Real Numbers

Every composite number has exactly one blueprint of prime factors, in exactly one combination, no matter how you go about finding it. That single fact of uniqueness turns out to be the one tool this whole chapter runs on -- it builds HCF and LCM directly, and it is also the hidden engine that proves numbers like the square root of 2 can never be written as a fraction.

Sonia takes 18 minutes per lap and Ravi 12. Both start together. Each returns to the start after a whole number of laps, so the first shared return is the Lowest Common Multiple. Sonia's return times are 18, 36, 54, ... minutes; Ravi's are 12, 24, 36, ... minutes. The first common time is 36 minutes. Prime factorisation confirms it: 18=2×3² and 12=2²×3; take the greatest power of each prime, so LCM=2²×3²=36. In 36 minutes Sonia completes two laps and Ravi three.

Factor 32760 into primes, following the book's large example. Divide by 2 three times to reach 4095; divide by 3 twice to reach 455; then 455=5×91=5×7×13. Thus 32760=2×2×2×3×3×5×7×13=2³×3²×5×7×13. The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes, and that prime-factor list is unique apart from its order. Reordering these same factors is not a different factorisation.

Use the book's first prime-factor example, 6=2×3 and 20=2²×5. HCF takes the smallest power of each prime common to both numbers: only 2 is shared, so HCF=2. LCM takes the greatest power of every prime present: 2²×3×5=60. For two positive numbers, HCF(a,b)×LCM(a,b)=a×b; here 2×60=120 and 6×20=120. The larger 96-and-404 example and the three-number case are handled separately, so this video stays on one idea.

The book's prime-square theorem says that if a prime p divides u², then p divides u: squaring repeats prime factors but creates no new prime. To prove √2 irrational, assume √2=a/b in lowest terms, where a and b share no factor greater than 1. Squaring gives 2b²=a², so 2 divides a² and therefore a; write a=2c. Substitute: 2b²=4c², hence b²=2c², so 2 divides b² and therefore b. Now a and b both have factor 2, contradicting lowest terms. The assumption was false; √2 is irrational.

The book has already established that √3 is irrational. To prove 5−√3 is irrational, assume the opposite: let 5−√3=a/b for integers a and b with b nonzero. Isolating the root gives √3=5−a/b=(5b−a)/b. The numerator and denominator are integers, so this would make √3 rational—contradicting the known result. Therefore the assumption is false and 5−√3 is irrational. This video stays on this one proof; the book's separate 3√2 example is taught separately.

The book first proves that √2 is irrational. Now suppose, for contradiction, that 3√2 is rational. Then 3√2=a/b for integers a,b with b nonzero. Dividing both sides by 3 gives √2=(a/b)/3=a/(3b). Since a and 3b are integers and 3b is nonzero, this would make √2 rational. That contradicts the proved result. Therefore 3√2 is irrational. This is the book's Example 7, kept separate from the preceding 5−√3 proof so each video has one clear question.