Use the Principle of Complete Induction to prove that fore each integer n>=12, t

Question

Explanation / Answer

first we check for n=12. 12 = 3*4 + 7*0 so there exists nonnegative integers such that n = 3a+7b now for any k>n assume that there exists nonnegative integers a and b such that k=3a+7b. now for k+1, k+1 = 3a+7b+1 there are two cases. i) k is odd and ii) k is even. i) if k is odd, one of a or b is even and other is odd. let a=2m and b = 2n+1 so k = 6m + 14n + 7 so now k+1 = 6m+14n+8 = 6m+14n+15-7 = 3(2m+5) + 7*(2n-1) so k+1 can also be written as 3a'+7b' where a' and b' are non negative. this is true when k is odd. ii) now when k is even, as k is even, and 3,7 are odd, both a and b should be even. so a = 2m, b = 2n. so k = 6m+14n. k+1 = 6m+14n +1 = 6m + 14n + 15 - 14 = 3*(2m+5) + 7*(2n-2) = 3a'+7b' hence even here k+1 can be written as 3a'+7b' where a' and b' are non negative. this is true when k is even. in both the cases we proved that k can be represented implies that k+1 can be represented. hence by the Principle of Complete Induction,each integer n>=12, can be written as n = 3a + 7b where a and b are nonnegative integers

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.