Answers

2016-04-08T16:18:15+05:30
Base Case: When n = 1 we have 111 − 6 = 5 which is divisible by 5. So P(1) is correct. Induction hypothesis: Assume that P(k) is correct for some positive integer k. That means 11k − 6 is divisible by 5 and hence 11k − 6 = 5m for some integer m. So 11k = 5m + 6. Induction step: We will now show that P(k + 1) is correct. Always keep in mind what we are aiming for and what we know to be true. In this case we want to show that 11k+1 − 6 can be expressed as a multiple of 5, so we will start with the formula 11k+1 − 6 and we will rearrange it into something involving multiples of 5. At some point we will also want to use the assumption that 
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2016-04-08T16:38:13+05:30
Suppose the positive integer is n.

∴ n = 2q or n = 2q + 1         where q∈Z.

CASE 1:-
 
n = 2q
∴ n² - n = (2q)² - 2q
             = 4q² - 2q
             = 2(2q² - q)

CASE 2:-

n = 2q + 1
∴n² - n = (2q + 1)² - (2q + 1)
           = 4q² + 4q + 1 - 2q - 1
           = 4q² + 2q
           = 2(2q² + q)

Thus, in any case, n² - n is divisible by 2.

Thus, n² - n is divisible by 2 for every positive integer n.
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