Consider whether 4^n can end with the digit 0 for any natural number n. If a number ends with 0, it must be divisible by 5, meaning 5 must be in its prime factorisation. Since 4^n = 2²n, the only prime factor is 2. By the uniqueness of the Fundamental Theorem of Arithmetic, 5 cannot be a factor, so 4^n can never end in 0.

Consider whether 4^n can end with the digit 0 for any natural number n. If a number ends with 0, it must be divisible by 5, meaning 5 must be in its prime factorisation. Since 4^n = 2²n, the only prime factor is 2. By the uniqueness of the Fundamental Theorem of Arithmetic, 5 cannot be a factor, so 4^n can never end in 0.

To prove that √2 is irrational, assume the contrary, that √2 = (a)/(b) where a and b are coprime integers. Squaring both sides gives 2b² = a², which implies that 2 divides a², and therefore 2 divides a. Writing a = 2c leads to b² = 2c², which implies 2 divides b. This contradicts the fact that a and b are coprime, proving √2 is irrational.

To prove that √2 is irrational, assume the contrary, that √2 = (a)/(b) where a and b are coprime integers. Squaring both sides gives 2b² = a², which implies that 2 divides a², and therefore 2 divides a. Writing a = 2c leads to b² = 2c², which implies 2 divides b. This contradicts the fact that a and b are coprime, proving √2 is irrational.

teach me the chapter real numbers

Answer: Chapter 1 (Real Numbers) focuses on two foundational ideas: (1) The Fundamental Theorem of Arithmetic for prime factorisation, finding HCF and LCM, and establishing divisibility properties; and (2) Revisiting irrational numbers using proof by contradiction.

Step-by-step solution

Step 1: State the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every composite number can be uniquely factored into prime factors, regardless of the order in which the prime factors occur. For example, $12 = 2^2 \times 3$ . This uniqueness allows us to solve divisibility problems and calculate greatest common divisor and least common multiple systematically.

Step 2: Compute HCF and LCM Using Prime Factorisation

To find the HCF of two or more numbers, identify all common prime factors and take the product of each raised to its smallest exponent. For the LCM, take every prime factor that appears in any of the factorisations and multiply them raised to their greatest exponents. For any two positive integers $a$ and $b$ , the relationship is $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ .

Step 3: Apply Uniqueness to Divisibility Conditions

Consider whether $4^n$ can end with the digit $0$ for any natural number $n$ . If a number ends with $0$ , it must be divisible by $5$ , meaning $5$ must be in its prime factorisation. Since $4^n = 2^{2n}$ , the only prime factor is $2$ . By the uniqueness of the Fundamental Theorem of Arithmetic, $5$ cannot be a factor, so $4^n$ can never end in $0$ .

Step 4: Prove Irrationality by Contradiction

To prove that $\sqrt{2}$ is irrational, assume the contrary, that $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are coprime integers. Squaring both sides gives $2b^2 = a^2$ , which implies that $2$ divides $a^2$ , and therefore $2$ divides $a$ . Writing $a = 2c$ leads to $b^2 = 2c^2$ , which implies $2$ divides $b$ . This contradicts the fact that $a$ and $b$ are coprime, proving $\sqrt{2}$ is irrational.

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