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7 × 11 × 13 + 13 = 13(7 × 11 + 1), so it has factors other than 1 and itself.
2ⁿ × 5ⁿ = (2 × 5)ⁿ = 10ⁿ.
Coprime integers have no common factor other than 1. Therefore, their HCF is 1.
Since HCF(a,b) = 1, HCF × LCM = ab gives LCM = ab.
HCF(18,60) = 6 and LCM(18,60) = 180. Hence 18 and 60 satisfy the conditions.
Since 2 is prime and 2 divides a², Theorem 1.2 gives 2 divides a. Therefore, a is even.
The proof assumes √2 is rational and writes it as a/b in lowest terms, where a and b are coprime.
Squaring √5 = a/b gives 5 = a²/b², so a² = 5b².
Since 3 is prime and divides a², Theorem 1.2 states that 3 must divide a.
If 2 + √3 were rational, subtracting the rational number 2 would make √3 rational, which is a contradiction.
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