show that the product of any two consecutive positive integers is always even.
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i) Let the two consecutive integers be n, (n+1)
ii) Then their product is P = n(n+1)
iii) For n = 1, P = 1*2 = 2, which is divisible by 2; so it is true for n = 1
Thus the statement is k(k+1) is divisible by 2 for k a positive integer ----------- (1)
P(k+1): (k+1)(k+1+1) = (k+1)(k+2) = k(k+1) + 2(k+1)
==> P(k+1) = P(k) + 2*(A positive integer) [From (1), P(k) = k(k+1)]
Thus both terms are divisible by 2
Hence P(k+1) is also divisible by 2 --------------- (2)
Thus from (1) & (2), for two consecutive arbitrary terms, it is proved that the product is divisible by 2. Hence the product of two consecutive positive integers is divisible by 2.
ii) Then their product is P = n(n+1)
iii) For n = 1, P = 1*2 = 2, which is divisible by 2; so it is true for n = 1
Thus the statement is k(k+1) is divisible by 2 for k a positive integer ----------- (1)
P(k+1): (k+1)(k+1+1) = (k+1)(k+2) = k(k+1) + 2(k+1)
==> P(k+1) = P(k) + 2*(A positive integer) [From (1), P(k) = k(k+1)]
Thus both terms are divisible by 2
Hence P(k+1) is also divisible by 2 --------------- (2)
Thus from (1) & (2), for two consecutive arbitrary terms, it is proved that the product is divisible by 2. Hence the product of two consecutive positive integers is divisible by 2.
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