(a) If 2 is given as an irrational number, then prove that (5−22) is an irrational number.
OR
(b) Check whether 6n can end with the digit 0 for any natural number n.
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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 5−22 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=0.
Step 2: Isolate 2
Now, we rearrange the equation to isolate 2. Subtract r from both sides and add 22 to both sides, then divide by 2. This gives us 2=25−r.
Step 3: Contradiction for part (a)
Since r is a rational number, 5−r is also rational, and 25−r 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.
Step 4: Conclusion for part (a)
Therefore, our initial assumption that 5−22 is rational must be false. Hence, 5−22 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 6n. We can write 6 as 2×3. So, 6n=(2×3)n=2n×3n.
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 6n are only 2 and 3. There is no factor of 5 in the prime factorization of 6n.
Step 7: Conclusion for part (b)
Since the prime factorization of 6n does not contain 5, 6n cannot be divisible by 10, and therefore, it cannot end with the digit 0 for any natural number n.