Proof by mathematical induction using a strong hypothesis
Occasionally a proof by mathematical induction is made easier by using a strong hypothesis:
To show P(n) [a statement form that depends on variable n], do:
(1) BASIS: Establish [P(m)vP(m+1)v...vP(m+r)] is true for some initial value of m
(2) IMPLICATION: Prove [P(m)vP(m+1)v...vP(m+r)v...vP(k)] Y P(k+1) for any k $ m+r
Then by strong induction, we know P(n) is true for every n $ m.
Note that r = 0 gives the simple principle of induction. Also it can be shown that the principle of strong
induction follows from simple induction.
Example:
Consider a sequence defined by a0 = 1, a1 = 2, a2 = 3; and an = a(n–2) + 2a(n–3) for n $ 3.
We compute the next five values as follows:
a3 = 2 + 2(1) = 4, a4 = 3 + 2(2) = 7, a5 = 4 + 2(3) = 10, a6 = 7 + 2(4) = 15, a7 = 10 + 2(7) = 24.
It appears that, after the first term, the values are increasing at a rate of about 150%. We may establish by
induction that P(n) : an > (3/2)n for n $ 1.
Because the definition of the general term requires knowledge not just of the term immediately preceding, but of
terms two and three preceding, it is useful to use a strong hypothesis for induction.
BASIS: a1 = 2 > 1.5 = (3/2) so P(1) is true.
a2 = 3 > 2.25 = (3/2)2 so P(2)is true
a3 = 4 > 3.375 = (3/2)3 so P(3) is true.
INDUCTION: Given P(j) is true for 1 # j # k where k $ 3; that is, aj > (3/2)j, we establish the desired result by a
sequence of substitutions and inequalities.
then a(k+1) = a(k–1) + 2(a(k–2)) and by the strong hypothesis this must be,
> (3/2)(k–1) + 2((3/2)(k–2)) and factoring out (3/2)(k–2) this
= (3/2)(k–2)[(3/2) + 2] which
= (3/2)(k–2)[3.5] which
> (3/2)(k–2)[3.375] which
= (3/2)(k–2)[27/8] which
= (3/2)(k+1) and the result is established.
Hence by strong induction P(n) is true for all n $ 1.
Another example:
Consider the sequence defined by a0 = 1, a1 = 2; and an =
We compute the next three values as follows:
a2 = 22/1= 4, a3 = 42/2 = 8, a4 = 82/4 = 16.
for n $ 2.
It appears that each term is a power of 2. We may establish by induction that P(n) : an = 2n for n $ 0.
BASIS: a0 = 1 = 20, and a1 = 2 = 21 so P(0) and P(1) are both true.
INDUCTION: Given P(j) is true for 0 # j # k where k $ 1; that is, aj = 2j, then
a(k+1) =
and the result is established.
Hence by strong induction P(n) is true for all n $ 0.
MAT251 Exercises 4.6, p170. 1 – 14
1. Prove 3 + 11 + ... + (8n – 5) = 4n2 – n for n 0 P; that is, p(n):
for n 0 P.
Only First Principle is needed: (B): (8*1 – 5) = 3 = 4 – 1 = 4*12 – 1 so p(1) is true.
(I): Given p(k – 1); that is,
then
which
equals 4k2 – 8k + 4 – k + 1 + 8k – 5 = 4k2 – k, so p(k – 1) implies p(k).
By (B) & (I), p(n) is true for all n in P.
2. For n 0 P, prove:
a)
; that is, p(n):
Only First Principle is needed: (B): 1(1+1) = 2 =
so p(1) is true.
(I): Given p(k – 1); that is,
, then
which =
;so p(k – 1) implies
p(k). By (B) & (I), p(n) is true for all n in P.
b)
Only First Principle is needed: (B):
(I): Given p(k – 1); that is,
; that is, p(n):
so p(1) is true.
, then
, which equals
; so p(k–1) implies p(k). By (B) & (I), p(n) is true for all n in
P.
3. Prove p(n): n5 – n is divisible by 10 for all n 0 P. [In Example 1, starting on page 137, the author relied on
two lemmas: “n5 – n is divisible by 2.” and “n5 – n is divisible by 5.”]
An alternate proof proceeds as follows: (B) 15 – 1 = 0 = 10*0, so p(1) is true.
(I): If k5 – k = 10q, then (k+1)5 – (k+1) = k5 + 5k4 + 10k3 + 10k2 + 5k + 1 – k – 1
which equals [k5 – k] + 5k4 + 10k3 + 10k2 + 5k which equals [10q] + 10k3 + 10k2 + 5k4 + 5k
which equals 10[q + 10k3 + 10k2] + 5k(k3 + 1) which equals 10[q + 10k3 + 10k2] + 5k(k + 1)(k2 – k + 1)
which is a multiple of 10 because either k or k+1 is even. Hence p(k) implies p(k+1). (B) & (I) Y p(n).
4. a) Define a sequence (bn) by: b0 = b1 = 1; bn = 2bn-1 + bn-2 for n $ 2.
b6 = 2b5 + b4 = 2[2b4 + b3] + b4 = 5b4 + 2b3 = 5[2b3 + b2] + 2b3 = 12b3 + 5b2 = 12[2b2 + b1] + 5b2 = 29b2 +
12b1 = 29[2b1 + b0] + 12b1 = 70b1 + 29b0 = 70 + 29 = 99.
b) Define a sequence (an) by a0 = a1 = a2 = 1; an = an-2 + an-3 for n $ 3. a9 = a7 + a6 = [a5 + a4] + a6 = [a4 + a3]
+ a5 + a4 = [a3 + a2] + 2a4 + a3 = 2[a2 + a1] + 2a3 + a2 = 2[a1 + a0] + 3a2 + 2a1 = 9.
5. Is the First Principle of Mathematical Induction adequate to prove the fact in Exercise 11(b) on page 234 [that
all the terms are odd in the sequence (an): defined by: a0 = a1 = 1; an = an-1 + 2an-2 for n $2]?
Yes, because only the oddness of an-1 matters in establishing the oddness of an, as 2an-2 is always even.
6. Recursively define a0 = 1, a1 = 2;
a) Calculate the first few terms of the sequence: 1, 2,
b) Guess a general formula for an. It appears that p(n): an = 2n for n $ 0.
c) Prove the guess in part b). We use the Second Principle of Mathematical Induction:
(B) 20 = 1 = a0 and 21 = 2 = a1 so Basis is established.
(I) If ar = 2r for all r such that 0 # r < k, then
for k $ 2.
(B) & (I) implies p(n) for n $ 0.
7. Recursively define a0 = a1 = 1;
a) Calculate the first few terms of the sequence:1, 1,
b) Guess a general formula for an. It appears that p(n): an = 1 for n $ 0.
c) Prove the guess in part b). We use the Second Principle of Mathematical Induction:
(B) is clearly established by definition.
(I) If ar = 1 for all r such that 0 # r < k, then
for k $ 2.
(B) & (I) implies p(n) for n $ 0.
8. Recursively define a0 = 1, a1 = 2;
for n $ 2.
a) Calculate the first few terms of the sequence. 1, 2,
b) Using part (a), guess a general formula for an. It seems an = n + 1 for n 0 ù.
c) Use the second principle of induction to prove p(n): an = n + 1 for n 0 ù.
(B): Need to check first two cases: 0 + 1 = 1 = a0 and 1 + 1 = 2 = a1, so basis is established.
(I): If ar = r + 1 for all r such that 0 # r < k, then
for k $2.
(B) & (I) implies p(n) for n $ 0.
9. Recursively define a0 = 0, a1 = 1; an = ¼(an-1 – an-2 + 3)2 for n $ 2.
a) Calculate the first few terms of the sequence. 0, 1, ¼(1 – 0 + 3)2 = 4, ¼(4 – 1 + 3)2 = 9, ...
b) Using part (a), guess a general formula for an. It seems an = n2 for n 0 ù.
c) Use the second principle of induction to prove p(n): an = n2 for n 0 ù.
(B): Need to check first two cases: 02 = 0 = a0 and 12 = 1 = a1, so basis is established.
(I): If ar = r2 for all r such that 0 # r < k, then ak = ¼(ak-1 – ak-2 + 3)2 = ¼[(k-1)2 – (k-2)2 + 3]2
which = ¼[k2 – 2k + 1 – k2 + 4k – 4 + 3]2 = ¼[2k]2 = k2 for k $ 2. That is, ak = k2 for k $ 2.
(B) & (I) implies p(n) for n $ 0.
10. Recursively define a0 = 1, a1 = 2, a2 = 3; an = an-2 + 2an-3 for n $ 3.
a) Calculate an for n = 3, 4, 5, 6, 7. a3 = 2 + 2(1) = 4, a4 = 3 + 2(2) = 7, a5 = 4 + 2(3) = 10,
a6 = 7 + 2(4) = 15, a7 = 10 + 2(7) = 24.
b) Prove that p(n): an >
for all n $ 1.
(B): Need to check 3 cases: a1 = 2 >
(I): If ar >
and a2 = 3 >
and a3 = 4 >
.
for all r such that 1 # r < k, where k $ 4, then
which equals
for k$4.
(B) & (I) implies p(n) for n $ 1.
11. Recursively define a0 = a1 = a2 = 1; an = an-1 + an-2 + an-3 for n $ 3.
a) Calculate the first few terms of the sequence: 1, 1, 1, 1+1+1=3, 3+1+1=5, 5+3+1=9, 9+5+3=17, ...
b) Prove that all the an’s are odd. [By second principle of induction]
(B): first three terms are odd;
(I): If ar is odd for all r such that 0 # r < k, where k $ 3, then ak is the sum of three odd numbers and is also
odd.
c) Prove that p(n): an #2n-1 for all n $1. [By second principle of induction]
(B): Need to check 3 cases: a1 = 1 # 20 = 21-1 , a2 = 1 # 21 = 22-1 , a3 = 3 # 22 = 23-1 ;
(I): If ar #2r-1 for all r such that 1 # r < k, where k $ 4, then ak = ak-1 + ak-2 + ak-3
which is #2k-1-1 + 2k-2-1 + 2k-3-1 = 2k-4 [22 + 2 + 1] = 2k-4 [7] < 2k-4 [23] = 2k-1 . So ak # 2k-1 for k$4.
(B) & (I) implies p(n) for n $ 1.
12. Recursively define a0 = 1, a1 = 3, a2 = 5; an = 3an-2 + 2an-3 for n $ 3.
a) Calculate an for n = 3, 4, 5, 6, 7. a3 = 3(3) + 2(1) = 11, a4 = 3(5) + 2(3) = 21,
a5 = 3(11) + 2(5) = 43, a6 = 3(21) + 2(11) = 85, a7 = 3(43) + 2(21) = 171.
b) Prove that p(n): an > 2n for n $ 1. [By second principle of induction]
(B): We need to check three cases: a1 = 3 > 2 = 21, a2 = 5 > 4 = 22, a3 = 11 > 8 = 23;
(I): If ar > 2r for all r such that 1 # r < k, where k $ 4, then ak = 3ak-2 + 2ak-3 > 3(2k-2 ) + 2(2k-3 )
which = 2k-3 [3(2) + 2] = 2k-3 [8] = 2k. So ak > 2k.
(B) & (I) implies p(n) for n $ 1.
c) Prove that p(n): an < 2n+1 for n $1. [By second principle of induction]
(B): We need to check three cases: a1 = 3 < 4 = 21+1, a2 = 5 < 8 = 22+1, a3 = 11 < 16 = 23+1;
(I): If ar < 2r+1 for all r such that 1 # r < k, where k $ 4, then ak = 3ak-2 + 2ak-3
which is < 3(2k-2+1) + 2(2k-3+1) = 2k-2 [3(2) + 2] = 2k-2 [8] = 2k+1. So ak < 2k+1.
(B) & (I) implies p(n) for n $ 1.
d) Prove that p(n): an = 2an-1 + (-1)n+1 for n $ 1. [By second principle of induction]
(B): We need to check three cases: a1 = 3 = 2(1) + (1) = 2(a0) + (-1)1+1,
a2 = 5 = 2(3) + (-1) = 2(a1) + (-1)2+1, a3 = 11 = 2(5) + (1) = 2(a2) + (-1)3+1.
(I): If ar = 2ar-1 + (-1)r+1 for all r such that 1 # r < k, where k $ 4, then ak = 3ak-2 + 2ak-3
which = 3[2ak-2-1 + (-1)k-2+1] + 2[2ak-3-1 + (-1)k-3+1] = 2[3ak-3 + 2ak-4 ] + (-1)k-2 [3(-1) + 2]
which = 2[ak-1 ] + (-1)k-2 [-1] = 2[ak-1 ] + (-1)k-1 = 2[ak-1 ] + (-1)k+1. So ak = 2ak-1 + (-1)k+1.
(B) & (I) implies p(n) for n $ 1.
13. Recursively define b0 = b1 = b2 = 1; bn = bn-1 + bn-3 for n $ 3.
a) Calculate bn for n = 3, 4, 5, 6. b3 = 1 + 1 = 2, b4 = 2 + 1 = 3, b5 = 3 + 1 = 4, b6 = 4 + 2 = 6.
b) Show that bn $2bn-2 for n $ 3. [By second principle of induction]
(B): We need to check three cases: b3 = 2 $ 2(1) = 2b1, b4 = 3 $ 2(1) = 2b2, b5 = 4 $ 2(2) = 2b3;
(I): If br $2br-2 for all r such that 3 # r < k, where k $6, then bk = bk-1 + bk-3 $2bk-1-2 + 2bk-3-2
which = 2[bk-3 + bk-5 ] = 2bk-2 . So bk $2bk-2 for k $6. (B) & (I) implies p(n) for n $ 3.
c) Prove the inequality bn $
for n $ 2. [By second principle of induction]
(B): We check three cases: b2 = 1 $1 =
(I): If br $
, b4 = 3 $2 =
for all r such that 2 # r < k, where k $5, then bk = bk-1 + bk-3 which
is $
So bk $
, b3 = 2 $
=
=
.
for k $5.
(B) & (I) implies p(n) for n $ 2.
14. Recursively define b0 = b1 = b2 = 1; bn = bn-1 + bn-3 for n $3. Show that
[By second principle of induction]
(B): Check three cases: b1 = 1 #
, b3 = 2 #
for all r such that 1 # r < k, where k $4, then bk = bk-1 + bk-3 which
(I): If
is #
, b2 = 1 #
=
(B) & (I) implies p(n) for n $ 1.
. So
for k $4.
for n$1.