Now, we rearrange the equation to isolate √2. Subtract r from both sides and add 2√2 to both sides, then divide by 2. This gives us √2 = (5 - r)/(2).

Since r is a rational number, 5-r is also rational, and (5-r)/(2) is rational. This means the right-hand side of the equation is rational. However, we are given that √2 is irrational. An irrational number cannot be equal to a rational number. This is a contradiction, which proves our initial assumption was false.

Since r is a rational number, 5-r is also rational, and (5-r)/(2) is rational. This means the right-hand side of the equation is rational. However, we are given that √2 is irrational. An irrational number cannot be equal to a rational number. This is a contradiction, which proves our initial assumption was false.

Therefore, our initial assumption that 5 - 2√2 is rational must be false. Hence, 5 - 2√2 is an irrational number.

(a) If $\sqrt{2}$ is given as an irrational number, then prove that $(5 - 2\sqrt{2})$ is an irrational number. OR (b) Check whether $6^n$ can end with the digit 0 for any natural number n.

Answer: (a) $5 - 2\sqrt{2}$ is an irrational number. (b) $6^n$ cannot end with the digit 0 for any natural number $n$ .

Step-by-step solution

Step 1: Assume the opposite for part (a)

For part (a), we will use the method of contradiction. We assume that $5 - 2\sqrt{2}$ is a rational number, denoted by $r$ . A rational number can be expressed as a fraction $p/q$ where $p$ and $q$ are integers and $q \neq 0$ .

Step 2: Isolate $\sqrt{2}$

Now, we rearrange the equation to isolate $\sqrt{2}$ . Subtract $r$ from both sides and add $2\sqrt{2}$ to both sides, then divide by 2. This gives us $\sqrt{2} = \frac{5 - r}{2}$ .

Step 3: Contradiction for part (a)

Since $r$ is a rational number, $5-r$ is also rational, and $\frac{5-r}{2}$ is rational. This means the right-hand side of the equation is rational. However, we are given that $\sqrt{2}$ is irrational. An irrational number cannot be equal to a rational number. This is a contradiction, which proves our initial assumption was false.

Step 4: Conclusion for part (a)

Therefore, our initial assumption that $5 - 2\sqrt{2}$ is rational must be false. Hence, $5 - 2\sqrt{2}$ is an irrational number.

Step 5: Analyze prime factors for part (b)

For part (b), a number ending with the digit 0 must have 10 as a factor, which means it must have both 2 and 5 as prime factors. Let's find the prime factorization of $6^n$ . We can write $6$ as $2 \times 3$ . So, $6^n = (2 \times 3)^n = 2^n \times 3^n$ .

Step 6: Check for prime factor 5

According to the Fundamental Theorem of Arithmetic, the prime factorization of any natural number is unique. The prime factors of $6^n$ are only 2 and 3. There is no factor of 5 in the prime factorization of $6^n$ .

Step 7: Conclusion for part (b)

Since the prime factorization of $6^n$ does not contain 5, $6^n$ cannot be divisible by 10, and therefore, it cannot end with the digit 0 for any natural number $n$ .

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(a) If 2\sqrt{2}2​ is given as an irrational number, then prove that (5−22)(5 - 2\sqrt{2})(5−22​) is an irrational number. OR (b) Check whether 6n6^n6n can end with the digit 0 for any natural number n.