How To Prove That A Number Is Irrational

The realm of numbers holds captivating secrets, and among them lies the distinction between rational and irrational numbers. While rational numbers can be expressed as a simple fraction p/q, where p and q are integers and q is not zero, irrational numbers defy this representation. Proving that a number is irrational often requires a blend of mathematical rigor and clever techniques.

Understanding Rational and Irrational Numbers

Before diving into the methods of proving irrationality, it's crucial to solidify our understanding of these number categories.

• Rational Numbers: These are numbers that can be written in the form p/q, where p and q are integers (whole numbers) and q ≠ 0. Examples include 2/3, -5/7, 6 (since 6 = 6/1), and 0.75 (since 0.75 = 3/4). Decimals that terminate or repeat are also rational.
• Irrational Numbers: These are numbers that cannot be expressed in the form p/q. Their decimal representations are non-terminating and non-repeating. Famous examples include √2, π (pi), and e (Euler's number).

The challenge arises when we encounter a number and need to definitively classify it as irrational. This is where proof techniques come into play.

Common Techniques for Proving Irrationality

Several methods exist to demonstrate that a given number is irrational. Here, we will explore some of the most widely used and effective approaches:

1. Proof by Contradiction: This is a cornerstone technique in mathematics. We begin by assuming that the number in question is rational. If this assumption leads to a logical contradiction, then our initial assumption must be false, implying that the number is indeed irrational.

2. The Irrational Root Theorem: This theorem provides a direct way to test for the irrationality of roots of polynomials.

3. Using Properties of Transcendental Numbers: Some numbers are transcendental, meaning they are not roots of any non-zero polynomial equation with rational coefficients. Proving a number is transcendental directly implies its irrationality.

4. Infinite Descent: A more specialized technique, infinite descent is particularly useful for proving the irrationality of certain algebraic numbers.

Let's examine each of these techniques in detail.

1. Proof by Contradiction: The Case of √2

The irrationality of √2 is a classic example often used to illustrate proof by contradiction. The steps are as follows:

• Assume √2 is rational: Suppose that √2 can be written as a fraction a/b, where a and b are integers with no common factors (i.e., the fraction is in its simplest form). This is a crucial assumption, as it implies the existence of such a simplified fraction.
• Write the equation: √2 = a/b
• Square both sides: 2 = a² / b²
• Rearrange: 2b² = a²
• Deduce that a² is even: Since 2b² is even, it follows that a² must also be even.
• Deduce that a is even: If a² is even, then a itself must be even. This is because the square of an odd number is always odd. Therefore, we can write a as 2k, where k is an integer.
• Substitute a = 2k into the equation: 2b² = (2k)² = 4k²
• Simplify: b² = 2k²
• Deduce that b² is even: Since 2k² is even, b² must also be even.
• Deduce that b is even: If b² is even, then b itself must be even.
• Contradiction: We have now shown that both a and b are even. This means they have a common factor of 2. However, we initially assumed that a/b was in its simplest form, with no common factors. This is a contradiction!
• Conclusion: Since our initial assumption leads to a contradiction, it must be false. Therefore, √2 cannot be written as a fraction a/b, meaning it is irrational.

This proof beautifully demonstrates how a simple assumption, when rigorously followed, can unravel its own falsehood, revealing the true nature of √2.

2. The Irrational Root Theorem

The Irrational Root Theorem provides a more general way to determine if certain numbers are irrational. It states:

If a polynomial with integer coefficients has an integer root, that root must be a factor of the constant term.

Conversely, if we can show that a polynomial with integer coefficients has no integer roots that are factors of the constant term, and if the polynomial has a root at all, then that root must be irrational.

Example: Proving the Irrationality of √3

Consider the polynomial x² - 3 = 0. The roots of this polynomial are x = ±√3.

1. Identify the constant term: The constant term is -3.
2. Find the factors of the constant term: The factors of -3 are ±1 and ±3.
3. Test the factors:
   • If x = 1, then (1)² - 3 = -2 ≠ 0.
   • If x = -1, then (-1)² - 3 = -2 ≠ 0.
   • If x = 3, then (3)² - 3 = 6 ≠ 0.
   • If x = -3, then (-3)² - 3 = 6 ≠ 0.

Since none of the integer factors of -3 are roots of the polynomial x² - 3 = 0, and since we know the polynomial does have roots (√3 and -√3), then those roots must be irrational. Therefore, √3 is irrational.

Generalization

The Irrational Root Theorem is easily generalized to show that if n is a positive integer that is not a perfect square, then √n is irrational. Similarly, if n is not a perfect cube, then ∛n is irrational, and so on. This provides a quick and easy way to identify many irrational numbers.

3. Using Properties of Transcendental Numbers

Transcendental numbers are a special class of irrational numbers. A number is transcendental if it is not a root of any non-zero polynomial equation with rational coefficients. This is a stricter condition than simply being irrational.

Examples of Transcendental Numbers:

• π (pi): Pi is perhaps the most famous transcendental number. It represents the ratio of a circle's circumference to its diameter. Ferdinand von Lindemann proved its transcendence in 1882.
• e (Euler's number): e is the base of the natural logarithm, approximately equal to 2.71828. Charles Hermite proved its transcendence in 1873.

Proving Irrationality via Transcendence:

If you can prove that a number is transcendental, you automatically prove that it is irrational. This is because all rational numbers are algebraic (i.e., they are roots of polynomial equations with rational coefficients).

Challenges:

Proving transcendence is notoriously difficult. The proofs for the transcendence of π and e are highly complex and require advanced mathematical techniques. However, if transcendence has already been established for a particular number, it provides a powerful shortcut to establishing its irrationality.

Example: Why Transcendence Implies Irrationality

Suppose we have a number, let's call it t, and we have proven that t is transcendental. By definition, this means that t is not the root of any polynomial equation with rational coefficients.

Now, let's assume, for the sake of contradiction, that t is rational. This means we can write t as p/q, where p and q are integers and q ≠ 0. But if t = p/q, then t is a root of the polynomial equation:

qx - p = 0

This is a polynomial equation with integer (and therefore rational) coefficients. Therefore, t is a root of a polynomial equation with rational coefficients, which contradicts our initial assertion that t is transcendental! This contradiction proves that t cannot be rational, and therefore t must be irrational.

4. Infinite Descent

Infinite descent is a specialized proof technique that relies on showing that if a solution exists, then a "smaller" solution also exists, leading to an infinite sequence of decreasing solutions, which is impossible in the integers.

How it Works:

1. Assume a solution exists: Suppose there exists a solution to a certain problem involving integers.
2. Show a smaller solution exists: Demonstrate that if the initial solution exists, then you can always find another solution that is "smaller" in some sense (e.g., has smaller values of the integers involved).
3. Reach a contradiction: This process can be repeated indefinitely, leading to an infinite sequence of decreasing positive integers. However, this is impossible because the positive integers are bounded below by 1.
4. Conclude no solution exists: The contradiction implies that the initial assumption of a solution was false.

Example: Proving the Irrationality of √2 (Using Infinite Descent - A More Complex Approach)

This example is a more complex application of infinite descent compared to the standard proof by contradiction.

1. Assume √2 is rational: Suppose √2 = a/b, where a and b are positive integers. We do not assume that a/b is in simplest form.

2. Manipulate the equation: √2 = a/b => √2 * b = a

3. Consider the following: Since 1 < √2 < 2, we have b < a < 2b. This is important.

4. Construct a "smaller" solution: Let a₁ = a - b and b₁ = b - (a - b) = 2b - a. Since b < a < 2b, both a₁ and b₁ are positive integers.

5. Show that √2 = a₁/b₁:

   • a₁/b₁ = (a - b) / (2b - a)
   • Multiply numerator and denominator by (2b + a):
   • a₁/b₁ = ((a - b) (2b + a)) / ((2b - a) (2b + a))
   • a₁/b₁ = (a² - b²) / (4b² - a²)
   • Since a² = 2b², we have:
   • a₁/b₁ = (2b² - b²) / (4b² - 2b²) = b²/2b² = 1/2 is incorrect, need to redo the algebra.
   • Starting from a1/b1 = (a-b)/(2b-a), multiply top and bottom by √2 + 1
   • (a-b) / (2b-a) = (a-b)(√2 + 1) / (2b-a)(√2 + 1)
   • = (a√2 + a - b√2 - b) / (2b√2 + 2b - a√2 -a)
   • since a = b√2
   • = (b√2 √2 + b√2 - b√2 -b) / (2b√2 + 2b - b√2 √2 -b√2)
   • = (2b - b) / (2b√2 + 2b - 2b - b√2)
   • = b / b√2 = 1/√2 = √2 / 2 which is not √2, so something is off.

6. A different approach to constructing a "smaller solution": Let's try a different construction. Assume √2 = a/b. Then also √2 = (2b-a) / (a-b) . To see why: If a/b = √2, then a = √2 b. Now consider (2b -a ) / (a-b). Substituting a = √2 b, we get (2b - √2 b) / (√2 b -b) = (2 - √2) b / (√2 -1)b = (2-√2) / (√2 -1). Multiplying top and bottom by (√2 +1), we get ((2-√2)(√2 +1)) / ((√2 -1)(√2+1)) = (2√2 + 2 -2 -√2) / (2-1) = √2 / 1 = √2. Therefore (2b-a) / (a-b) = √2.

7. Define a1 = 2b - a and b1 = a-b. Now we need to show that a1 < a and b1 <b which means these are "smaller" solutions.

8. Showing a1 < a: Since a < 2b, we know that a1 = 2b - a > 0. We want to show a1 < a or 2b - a < a which implies 2b < 2a or b < a which is true since sqrt(2) = a/b > 1

9. Showing b1 < b: We want to show that b1 = a - b < b, which implies a < 2b, which we already know to be true.

10. The infinite descent: We have shown that if √2 = a/b, then we can find "smaller" integers a1 and b1 such that √2 = a1/b1. We can repeat this process infinitely, finding smaller and smaller positive integers a2, b2, a3, b3, and so on, such that √2 = a2/b2 = a3/b3 = ... . This leads to an infinite sequence of decreasing positive integers: a > a1 > a2 > a3 > ... and b > b1 > b2 > b3 > .... This is impossible because the positive integers are bounded below by 1.

11. Conclusion: Our initial assumption that √2 is rational must be false. Therefore, √2 is irrational.

This application of infinite descent is more involved, but it highlights the power of the technique in demonstrating the non-existence of solutions and, consequently, the irrationality of certain numbers.

Examples of Proving Irrationality of Specific Numbers

Now, let's look at how these techniques can be applied to prove the irrationality of other numbers.

1. Proving the Irrationality of √5

We can use proof by contradiction, similar to the √2 example.

• Assume √5 = a/b, where a and b are integers with no common factors.
• Square both sides: 5 = a²/b²
• Rearrange: 5b² = a²
• a² is divisible by 5, so a is divisible by 5. Let a = 5k.
• Substitute: 5b² = (5k)² = 25k²
• Simplify: b² = 5k²
• b² is divisible by 5, so b is divisible by 5.
• Contradiction: Both a and b are divisible by 5, contradicting the assumption that a/b is in simplest form.
• Therefore, √5 is irrational.

2. Proving the Irrationality of √[3]2

We can again use proof by contradiction.

• Assume ∛2 = a/b, where a and b are integers with no common factors.
• Cube both sides: 2 = a³/b³
• Rearrange: 2b³ = a³
• a³ is even, so a is even. Let a = 2k.
• Substitute: 2b³ = (2k)³ = 8k³
• Simplify: b³ = 4k³
• b³ is even, so b is even.
• Contradiction: Both a and b are even, contradicting the assumption that a/b is in simplest form.
• Therefore, ∛2 is irrational.

3. Proving the Irrationality of e

The proof of the irrationality of e is more complex, often employing the series representation of e:

e = 1 + 1/1! + 1/2! + 1/3! + ...

We can use a proof by contradiction, assuming e = a/b and then manipulating the series to arrive at a contradiction. The full proof is beyond the scope of this discussion, but it involves showing that if e were rational, a certain integer would have to be equal to a non-integer, which is impossible.

Advanced Considerations

• Algebraic vs. Transcendental Numbers: All rational numbers are algebraic. Irrational numbers can be either algebraic (like √2) or transcendental (like π and e).
• Liouville Numbers: These are transcendental numbers that can be very closely approximated by rational numbers. They are constructed specifically to be transcendental.
• The Gelfond-Schneider Theorem: This powerful theorem states that if a and b are algebraic numbers with a ≠ 0, a ≠ 1, and b irrational, then aᵇ is transcendental. This can be used to prove the transcendence (and therefore irrationality) of numbers like 2√².

Conclusion

Proving the irrationality of a number is a fundamental exercise in mathematical rigor. While some cases, like √2, can be tackled with relatively straightforward proof by contradiction, others require more sophisticated techniques like the Irrational Root Theorem, understanding of transcendental numbers, or the ingenious method of infinite descent. The journey to demonstrate irrationality is not merely about arriving at a conclusion; it's about honing our logical reasoning and deepening our appreciation for the intricate structure of the number system. The techniques discussed here provide a solid foundation for exploring the fascinating world of irrational numbers and their profound implications in mathematics.