Chapter 1.5 — Exercise 1.4 — Irrationality Proofs

What Makes a Number Irrational

A number is irrational if it cannot be written as p/q for any integers p and q with q ≠ 0. Its decimal expansion is non-terminating and non-repeating — it runs forever without ever settling into a repeating block, which is what separates it from every rational number in the previous exercise. √2, √3, √5, ∛5, and π are all irrational.

Two facts about combining rational and irrational numbers get used repeatedly in this exercise: adding or subtracting a rational number from an irrational one always gives an irrational result (3 + √7, 5 − √8, and 5 − √3 are all irrational), and multiplying or dividing a non-zero rational number by an irrational one is also always irrational (3√7 and 5/√2 are both irrational). Both facts follow the same logic used throughout this exercise — if the combined result were rational, a little rearranging would isolate the irrational part as equal to some ratio of integers, which is exactly the contradiction every proof below is built around.

The Theorem Underneath Every Proof Here

Before proving any specific number irrational, the chapter establishes one supporting result: if a prime number p divides a², then p must also divide a.

The reasoning traces back to unique prime factorisation from the last exercise: write a as a product of primes, p₁·p₂·…·pₙ. Then a² is the product of those same primes, each appearing twice. If p divides a², p must be one of p₁ through pₙ — which means p divides a directly, since a is just the product of those same primes. This one fact is what lets every proof below turn a squared equation into a genuine contradiction.

The Model Proof: √2 Is Irrational

Every other proof in this exercise follows the shape of this one. Assume √2 is rational, so √2 = a/b for some integers a and b with no common factor other than 1 (co-prime), and b ≠ 0.

• Squaring both sides: a² = 2b², so 2 divides a².
• By the theorem above, 2 must divide a — so write a = 2c for some integer c.
• Substituting back: (2c)² = 2b², which simplifies to 2c² = b², so 2 divides b² — and therefore 2 divides b as well.
• But now 2 divides both a and b, contradicting the assumption that a and b are co-prime.

The contradiction means the original assumption was false. √2 is irrational.

Five More Numbers, Same Method

Question 1 asks for five separate irrationality proofs. Each one reaches its contradiction slightly differently, depending on where the known irrational number is hiding in the expression.

• 1/√2 — assume 1/√2 = a/b. Rearranging gives √2 = b/a, a ratio of two integers — which would make √2 rational, directly contradicting the model proof above.
• 3 + √5 — assume 3 + √5 = a/b. Isolating the root gives √5 = a/b − 3, then squaring and simplifying eventually expresses √5 as a ratio of integers — a contradiction, since √5 is irrational by the same argument used for √2.
• 6 + √2 — assume 6 + √2 = a/b. Then √2 = (a − 6b)/b, and since a and b are integers, the right-hand side is rational. This contradicts the irrationality of √2 proved above.
• √5 — identical in structure to the √2 proof: assume √5 = a/b (co-prime), square to get a² = 5b², use the supporting theorem to show 5 divides both a and b, and reach the same co-prime contradiction.
• 3 + 2√5 — assume 3 + 2√5 = a/b. Then 2√5 = (a − 3b)/b, so √5 = (a − 3b)/(2b) — a ratio of integers, contradicting √5's irrationality.

Notice the pattern across all five: whatever the exact expression, the algebra is always steered toward isolating the one known-irrational term (√2 or √5) on its own, so that the contradiction lands on a result already proved.

One of the Five, Written Out in Full

The 3 + √5 proof is worth seeing in complete detail, since it's the one most often mis-copied as "√3 + √5" — a genuinely different, harder expression. The actual question is a rational integer plus a single square root, and it turns out to need no squaring at all.

• Assume 3 + √5 is rational, so 3 + √5 = a/b for co-prime integers a, b with b ≠ 0.
• Rearranging: √5 = a/b − 3.
• Since a, b, and 3 are all integers, the right-hand side a/b − 3 is a ratio of integers — a rational number.
• So the equation says √5 equals a rational number — but √5 is already known to be irrational.

A rational number can never equal an irrational one, so the assumption fails: 3 + √5 is irrational. Unlike the model √2 proof, this one never needs to square anything — isolating √5 alone on one side is already enough to expose the contradiction, since everything on the other side is visibly built only from integers.

Generalising to Any Two Primes

Question 2 asks for something more sweeping: prove that √p + √q is irrational, where p and q are any two primes — not just 2, 3, or 5. Assume √p + √q = a/b, with a and b co-prime and b ≠ 0.

The right-hand side of that final equation is a ratio of integers — rational. But p is prime, so √p is irrational by the same theorem used throughout this exercise, and a rational expression cannot equal an irrational one. The contradiction proves that √p + √q is irrational for every possible pair of primes, all at once — a single argument that covers √2 + √3, √5 + √7, √11 + √13, and infinitely many other pairs without needing to repeat the proof for each.

The Five-Step Shape

Stripped of the specific numbers, every proof in this exercise runs through the same sequence of moves:

Carrying This Forward

This exercise closes the loop on how numbers were classified back in the chapter introduction — every proof here rests on the "p divides a² ⟹ p divides a" theorem, which is itself a direct consequence of the unique prime factorisation covered in Exercise 1.2. The last exercise in the chapter, Exercise 1.5, moves on to logarithms — a different tool entirely, though Question 9 there circles back to irrationality one more time, asking whether log 2 is rational or not.