
| 5 | √3 |
| −7 | π |
| 0.75 | √5 |
Every irrational number has a non-terminating and non-repeating decimal expansion.
Why Do We Need Another Theorem?
If √2 is irrational, why can't we simply say so?
Because we need a mathematical proof. To construct that proof, we first need an important theorem about prime numbers and divisibility.
Important Theorem
Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
Understanding the Theorem
Before proving the theorem, let us observe the pattern through a few simple examples.
Problem
Suppose 3 divides 18². Does 3 divide 18?
- 1.18² = 324.
- 2.3 divides 324.
- 3.Also, 3 divides 18.
- 4.Therefore, the statement is true in this case.
Problem
Suppose 5 divides 20². Does 5 divide 20?
- 1.20² = 400.
- 2.5 divides 400.
- 3.Also, 5 divides 20.
- 4.Therefore, the statement is true in this case.
Problem
Suppose 7 divides 35². Does 7 divide 35?
- 1.35² = 1225.
- 2.7 divides 1225.
- 3.Also, 7 divides 35.
- 4.Therefore, the statement is true in this case.
Proof of the Theorem
Let p be a prime number and let a be a positive integer. Suppose p divides a².
By the Fundamental Theorem of Arithmetic, a can be written as a product of prime factors. When a is squared, every prime factor occurring in a appears twice in a².
Since p divides a², p must appear in the prime factorisation of a². But the prime factors of a² are exactly the same prime factors that occur in a. Therefore, p must also divide a.
Hence, if p divides a², then p divides a.
This completes the proof.
Why Is This Theorem Useful?
- It helps us prove that √2 is irrational.
- It helps us prove that √3 is irrational.
- The same reasoning can be used for √5, √7 and many other square roots.
- It connects prime factorisation with proofs about irrational numbers.
Two positive integers are called coprime (or relatively prime) if their only common factor is 1. In other words, their Highest Common Factor (HCF) is 1. Example: Consider the numbers 8 and 15. Factors of 8 are 1, 2, 4 and 8. Factors of 15 are 1, 3, 5 and 15. The only common factor is 1. Therefore, 8 and 15 are coprime numbers.
Proof That √2 Is Irrational
The proof that √2 is irrational is one of the most important proofs in this lesson. Read every step carefully and notice where the contradiction occurs.
We will prove this result by contradiction. This means we will first assume that √2 is rational and then show that this assumption leads to an impossible conclusion.
Assume that √2 is rational.
We choose p/q in its lowest form. Therefore, p and q are coprime.
Since 2 divides p² and 2 is a prime number, the theorem tells us that 2 divides p. Therefore, p is even.
Since 2 divides q² and 2 is prime, the theorem tells us that 2 divides q. Therefore, q is also even.
We have shown that both p and q are divisible by 2. Therefore, p and q have a common factor 2. But we began by assuming that p and q have no common factor other than 1. These two statements cannot both be true.
Our assumption that √2 is rational is false. Therefore, √2 is irrational.
Why Did We Reach a Contradiction?
At the beginning, we wrote √2 = p/q in lowest terms. This means p and q had no common factor other than 1. However, the proof showed that both p and q are divisible by 2. Therefore, they share the common factor 2. This is impossible because the fraction was already assumed to be in lowest terms.
- Assume √2 is rational.
- Write √2 = p/q in lowest terms.
- Show that p² is divisible by 2.
- Use the theorem to show that p is divisible by 2.
- Write p = 2k and substitute it.
- Show that q² is divisible by 2.
- Use the theorem to show that q is divisible by 2.
- Both p and q are even, which contradicts the lowest-terms assumption.
- Therefore, √2 is irrational.
Proof That √3 Is Irrational
The proof for √3 follows the same method used for √2.
Assume that √3 is rational. Then it can be written in the form p/q, where p and q are coprime integers and q ≠ 0.
Since 3 divides p² and 3 is prime, the theorem tells us that 3 divides p. Therefore, p = 3k for some integer k.
Since 3 divides q² and 3 is prime, the theorem tells us that 3 divides q.
Both p and q are divisible by 3. Therefore, they have a common factor 3. This contradicts the assumption that p and q are coprime.
Our assumption that √3 is rational is false. Therefore, √3 is irrational.
Other Irrational Numbers
The same reasoning can be used to prove that √5, √7, √11, √13 and many other square roots of prime numbers are irrational.
| Number | Type | Reason |
|---|---|---|
| √4 = 2 | Rational | 4 is a perfect square. |
| √9 = 3 | Rational | 9 is a perfect square. |
| √16 = 4 | Rational | 16 is a perfect square. |
| √25 = 5 | Rational | 25 is a perfect square. |
| √2 | Irrational | 2 is not a perfect square. |
| √3 | Irrational | 3 is not a perfect square. |
| √5 | Irrational | 5 is not a perfect square. |
| √7 | Irrational | 7 is not a perfect square. |
The square root of a perfect square is rational. The square root of a positive integer that is not a perfect square is irrational.
• Forgetting to assume that p/q is in lowest terms. • Failing to state that p and q are coprime. • Saying that p is even without using the theorem. • Skipping the substitution p = 2k or p = 3k. • Ending the proof without clearly identifying the contradiction. • Assuming that every square root is irrational. • Forgetting that square roots of perfect squares are rational.
Quiz
Which statement correctly defines an irrational number?
If a prime number p divides a², what does the theorem imply?
In the proof that √2 is irrational, why is p even?
What contradiction is reached in the proof that √2 is irrational?
Which of the following is rational?
Practice Problems
- State and prove the theorem: If p is a prime number and p divides a², then p divides a.
- Prove that √5 is irrational.
- Prove that √7 is irrational.
- Explain why the assumption that p and q are coprime is necessary in the proof of the irrationality of √2.
- Is √49 irrational? Give a reason.
- Is √64 irrational? Give a reason.
- Explain why √25 is rational.
- Classify √11, √36, √13 and √81 as rational or irrational.
Key Takeaways
• An irrational number cannot be expressed in the form p/q, where p and q are integers and q ≠ 0. • Irrational numbers have non-terminating and non-repeating decimal expansions. • If a prime number p divides a², then p divides a. • The proof that √2 is irrational uses the method of contradiction. • In the proof, assuming √2 = p/q leads to both p and q being even. • This contradicts the assumption that p/q is in lowest terms. • The same reasoning proves that √3 and the square roots of other prime numbers are irrational. • Square roots of perfect squares are rational.