January 9th , 2016
DM - Quiz 5 (BCSF15A)
Name:
Roll number:
1. 10 points For all integers n ≥ 2, prove the following by induction. Clearly state the inductive
hypothesis and what you want to prove in the inductive step.
1
1
1
n+1
1
1− 2
1 − 2 ··· 1 − 2 =
1− 2
2
3
4
n
2n
Solution:
Basis Step: For n = 2, the L.H.S. is 1 −
base case.
1
22
= 43 , and the R.H.S is
2+1
2∗2
= 34 . This establishes the
Inductive Hypothesis: Assume, for some k ≥ 2, that
1
1− 2
2
1
1
1
k+1
1− 2
1 − 2 ··· 1 − 2 =
3
4
k
2k
Inductive Step: We need to prove that, for k ≥ 2,
1
1
1
1
1
k+1+1
1− 2
1− 2
1 − 2 ··· 1 − 2
1−
=
2
3
4
k
(k + 1)2
2(k + 1)
Starting from the L.H.S, we have:
1
1
1
1
1
1− 2
1− 2
1 − 2 ··· 1 − 2
1−
2
3
4
k
(k + 1)2
k+1
1
=
1−
By inductive hypothesis
2k
(k + 1)2
k+1
(k + 1)2 − 1
=
2k
(k + 1)2
2
1
k + 2k + 1 − 1
=
2k
k+1
2
1
k + 2k
=
2k
k+1
1
k+2
=
k
2k
k+1
k+2
=
2(k + 1)
Discrete Mathematics (BCSF15A)
Quiz 5
January 9th , 2017
2. 10 points Use strong induction to prove that postage of Rs. 6 or more can be formed using just Rs. 3
and Rs. 4 stamps. Clearly state the inductive hypothesis and what you want to prove in the inductive
step.
Solution: We prove this using strong induction.
Basis Step: For n = 6, we note that this amount can be formed by using two Rs. 3 stamps; also, for
n = 7, we note that this amount can be formed by using one Rs. 3 and one Rs. 4 stamps.
Inductive Hypothesis: Assume that postage of Rs. j can be formed using just Rs. 3 and Rs. 4
stamps, for every integer j in the range 6 ≤ j ≤ k, for some k ≥ 8.
Inductive Step: We need to prove that postage of Rs. k + 1 can be formed using just Rs. 3 and Rs.
4 stamps, for k ≥ 8.
By inductive hypothesis we know that we can form the postage of Rs. k − 2 (because 6 ≤ k − 2 ≤ k),
so we take this postage, and add a Rs. 3 stamp to obtain the postage of Rs. k + 1.
3. 5 points A college received 300 applications for admissions into their different programs. 175 of these
applications were for admission in their CS program, and 150 for the EE program. 75 students applied
to both CS and EE programs. How many of the students applied in neither of the two programs?
Solution: Let C be the set of students who applied in CS, E be the set of students who applied in
EE. Then |C| = 175, |E| = 150 and |C ∩ E| = 75. So total applicants in these two programs are
|C ∪ E| = |C| + |E| − |C ∩ E| = 175 + 150 − 75 = 250
So 300 − 250 = 50 students did not apply in any of these two programs.
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