Tenth Grade Mathematics: Revisiting Irrational Numbers and Formal Proofs

In earlier grades, irrational numbers were introduced as numbers whose decimal expansions are non-terminating and non-recurring. While this descriptive property helps identify numbers like sqrt(2), sqrt(3), or pi, it does not provide an algebraic proof of why these numbers can never be expressed as fractions. In tenth-grade mathematics, we establish the existence of irrational numbers through rigorous deductive logic, employing one of the most powerful tools in pure mathematics: proof by contradiction (reductio ad absurdum).

The Logic of Proof by Contradiction

Proof by contradiction operates by assuming that the exact negation of the proposition you wish to prove is true.

Logical Flow of Reductio Ad Absurdum:
1. Target Statement: Proposition P is TRUE.
2. Contradiction Assumption: Suppose Proposition P is FALSE (not-P is TRUE).
3. Deductive Chain: Follow valid mathematical deductions from this assumption.
4. Absurdity: Arrive at a result that contradicts known axioms or the assumption itself.
5. Conclusion: Because valid deduction cannot produce a false result from a true premise,
               the assumption must be FALSE, proving Proposition P is TRUE.

To apply this logic to square roots, we rely on a key theorem derived directly from Euclid's Lemma and the Fundamental Theorem of Arithmetic.

Fundamental Lemma: Let p be a prime number. If p divides a^2 (where a is a positive
integer), then p must also divide a.

To see why this lemma holds, consider the unique prime factorization of a: a = p1 * p2 * ... * pk. Squaring both sides yields a^2 = p1^2 * p2^2 * ... * pk^2. If p divides a^2, then p must be one of the prime factors of a^2. But the only prime factors of a^2 are p1, p2, ..., pk, each repeated twice. Therefore, p must be one of the prime factors of a, which means that p divides a.

Formal Proof: The Irrationality of sqrt(2)

We now execute the classic Euclidean proof that sqrt(2) is an irrational number.

Theorem: The square root of 2 is irrational.

Assume, to the contrary, that sqrt(2) is a rational number.

By definition of a rational number, there exist two integers a and b with b != 0 such that:

sqrt(2) = a / b

Assume that a and b are co-prime, meaning they share no common factors other than 1. (If they have common factors, we divide both by their highest common factor to reduce the fraction to lowest terms).

Squaring both sides of the equation:

2 = a^2 / b^2
2 * b^2 = a^2      --- (Equation 1)

This equation states that 2 divides a^2. By our Fundamental Lemma, because 2 is a prime number and 2 divides a^2, it follows that 2 must also divide a.

Since 2 divides a, we can express a as:

a = 2 * c,   for some integer c

Substitute a = 2 * c back into Equation 1:

2 * b^2 = (2 * c)^2
2 * b^2 = 4 * c^2
Divide both sides by 2:
b^2 = 2 * c^2      --- (Equation 2)

Equation 2 shows that 2 divides b^2. Applying the Fundamental Lemma once more, because 2 is a prime number and 2 divides b^2, it follows that 2 must also divide b.

We have now deduced that 2 divides a and 2 divides b. This means that a and b have at least 2 as a common factor.

However, this directly contradicts our initial established premise that a and b are co-prime (having no common factors other than 1).

This contradiction arose solely because of our incorrect assumption that sqrt(2) is rational. Therefore, the assumption is false, and we conclude that sqrt(2) is irrational.

Generalizing to Any Non-Square Integer

The identical logical argument proves that sqrt(p) is irrational for any prime number p.

Proof Framework for sqrt(p) where p is prime:
1. Let sqrt(p) = a/b where HCF(a, b) = 1.
2. p * b^2 = a^2  ==> p divides a^2 ==> p divides a ==> a = p*c.
3. p * b^2 = p^2 * c^2 ==> b^2 = p * c^2 ==> p divides b^2 ==> p divides b.
4. p divides both a and b, contradicting HCF(a, b) = 1.
5. Therefore, sqrt(p) is irrational.

More generally, if n is any positive integer that is not a perfect square, sqrt(n) is irrational.

Arithmetic Combinations of Rational and Irrational Numbers

When rational and irrational numbers are combined through standard arithmetic operations, their resulting nature follows precise rules:

1. Rational + Irrational = Irrational
2. Rational - Irrational = Irrational
3. Non-zero Rational * Irrational = Irrational
4. Non-zero Rational / Irrational = Irrational

Proving the Irrationality of Compound Expressions

To prove that an expression such as 5 - sqrt(3) or 3 * sqrt(2) + 7 is irrational, we do not need to re-prove the irrationality of the square root from first principles. Instead, we isolate the radical term algebraically.

Example: Prove that 3 + 2*sqrt(5) is irrational, given that sqrt(5) is irrational.
1. Assume 3 + 2*sqrt(5) is rational, equal to a/b where a, b are integers and b != 0.
2. 3 + 2*sqrt(5) = a/b
3. 2*sqrt(5) = (a/b) - 3 = (a - 3*b) / b
4. sqrt(5) = (a - 3*b) / (2*b)

Since a and b are integers, the expression (a - 3b) is an integer, and (2b) is a non-zero integer. Therefore, the right-hand side is a rational number. This implies that sqrt(5) is rational, which directly contradicts the known fact that sqrt(5) is irrational. Hence, 3 + 2*sqrt(5) must be irrational.

Chapter Exercises

Exercise 1: Prove rigorously by contradiction that the square root of 3 is an irrational number.

Exercise 2: Prove that 7 * sqrt(5) is an irrational number, given that sqrt(5) is irrational.

Exercise 3: Prove that the expression sqrt(2) + sqrt(3) is irrational using algebraic manipulation and proof by contradiction.

Solutions and Explanations

Solution 1: Assume to the contrary that sqrt(3) is rational. Then there exist co-prime integers a and b (b != 0, HCF(a, b) = 1) such that sqrt(3) = a / b. Squaring both sides yields 3 = a^2 / b^2, which gives 3 * b^2 = a^2. This means that 3 divides a^2. Since 3 is prime, 3 must also divide a. Therefore, we can write a = 3 * c for some integer c. Substituting this into our equation gives 3 * b^2 = (3 * c)^2 = 9 * c^2. Dividing by 3 yields b^2 = 3 * c^2. This implies that 3 divides b^2, and consequently 3 divides b. Thus, 3 is a common factor of both a and b, which contradicts the fact that a and b are co-prime. Therefore, sqrt(3) must be irrational.

Solution 2: Assume that 7 * sqrt(5) is rational. Then there exist integers a and b with b != 0 such that 7 * sqrt(5) = a / b. Dividing both sides by 7 gives sqrt(5) = a / (7 * b). Since a is an integer and b is an integer, 7 * b is also an integer (and non-zero). Hence, a / (7 * b) is a rational number by definition. This implies that sqrt(5) is rational, which contradicts the given fact that sqrt(5) is irrational. Therefore, our assumption is false, and 7 * sqrt(5) is irrational.

Solution 3: Assume to the contrary that sqrt(2) + sqrt(3) is a rational number r. Because sqrt(2) and sqrt(3) are positive, r > 0. Squaring both sides of r = sqrt(2) + sqrt(3) gives r^2 = (sqrt(2) + sqrt(3))^2 = 2 + 2 * sqrt(6) + 3 = 5 + 2 * sqrt(6). Rearranging this equation to isolate the radical yields 2 * sqrt(6) = r^2 - 5, which means sqrt(6) = (r^2 - 5) / 2. Since r is rational, r^2 is rational, and (r^2 - 5) / 2 is also a rational number. This implies that sqrt(6) is rational. However, 6 is not a perfect square (its prime factors are 2 * 3, both with exponent 1), making sqrt(6) irrational. This contradiction shows that the assumption was false. Hence, sqrt(2) + sqrt(3) is irrational.