Real Numbers — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)

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 are written. For example, 60 can be factorised as 2 × 2 × 3 × 5 — and no matter how you approach the factorisation, you will always arrive at exactly these same four prime factors (two 2s, one 3, one 5), just possibly written in a different order. This uniqueness is precisely what makes prime factorisation such a powerful and reliable tool — it is the basis for finding the HCF and LCM of numbers, since once every number is broken into its unique prime "building blocks," HCF and LCM can be read off directly by comparing which prime factors the numbers share and which they don't.

This is proved by contradiction. Step 1: Assume √2 is rational. Then it can be written as √2 = p/q, where p and q are integers with no common factor other than 1 (the fraction is in lowest terms), and q ≠ 0. Step 2: Squaring both sides: 2 = p²/q², so p² = 2q². Step 3: This means p² is even (since it equals 2 times an integer). An even square can only come from an even number, so p itself must be even. Let p = 2k for some integer k. Step 4: Substituting: (2k)² = 2q², so 4k² = 2q², giving q² = 2k². Step 5: By the same reasoning, q² is even, so q must be even too. Step 6: But if both p and q are even, they share a common factor of 2 — directly contradicting our Step 1 assumption that p and q have no common factor. Step 7: Since the assumption leads to a contradiction, it must be false. Therefore, √2 is irrational.

Euclid's Division Algorithm repeatedly applies: for two positive integers a and b (a > b), a = bq + r, where 0 ≤ r < b, and then replaces (a, b) with (b, r), continuing until the remainder is 0. The last non-zero remainder is the HCF. Step 1: 252 = 1 × 180 + 72 Step 2: 180 = 2 × 72 + 36 Step 3: 72 = 2 × 36 + 0 Since the remainder is now 0, the algorithm stops. The last non-zero remainder was 36. HCF(180, 252) = 36.

Step 1: Express both numbers as a product of prime factors. 24 = 2 × 2 × 2 × 3 = 2³ × 3¹ 36 = 2 × 2 × 3 × 3 = 2² × 3² Step 2: HCF is the product of the LOWEST power of each prime factor common to both numbers. HCF = 2² × 3¹ = 4 × 3 = 12 Step 3: LCM is the product of the HIGHEST power of every prime factor appearing in either number. LCM = 2³ × 3² = 8 × 9 = 72 Verification: HCF × LCM should equal the product of the two original numbers. 12 × 72 = 864, and 24 × 36 = 864 ✓ HCF = 12, LCM = 72.

Let x = 0.454545... Step 1: Since two digits (45) repeat, multiply both sides by 100 to shift the decimal point past one full repeating block: 100x = 45.454545... Step 2: Subtract the original equation from this new one — the repeating parts cancel exactly: 100x − x = 45.454545... − 0.454545... 99x = 45 Step 3: Solve for x x = 45/99 Step 4: Simplify by dividing both numerator and denominator by their HCF (which is 9) x = 5/11 Verification: 5 ÷ 11 = 0.454545... ✓ 0.454545... = 5/11.

To find the HCF of three numbers, first find the HCF of any two, then find the HCF of that result with the third number. Step 1: Prime factorise all three numbers. 96 = 2⁵ × 3 144 = 2⁴ × 3² 180 = 2² × 3² × 5 Step 2: Identify the lowest power of each prime factor common to ALL THREE numbers. The prime 2 appears in all three: lowest power is 2² (from 180). The prime 3 appears in all three: lowest power is 3¹ (from 96). The prime 5 appears only in 180, so it is NOT common to all three and is excluded. Step 3: HCF = 2² × 3 = 4 × 3 = 12 HCF(96, 144, 180) = 12.

Let the number be N. From the given information: N = 342k + 47, for some integer k (the quotient). Step 1: Since 342 = 18 × 19, rewrite N in terms of 18: N = 18 × 19k + 47 Step 2: Break 47 down in terms of 18: 47 = 18 × 2 + 11 Step 3: Substitute back: N = 18 × 19k + 18 × 2 + 11 N = 18(19k + 2) + 11 Since 18(19k + 2) is exactly divisible by 18, the remainder when N is divided by 18 is simply what's left over: 11. The remainder is 11.

For any two positive integers a and b, a fundamental relationship always holds: LCM(a, b) × HCF(a, b) = a × b Given: LCM = 504, HCF = 14, one number (a) = 56. Find the other number (b). Step 1: Substitute into the relationship 504 × 14 = 56 × b Step 2: Solve for b b = (504 × 14) / 56 b = 7056 / 56 b = 126 Verification: 56 = 2³ × 7, and 126 = 2 × 3² × 7. HCF: lowest common powers = 2¹ × 7¹ = 14 ✓ LCM: highest powers of every prime present = 2³ × 3² × 7 = 8 × 9 × 7 = 504 ✓ The other number is 126.

This is proved by contradiction, using the already-established fact that √3 is irrational. Step 1: Assume 4 + √3 IS rational. Then it can be written as 4 + √3 = p/q, for some integers p and q with q ≠ 0. Step 2: Rearrange to isolate √3: √3 = p/q − 4 = (p − 4q)/q Step 3: Since p and q are integers, (p − 4q) is also an integer, and q is a non-zero integer. This means (p − 4q)/q is a ratio of two integers — by definition, a RATIONAL number. Step 4: This gives us √3 = (a rational number), which directly contradicts the known fact that √3 is irrational. Step 5: Since the assumption leads to a contradiction, it must be false. Therefore, 4 + √3 is irrational. (This same proof technique — isolating the known irrational term and showing it would have to equal a rational expression — works for proving any expression of the form "integer + irrational number" is itself irrational.)