Real Numbers — Class 10 CBSE Mathematics MCQs with Solutions

Solution: Euclid's division lemma states that for any two positive integers $a$ and $b$, we can express $a = bq + r$ where $q$ is the quotient and $r$ is the remainder. The crucial condition is that the remainder must satisfy $0 \leq r < b$ — it cannot be negative and must be strictly less than the divisor. This form is unique for the given $a$ and $b$.

Solution: When $r = 0$, the equation becomes $a = bq + 0 = bq$. This means $a$ is exactly divisible by $b$, or in other words, $b$ is a divisor of $a$. The remainder being zero is the defining condition for divisibility.

Solution: Uniqueness of prime factorization means that if you factor a number into primes, you will always get the same set of primes with the same frequencies, no matter which method you use. For example, $12 = 2 \times 2 \times 3 = 2^2 \times 3$ is the *only* way to express 12 as a product of primes. You cannot express 12 as $2 \times 5 + \text{something}$ or any other prime combination.

Solution: The Fundamental Theorem of Arithmetic covers all integers greater than 1. A number is either: (1) prime (has no prime factorization other than itself), or (2) composite (can be uniquely factorized into primes). This theorem guarantees that every number falls into one of these two categories, and composite numbers have a unique factorization.

Solution: An irrational number is precisely one that cannot be written in the form $\frac{p}{q}$ where $p$ and $q$ are integers with $q \neq 0$. This is the defining characteristic. Note that 'not an integer' is too broad (includes fractions like $\frac{1}{2}$, which are rational), and negative numbers can be rational or irrational.

Solution: $\sqrt{2}$ is irrational. We can prove this: assume $\sqrt{2} = \frac{p}{q}$ for some integers $p$ and $q$. Then $2q^2 = p^2$. By prime factorization, the left side has an even power of 2, but the right side has an odd power, a contradiction. Therefore, $\sqrt{2}$ cannot be rational; it must be irrational.

Solution: $\sqrt{3}$ is irrational (similar proof to $\sqrt{2}$: assuming $\sqrt{3} = \frac{p}{q}$ leads to a contradiction using prime factorization). $\frac{22}{7}$ is rational by definition. $0.333... = \frac{1}{3}$ is rational (repeating decimal). $\sqrt{16} = 4$ is an integer, hence rational.

Solution: $\frac{1}{6}$ has denominator $6 = 2 \times 3$. Since 3 is a prime factor (not 2 or 5), the decimal will not terminate. Computing: $1 \div 6 = 0.1\overline{6}$ (the digit 6 repeats). In contrast, $\frac{1}{8} = \frac{1}{2^3}$ terminates as $0.125$, and $\frac{1}{5}$ terminates as $0.2$.

Solution: Uniqueness is critical because it means Euclid's division lemma gives us exactly one answer every time. This allows us to build algorithms that always work correctly (like the Euclidean algorithm for GCD), and it provides a rigorous foundation for mathematical proofs. Without uniqueness, we couldn't rely on the lemma for logical deduction.

Solution: Euclid's division lemma can be extended to negative integers while maintaining the constraint $0 \leq r < |b|$. For $-17$ and $5$: we express $-17 = 5 \times (-4) + 3$. Notice that $q = -4$ (not $-3$) so that the remainder stays non-negative. This extension preserves uniqueness and the division property, making it a powerful tool even for negative integers.

Solution: Uniqueness means that once we choose specific values for $a$ and $b$, there is only one pair $(q, r)$ that can satisfy $a = bq + r$ with $0 \leq r < b$. We cannot find two different valid pairs. This is what makes the lemma a fundamental theorem — it guarantees that the division process has a predictable, single outcome.

Solution: The Fundamental Theorem of Arithmetic guarantees that there is *only one* way to express 60 as a product of primes. If someone claims to have found a different factorization, either (1) their factorization is wrong, (2) it includes composite factors (not just primes), or (3) it is the same factorization written in a different order. The theorem rules out any genuinely different factorization.

Solution: Since the prime factorization is unique, the set of all divisors of a number is completely determined by its primes. For $12 = 2^2 \times 3$, the divisors are exactly those formed by taking powers of 2 from $0$ to $2$ and powers of 3 from $0$ to $1$: {1, 2, 3, 4, 6, 12}. Uniqueness means there are no hidden divisors outside this system. This makes the divisibility structure predictable and complete.

Solution: For any prime $p$: assume $\sqrt{p} = \frac{a}{b}$ where $\gcd(a, b) = 1$. Squaring gives $pb^2 = a^2$. By the Fundamental Theorem of Arithmetic, the prime $p$ appears an odd number of times (exactly once) on the left, but an even number of times on the right (as part of $a^2$). This is impossible, so $\sqrt{p}$ is irrational. This applies to every prime.

Solution: While the product of a rational and irrational is always irrational, the product of two irrationals can be either rational or irrational. The key example: $\sqrt{2} \times \sqrt{2} = 2$, which is rational. Another example: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$, which is irrational. So there is no universal rule — it depends on the specific irrationals.