Real Numbers Mind Map

If one integer a is divided by another positive integer b, we can write a = bq + r, where q is the quotient and r is the remainder, with 0 <= r < b.

Always write the condition 0 <= r < b after the division form. Many CBSE answers expect both the equation and the remainder condition.

Iterative division for HCF is a step-by-step method where the larger number is divided by the smaller number, then the divisor and remainder are used again until the remainder becomes 0. The last non-zero remainder is the HCF.

Write each division clearly and stop only when the remainder becomes 0. The last non-zero remainder is the answer.

Every composite number can be written as a product of prime numbers, and this prime factorisation is unique apart from the order of the primes.

Always break numbers completely into primes before using them in HCF or LCM questions.

To find the HCF by prime factorisation, write both numbers as products of primes and take only the common prime factors with the smallest powers.

For HCF, remember the phrase: common primes with minimum powers.

To find the LCM by prime factorisation, write both numbers as primes and take every prime that appears in either number, using the highest power needed.

For LCM, remember the phrase: all primes with maximum powers.

For two positive integers a and b, the product of the numbers equals the product of their HCF and LCM.

Use this relation as a shortcut after finding either the HCF or the LCM by factorisation.

A fraction in lowest form has a terminating decimal expansion only when its denominator has prime factors 2 and/or 5 and no other prime factors.

Always simplify the fraction first, then inspect only the denominator's prime factors.

If a fraction in lowest form has any prime factor in its denominator other than 2 or 5, its decimal expansion is non-terminating recurring.

Remember the pair: repeating decimal means rational; endless non-repeating decimal means irrational.

An irrational number cannot be written in the form p/q, where p and q are integers and q != 0.

Check both conditions: the decimal should not end and should not repeat.

Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms.

In a proof by contradiction, always end with the words 'this contradicts lowest terms'.

Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms.

Mention the prime 3 clearly at the divisibility step so the logic stays correct.

Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms.

Keep the same proof steps and only change the prime factor to 5.