Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Discrete Mathematics
Strong Induction
Principle of Strong Mathematical Induction: Let n be a natural number. We denote by n a statement which is related to n.
0 , 1 , 2 , ..., n , ...
A proof by strong induction encloses
• Basis case: Proof of the first statement 0
• Strong induction hypothesis: ”If 0 , 1 , 2 , . . . , k , then k+1 ”
By the principle of strong induction, it follows that we have
proved:
Statement 0 (basis case).
Statement ”If 0 , then 1 ”.
Statement ”If 0 , 1 , then 2 ”.
Statement ”If 0 , 1 , 2 , then 3 ”.
Statement ”If 0 , 1 , 2 , 3 , then 4 ”.
..
.
Statement ”If 0 , 1 , 2 , 3 , . . . , k then k+1 ”.
Principle of Strong Mathematical Induction for a special
case: For each natural number n the statements 0 , 1 , 2 , . . . ,
n , . . . related to each other as follows:
• Statements 0 , 1 are ”independent”
• The veracity of n depends of the veracity of n-1 and n-2 for
all n ≥ 2
then the proof by strong induction encloses
• Basis case:
(a) Proof of the first statement 0
(b) Proof of the first statement 1
• Strong induction hypothesis: ”If k-1 , k , then k+1 ”
It follows that we have proved:
Statements 0 , 1 (basis case)
Statement ”If 0 , 1 , then 2 ”.
Statement ”If 1 , 2 , then 3 ”.
Statement ”If 2 , 3 , then 4 ”.
..
.
Statement ”If k-1 , k , then k+1 ”.
22.16. (d) Let d0 = 2, d1 = 5 and for n > 1, let
dn = 5dn−1 − 6dn−2 .
Prove: dn = 2n + 3n .
Using lecture notations we have following statements which are
related to n:
0
d0 = 20 + 30 ,
1
d1 = 21 + 31 ,
2
d2 = 22 + 32 ,
3
d3 = 23 + 33 ,
..
.
n
dn = 2n + 3n ,
By the definition of dn , the veracity of n depends on the veracity of n-1 and veracity of n-2 . In other words the proof of the
statement n bases on the veracity of n-1 and n-2 . Moreover the
statements 0 , 1 do not depend on any other statement. Thus for
the proof we have to show
• Basis case:
(a) Proof of the first statement 0
(b) Proof of the first statement 1
• Strong induction hypothesis: ”If k-1 , k , then k+1 ”
2
Proof. Basis case: When n = 0 then dn = d0 = 2 and 2n + 3n =
20 + 30 = 1 + 1 = 2, as required. Furthermore, when n = 1,
dn = d1 = 5 and 2n + 3n = 21 + 31 = 2 + 3 = 5, as required.
Induction hypothesis (strong): Suppose the formula holds
for values k − 1 and k, i.e. it holds
dk−1 = 2k−1 + 3k−1
and dk = 2k + 3k
To prove: dk+1 = 2k+1 + 3k+1 . We calculate,
dk+1 = 5dk − 6dk−1 (by the definition of dk+1 )
= 5(2k + 3k ) − 6(2k−1 + 3k−1 ) (by induction hypothesis )
= 5 · 2k + 5 · 3k − |{z}
3 · 2 ·2k−1 − |{z}
2 · 3 ·3k−1 )
6
= 5 · 2k + 5 · 3k − 3 · 2k − 2 · 3k
= 2 · 2k + 3 · 3k
= 2k+1 + 3k+1
6
as required.
22.16. (e) Let e0 = 1, e1 = 4 and, for n > 1, let
en = 4(en−1 − en−2 ).
Prove: en = 2n (n + 1).
Using lecture notations we have following statements which are
related to n:
0
e0 = 20 (0 + 1),
1
e1 = 21 (1 + 1),
2
e2 = 22 (2 + 1),
3
e3 = 23 (3 + 1),
..
.
n
en = 2n (n + 1),
By the definition of en , the veracity of n depends on the veracity of n-1 and veracity of n-2 . In other words the proof of the
statement n bases on the veracity of n-1 and n-2 . Moreover the
statements 0 , 1 do not depend on any other statement. Thus for
the proof we have to show
3
• Basis case:
(a) Proof of the first statement 0
(b) Proof of the first statement 1
• Strong induction hypothesis: ”If k-1 , k , then k+1 ”
Proof. Basis case: When n = 0 then en = e0 = 1 and 20 (0+1) =
1 · 1 = 1, as required. Furthermore, when n = 1, en = e1 = 4 and
21 (1 + 1) = 2 · 2 = 4, as required.
Induction hypothesis (strong): Suppose the formula holds
for values k − 1 and k, i.e. it holds
ek−1 = 2k−1 (k − 1 + 1) = 2k−1 k, and ek = 2k (k + 1),
To prove: ek+1 = 2k+1 (k + 1 + 1) = 2k+1 (k + 2). We calculate,
ek+1 =
=
=
=
=
=
=
4(ek+1−1 − ek+1−2 ) = 4(ek − ek−1 ) (by the definition of ek+1 )
4(2k (k + 1) − 2k−1 k) (by induction hypothesis )
22 (2k (k + 1) − 2k−1 k)
22 · 2k (k + 1) − 22 · 2k−1 k
2 · 2k+1 (k + 1) − 2k+1 k
2k+1 (2(k + 1) − k)
2k+1 (k + 2)
as required.
4