Real Numbers

Concept: Euclid's Division Lemma states that for any two positive integers a and b, there exist unique whole numbers q (quotient) and r (remainder) such that a = bq + r, where 0 <= r < b. This is just the familiar idea of division written as an equation. Euclid's Division Algorithm uses this lemma repeatedly to find the HCF (Highest Common Factor) of two positive integers.

Method to find HCF(a, b) with a > b: (1) Apply the lemma to write a = bq + r. (2) If r = 0, then b is the HCF. (3) If r is not 0, apply the lemma again to b and r. (4) Repeat using the previous divisor and remainder until the remainder becomes 0. The divisor at that final stage is the HCF.

Worked example: Find HCF(420, 130). 420 = 130 x 3 + 30. Now take 130 and 30: 130 = 30 x 4 + 10. Now take 30 and 10: 30 = 10 x 3 + 0. The remainder is 0, and the last divisor is 10, so HCF(420, 130) = 10.

Common mistake: Students often stop too early at a non-zero remainder, or they pick the wrong number as the answer. Remember: always continue until the remainder is exactly 0, and the HCF is the LAST DIVISOR, not the last remainder. Also note the remainder must satisfy 0 <= r < b; r can never be negative or equal to or larger than the divisor.

Step 1: Identify the larger number as a and the smaller as b. Here a = 867, b = 255.

Step 2: Apply Euclid's division lemma to 867 and 255.
867 = 255 x 3 + 102 (since 255 x 3 = 765 and 867 - 765 = 102)
The remainder is 102, which is not 0, so continue.

Step 3: Apply the lemma to the previous divisor 255 and remainder 102.
255 = 102 x 2 + 51 (since 102 x 2 = 204 and 255 - 204 = 51)
The remainder is 51, not 0, so continue.

Step 4: Apply the lemma to 102 and 51.
102 = 51 x 2 + 0
The remainder is now 0.

Step 5: The last divisor is 51. Therefore HCF(867, 255) = 51.

Verification: 867 = 51 x 17 and 255 = 51 x 5. Since 17 and 5 share no common factor other than 1, the HCF is indeed 51.

Goal: Use Euclid's division lemma to prove the given statement.

Step 1: Let a be any positive integer and take b = 4. By Euclid's division lemma, there exist whole numbers q and r such that a = 4q + r, where 0 <= r < 4.

Step 2: Since 0 <= r < 4, the only possible values of r are 0, 1, 2 and 3. So a can be written in exactly one of these forms: 4q, 4q + 1, 4q + 2, or 4q + 3.

Step 3: Check which forms are even and which are odd.
4q = 2(2q), which is even.
4q + 2 = 2(2q + 1), which is even.
4q + 1 is even-plus-one, so it is odd.
4q + 3 is even-plus-one, so it is odd.

Step 4: Conclusion. A positive odd integer cannot be of the form 4q or 4q + 2 (those are even). Therefore every positive odd integer must be of the form 4q + 1 or 4q + 3. Proved.

Concept: The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur. For example, 12 = 2 x 2 x 3, and no other set of primes can multiply to give 12.

Method (factor tree / prime factorisation): Keep splitting the number into a product of two factors until every branch ends in a prime. Then collect the primes and write them using exponents. This prime factorisation is the key tool for finding HCF and LCM of numbers.

Worked example: Factorise 60 using a factor tree.

Reading the primes off the tree: 60 = 2 x 2 x 3 x 5 = 2^2 x 3 x 5.

A handy use: once a number is in prime form, you can immediately list its factors and check divisibility. For instance, since 60 = 2^2 x 3 x 5 has no extra prime, 60 is not divisible by 7. This uniqueness of prime factorisation is exactly why HCF and LCM calculations always give one definite answer.

Common mistake: Writing a factorisation that still contains a composite number (for example leaving 60 = 4 x 15 and stopping). Always continue until every factor is prime, and remember 1 is NOT a prime, so it should never appear in the prime factorisation.

Step 1: Write the prime factorisation of each number.
96 = 2 x 48 = 2 x 2 x 24 = 2 x 2 x 2 x 12 = 2 x 2 x 2 x 2 x 6 = 2 x 2 x 2 x 2 x 2 x 3 = 2^5 x 3.
404 = 2 x 202 = 2 x 2 x 101 = 2^2 x 101 (101 is prime).

Step 2: HCF = product of the SMALLEST powers of the COMMON prime factors.
The only common prime is 2. Smallest power of 2 is 2^2.
HCF(96, 404) = 2^2 = 4.

Step 3: LCM = product of the GREATEST powers of ALL prime factors that appear.
Primes appearing: 2 (greatest power 2^5), 3 (power 3^1), 101 (power 101^1).
LCM(96, 404) = 2^5 x 3 x 101 = 32 x 3 x 101 = 96 x 101 = 9696.

Step 4: Verify using HCF x LCM = product of the two numbers.
HCF x LCM = 4 x 9696 = 38784.
Product = 96 x 404 = 38784.
Both sides are equal, so the answer is verified. HCF = 4 and LCM = 9696.

Step 1: Prime factorise all three numbers.
6 = 2 x 3 = 2^1 x 3^1.
72 = 2 x 36 = 2 x 2 x 18 = 2 x 2 x 2 x 9 = 2^3 x 3^2.
120 = 2 x 60 = 2 x 2 x 30 = 2 x 2 x 2 x 15 = 2^3 x 3 x 5 = 2^3 x 3^1 x 5^1.

Step 2: HCF = product of smallest powers of common primes.
Common primes to all three: 2 and 3.
Smallest power of 2 across the three numbers is 2^1; smallest power of 3 is 3^1.
HCF(6, 72, 120) = 2^1 x 3^1 = 6.

Step 3: LCM = product of greatest powers of all primes that appear.
Primes appearing: 2 (greatest power 2^3), 3 (greatest power 3^2), 5 (greatest power 5^1).
LCM(6, 72, 120) = 2^3 x 3^2 x 5 = 8 x 9 x 5 = 360.

Step 4: Note. For THREE or more numbers, the identity HCF x LCM = product of numbers does NOT hold in general (here 6 x 360 = 2160, but 6 x 72 x 120 = 51840). That shortcut works only for TWO numbers. Final answer: HCF = 6, LCM = 360.

1. Fundamental Theorem of Arithmetic: every composite number = a unique product of primes (apart from order). Example: 540 = 2^2 x 3^3 x 5.

2. HCF (Highest Common Factor) = product of the SMALLEST power of each COMMON prime factor.

3. LCM (Lowest Common Multiple) = product of the GREATEST power of each prime factor that appears in any of the numbers.

4. Relationship for TWO positive integers a and b:
   HCF(a, b) x LCM(a, b) = a x b.
   Equivalently: LCM(a, b) = (a x b) / HCF(a, b), and HCF(a, b) = (a x b) / LCM(a, b).
   Important: this product rule holds ONLY for two numbers, not for three or more.

5. Decimal-expansion test: a rational p/q in lowest terms terminates if and only if q = 2^n x 5^m (n, m are whole numbers); otherwise it is non-terminating repeating.

6. Euclid's Division Lemma: for positive integers a and b, a = bq + r with 0 <= r < b (q and r unique).

1. Euclid's Division Lemma: for any positive integers a and b, there exist unique whole numbers q and r with a = bq + r and 0 <= r < b. Applying it repeatedly (Euclid's Division Algorithm) gives the HCF of two numbers; the HCF is the last divisor when the remainder becomes 0.

2. Fundamental Theorem of Arithmetic: every composite number can be factorised into primes uniquely, apart from the order of the factors. Prime factorisation (factor trees) is the main tool of the chapter.

3. HCF and LCM by prime factorisation: HCF uses the smallest powers of common primes; LCM uses the greatest powers of all primes. For two numbers, HCF x LCM = product of the numbers.

4. Irrational numbers: numbers that cannot be written as p/q with integer p, q and q not 0; their decimals are non-terminating and non-repeating. Numbers like root 2, root 3 and root 5 are proved irrational by contradiction, using 'if a prime p divides a^2 then p divides a'. Also, rational +/- irrational is irrational, and a non-zero rational times an irrational is irrational.

5. Decimal expansions of rationals: a rational p/q in lowest terms has a terminating decimal exactly when q = 2^n x 5^m; otherwise the decimal is non-terminating and repeating. Always reduce the fraction to lowest terms before testing the denominator.

Concept: A number is rational if it can be written as p/q where p and q are integers and q is not 0. A number is irrational if it CANNOT be written in this form; its decimal expansion is non-terminating and non-repeating. Examples include the square root of 2, the square root of 3, and the square root of 5.

Key theorem used in proofs: If a prime number p divides a^2 (where a is a positive integer), then p also divides a. This follows from the Fundamental Theorem of Arithmetic and is essential for irrationality proofs.

Method (proof by contradiction): To prove a number is irrational, first ASSUME the opposite, that it is rational and can be written as a fraction p/q in lowest terms (so p and q have no common factor other than 1). Then derive a logical contradiction, usually by showing p and q share a common factor after all. The contradiction means the assumption was false, so the number must be irrational.

Worked example (outline): To prove the square root of 3 is irrational, assume root 3 = p/q in lowest terms. Squaring gives 3q^2 = p^2, so 3 divides p^2, hence 3 divides p. Write p = 3m, then 3q^2 = 9m^2, so q^2 = 3m^2, meaning 3 divides q too. But then 3 is a common factor of p and q, contradicting lowest terms. So root 3 is irrational.

Common mistake: Forgetting to state that p/q is in lowest terms (with no common factors) at the start, which is the very assumption the contradiction breaks. Another error is claiming a number is irrational just because it looks messy. Useful fact: rational + irrational = irrational, and (non-zero rational) x irrational = irrational.

We use proof by contradiction.

Step 1: Assume the opposite. Suppose root 2 is rational. Then we can write root 2 = p/q, where p and q are integers, q is not 0, and the fraction is in its lowest terms (p and q have no common factor other than 1).

Step 2: Rearrange and square both sides.
root 2 = p/q => q x root 2 = p => squaring gives 2q^2 = p^2. ...(i)

Step 3: Deduce a factor of p. Equation (i) shows that 2 divides p^2. By the theorem 'if a prime divides a^2 then it divides a', it follows that 2 divides p.

Step 4: Substitute p = 2m for some integer m. Put this into (i):
2q^2 = (2m)^2 = 4m^2 => q^2 = 2m^2. ...(ii)

Step 5: Deduce a factor of q. Equation (ii) shows 2 divides q^2, so by the same theorem 2 divides q.

Step 6: Reach the contradiction. From Steps 3 and 5, both p and q are divisible by 2, so 2 is a common factor of p and q. This contradicts our assumption that p/q is in lowest terms.

Step 7: Conclusion. The assumption that root 2 is rational must be false. Therefore the square root of 2 is irrational. Proved.

We use proof by contradiction, and we are allowed to use the known fact that root 3 is irrational.

Step 1: Assume the opposite. Suppose 5 - root 3 is rational. Then we can write 5 - root 3 = a/b, where a and b are integers and b is not 0.

Step 2: Isolate root 3.
5 - root 3 = a/b
=> root 3 = 5 - a/b
=> root 3 = (5b - a)/b.

Step 3: Examine the right-hand side. Since a and b are integers, 5b - a is an integer and b is a non-zero integer, so (5b - a)/b is a ratio of two integers, that is, a rational number.

Step 4: Reach the contradiction. The equation now says root 3 equals a rational number. But it is given that root 3 is irrational. A number cannot be both rational and irrational at the same time. This is a contradiction.

Step 5: Conclusion. Our assumption was wrong, so 5 - root 3 cannot be rational. Therefore 5 - root 3 is irrational. Proved. (The same reasoning shows that any rational number minus an irrational number is irrational.)

Concept: Every rational number has a decimal expansion that is either terminating (ends after finitely many digits) or non-terminating but repeating (a block of digits repeats forever). Which one it is can be decided WITHOUT actually dividing, by looking at the prime factors of the denominator.

Key theorem: Let x = p/q be a rational number in its lowest terms (p and q have no common factor). Then x has a TERMINATING decimal expansion if and only if the denominator q can be written in the form 2^n x 5^m, where n and m are whole numbers (zero or more). If q has any prime factor other than 2 or 5, the expansion is non-terminating and repeating.

Method: (1) Reduce p/q to lowest terms. (2) Prime factorise the denominator q. (3) If only 2s and 5s appear, it terminates; otherwise it does not.

Worked example: Decide for 7/80. Here 80 = 2^4 x 5, which is of the form 2^n x 5^m, so 7/80 terminates. (Indeed 7/80 = 0.0875.) Now check 7/30: 30 = 2 x 3 x 5 contains the prime 3, so 7/30 is non-terminating repeating (7/30 = 0.2333...).

Common mistake: Forgetting to reduce the fraction to lowest terms first. For example 6/15 looks like it has denominator 15 = 3 x 5 (suggesting non-terminating), but 6/15 reduces to 2/5, whose denominator is just 5, so it actually terminates (2/5 = 0.4). Always simplify before testing the denominator.

We use the rule: in lowest terms, a fraction terminates exactly when the denominator is of the form 2^n x 5^m.

Part A: 13/3125.
Step 1: 13 is prime and does not divide 3125, so the fraction is already in lowest terms.
Step 2: Factorise the denominator. 3125 = 5 x 5 x 5 x 5 x 5 = 5^5. This is 2^0 x 5^5, only the prime 5, so the decimal TERMINATES.
Step 3: Find the decimal by making the denominator a power of 10. Multiply top and bottom by 2^5 = 32: (13 x 32)/(5^5 x 2^5) = 416/10^5 = 416/100000 = 0.00416.

Part B: 17/8.
Step 1: 17 is prime and does not divide 8, so 17/8 is in lowest terms.
Step 2: 8 = 2^3 = 2^3 x 5^0, only the prime 2, so the decimal TERMINATES.
Step 3: Multiply top and bottom by 5^3 = 125: (17 x 125)/(2^3 x 5^3) = 2125/10^3 = 2125/1000 = 2.125.

Conclusion: 13/3125 = 0.00416 and 17/8 = 2.125; both are terminating decimals.

We use the rule: in lowest terms, a fraction is non-terminating repeating exactly when its denominator has a prime factor other than 2 or 5.

Step 1: Reduce to lowest terms. Factorise numerator and denominator.
64 = 2^6.
455 = 5 x 91 = 5 x 7 x 13.
The numerator's only prime is 2, while the denominator has primes 5, 7 and 13. They share no common factor, so 64/455 is already in lowest terms.

Step 2: Apply the test to the denominator 455 = 5 x 7 x 13. Besides the allowed prime 5, it also contains the primes 7 and 13. Since these are NOT 2 or 5, the denominator is NOT of the form 2^n x 5^m.

Step 3: Conclusion by the theorem. Because the reduced denominator has prime factors other than 2 and 5, the decimal expansion of 64/455 is non-terminating and repeating (recurring). We reached this conclusion purely from the prime factors, without performing any long division.