Number Theory ArielU 2016
Practical Session 2
Practical session 3
§1. PWI, PSI, and WOP
We have met two principles of induction in the lectures to which we referred as weak and strong
inductions. Here are two formulations for weak induction.
Theorem 1.1. (PWI) Let S ⊆ N be a set satisfying:
(W.1) 0 ∈ S; and
(W.2) if k ∈ S then k + 1 ∈ S.
Then S = N.
A proof of Theorem 1.1 was given in the lecture for Z+ . Theorem 1.1 is commonly referred to as the first
principle of finite induction. We will abbreviate this with PWI (representing principle of weak induction).
The more traditional formulation of the PWI is the following.
Theorem 1.2. (Weak mathematical induction) Let S(n) denote a mathematical statement that
depends on n ∈ Z+ .
(Base) If S(1) is true; and
(Step) if whenever S(k) is true then S(k + 1) is true as well,
then S(n) is true for all n ∈ Z+ .
We proceed to strong or complete induction.
Theorem 1.3. (PSI) Let S ⊆ N be a set satisfying:
(S.1) 0 ∈ S; and
(S.2) if {0, 1, . . . , n} ⊆ S then n + 1 ∈ S.
Then S = N.
Theorem 1.3 is commonly referred to as the second principle of finite induction. We will abbreviate
this with PSI (representing principle of strong induction). A more familiar form of this principle is the
following.
Theorem 1.4. (Strong/Complete mathematical induction) Let S(n) denote a mathematical statement that depends on n ∈ Z+ . In addition, let n0 , n1 ∈ Z+ satisfy n0 ≤ n1 .
(Base) If S(n0 ), S(n0 + 1), . . . , S(n1 ) are all true; and
(Step) if whenever S(n0 ), S(n0 + 1), . . . , S(k − 1), S(k) are true then S(k + 1) is true as well,
then S(n) is true for all n ≥ n0 .
Another principle we have met throughout out lectures and sessions so far is the WOP formulated
next.
Theorem 1.5. (WOP) Every non-empty A ⊆ N has a least element.
The PWI, the PSI, and the WOP are related to one another in the following sense.
Theorem 1.6. the PWI, the PSI, and the WOP are all equivalent.
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Number Theory ArielU 2016
Practical Session 2
In the lecture we proved the following implication.
Lemma 1.7. The WOP implies the PWI.
We mentioned in class that the PWI and PSI are equivalent but provided no proof. Here we prove another
piece of the puzzle showing that the PSI implies the WOP.
Lemma 1.8. The PSI implies the WOP.
Proof. Assume towards a contradiction that there exists a set S ⊆ N admitting no least element. We
show that the set T = N \ S, i.e., the complement of S satisfies both (S.1) and (S.2); this would then mean
that T = N, by Theorem 1.3, which in turn implies that S = ∅.
As 0 is the least element in N it follows that 0 ∈
/ S as S is assumed to not have a least element. Then
0 ∈ T verifying (S.1) for T . To verify (S.2) for T we have to show that if {0, 1, . . . , n} ⊆ T then n + 1 ∈ T .
The assumption that {0, 1, . . . , n} ⊆ T means that i ∈
/ S for every i ∈ [0, n]. So if n + 1 ∈ S that would
mean that n + 1 is the least element in S. As S is assumed to not have a least element n + 1 ∈
/ S so that
n + 1 ∈ T . We have thus verified (S.2) for T .
§2. Applying the induction technique
Lemma 2.1.
Pn
i
i=1 2i
≤ 2 for every n ≥ 1.
Pn 1
Before provingPthis inequality let us recall our proof that
i=1 i2 ≤ 2. There we proved a stronger
assertion that ni=1 i12 ≤ 2 − n1 . Based on this experience let us try to prove that
n
X
1
i
≤ 2 − n.
2i
2
(2.2)
i=1
For n = 1 we have that 211 = 12 ≤ 2 − 12 = 32 . We proceed to the induction step. Suppose that the claim
holds for n we shall try to prove that it holds for n + 1.
n+1
X
i=1
n
X i
i
n+1
1
n+1
=
+ n+1 ≤ 2 − n + n+1 .
i
i
2
2
2
2
2
i=1
1
We need to show that 2 − 21n + 2n+1
n+1 ≤ 2 − 2n+1 which is equivalent to showing that
to hold we require that (n + 2)2n ≤ 2n+1 = 2 · 2n which is impossible.
Let us analyse this failure. Let us rewrite (2.2) as follows
n+2
2n+1
≤
1
2n .
For this
n
X
i
≤ 2 − α(n),
2i
i=1
where α(n) is some function of n yet to be determined. How can we determine α(n). From the above
failed proof we notice that we emerge out of the induction step with the requirement
2 − α(n) +
n+1
≤ 2 − α(n + 1).
2n+1
That is we need an α(n) for which we can have
α(n + 1) +
n+1
≤ α(n).
2n+1
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Number Theory ArielU 2016
Practical Session 2
This immediately implies that α(n + 1) must be much smaller than α(n), i.e., α(n) must tend to zero as
n tends to infinity. Our initial choice of α(n) = 2−n satisfies this property. However, we have seen that
1
such an α(n) is too small to follow through as we reached the false inequality 2n+2
n+1 ≤ 2n . The lesson is
that we must increase the numerator of 21n . By how much? So suppose we wish to have α(n) =
we now set out to determine β(n). Consider the induction step again
2−
β(n)
2n
and
β(n) n + 1
β(n + 1)
+ n+1 ≤ 2 −
.
n
2
2
2n+1
That is we need
β(n + 1) + n + 1
β(n)
2β(n)
≤ n = n+1 .
n+1
2
2
2
This in turn means that we require
β(n + 1) + n + 1 ≤ 2β(n).
We see that that β(n) = n + 2 does the job as with such a choice we arrive at
β(n + 1) + n + 1 = 2n + 3 ≤ 2β(n) = 2n + 4.
We now prove Lemma 2.1.
Proof of Lemma 2.1. We prove that
n
X
n+2
i
≤ 2 − n , for every n ≥ 1.
i
2
2
(2.3)
i=1
For n = 1 we have that
holds for n + 1.
n+1
X
i=1
1
2
≤ 2−
3
2
= 21 . Suppose that the claim holds for n we shall try to prove that it
n
X i
i
n+1
n+2 n+1
2n + 4 + n + 1
3n + 5
n+3
=
+ n+1 ≤ 2 − n + n+1 = 2 −
= 2 − n+1 ≤ 2 − n+1 .
2i
2i
2
2
2
2n+1
2
2
i=1
Lemma 2.4. Let
a1 = 11,
a2 = 21,
an = 3an−1 − 2an−2 , n ≥ 3.
Then an = 5 · 2n + 1 for every n ≥ 1.
Proof. The proof is by induction on n. For the induction basis we consider n = 1, 2. Here we note that
a1 = 5 · 2 + 1 = 11,
a2 = 5 · 22 + 1 = 21.
We proceed to the induction hypothesis. Assume that the claim holds for all positive integers up to k.
That is, ai = 5 · 2i + 1 for all i ∈ [1, k]. We prove the same equality for k + 1.
ak+1 = 3 · ak − 2 · ak−1
= 3(5 · 2k + 1) − 2(5 · 2k−1 + 1)
= 10 · 2k + 1
= 5 · 2 · 2k + 1
= 5 · 2k+1 + 1
Exercise 2.5. Prove by induction the following equalities:
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Number Theory ArielU 2016
n(n+1)
,
2
1.
Pn
=
2.
P2n−1
1
i(i+1)
3.
Pn
4.
Pn
i=1 i
i=n
3
i=1 i =
i=0 ar
i
=
=
Practical Session 2
for every n ∈ Z+ .
1
2n ,
n(n+1)
2
for every n ∈ Z+ .
2
, for every n ∈ Z+ .
a(rn+1 −1)
,
r−1
for every n ∈ N, a, r ∈ R, r 6= 1.
Proof.
1. The proof is by induction on n. For the induction basis we consider n = 1. Here we have that 1 = 1·2
2
holds.
We
proceed
to
the
induction
step.
Assume
the
claim
holds
for
n
≥
1,
that
is
we
assume
that
Pn
n(n+1)
. We prove that the claim holds for n + 1.
i=1 i =
2
n+1
X
i=
i=1
n
X
i+n+1=
i=1
n2 + 3n + 2
(n + 1)(n + 2)
n(n + 1)
+n+1=
=
2
2
2
2. The proof is by induction on n. For the induction basis we consider n = 1. Here we have that
1
1
We proceed to the induction step. Assume the claim holds for n ≥ 1, that is we
1·2 = 2 holds.
P2n−1 1
1
assume that i=n i(i+1) = 2n
. We prove that the claim holds for n + 1.
2n+1
X
i=n+1
2n−1
X
1
1
1
1
1
=
−
+
+
=
i(i + 1)
i(i + 1) n(n + 1) 2n(2n + 1) (2n + 1)(2n + 2)
i=n
1
2n + 1
(2n + 1)(n + 1) − 2n − 1
2n2 + n
1
−
=
=
=
.
2n 2n(2n + 1)(n + 1)
2n(n + 1)(2n + 1)
2n(n + 1)(2n + 1)
2(n + 1)
3. The proof is by induction on n. For the induction basis we consider n = 1. Here we have that
2
13 = 1·2
holds. We proceed to the induction step. Assume the claim holds for n ≥ 1, that is we
2
2
P
assume that ni=1 i3 = n(n+1)
. We prove that the claim holds for n + 1.
2
n+1
X
n
X
n(n + 1) 2
+ (n + 1)3 =
i =
i + (n + 1) =
2
i=1
i=n
2
2
(n + 1)2 (n + 2)2
2 n
2 n + 4n + 4
(n + 1)
+ n + 1 = (n + 1)
=
.
4
4
4
3
3
3
4. The proof is by induction on n. For the induction basis we consider n = 0. Here we have that
1 −1)
ar0 = a(rr−1
holds. We proceed to the induction step. Assume the claim holds for n ≥ 0, that is
n+1
P
we assume that ni=0 ari = a(r r−1−1) . We prove that the claim holds for n + 1.
n+1
X
i=0
ari =
n
X
i=0
ari + arn+1 =
a(rn+1 − 1)
a(rn+1 − 1 + rn+1 (r − 1))
a(rn+2 − 1)
+ arn+1 =
=
.
r−1
r−1
r−1
Exercise 2.6. Prove by induction the following inequalities:
1. (1 + a)n ≥ 1 + na, for every n ∈ Z+ , a > −1.
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Number Theory ArielU 2016
Practical Session 2
2. 2 · 5n ≤ 4n + 6n , for every n ∈ N.
Proof.
1. The proof is by induction on n. For the induction basis we consider n = 1. Here we have that
(1 + a)1 ≥ 1 + a holds. We proceed to the induction step. Assume the claim holds for n ≥ 1, that
is we assume that (1 + a)n ≥ 1 + na. We prove that the claim holds for n + 1.
(1 + a)n+1 = (1 + a)(1 + a)n ≥ (1 + a)(1 + na) = 1 + (n + 1)a + na2 ≥ 1 + (n + 1)a.
2. The proof is by induction on n. For the induction basis we consider n = 0. Here we have that
2 · 50 ≤ 40 + 60 holds. We proceed to the induction step. Assume the claim holds for n ≥ 0, that is
we assume that 2 · 5n ≤ 4n + 6n . We prove that the claim holds for n + 1.
2 · 5n+1 = 5 · (2 · 5n ) ≤ 5 · (4n + 6n ) =
4 · 4n + 4n + 6 · 6n − 6n = 4n+1 + 6n+1 + 4n − 6n ≤ 4n+1 + 6n+1 .
Exercise 2.7. Prove by induction the following divisibility statements:
1. 6 | (2n3 + 3n2 + n), for every n ∈ N.
2. 6 | (n3 + 5n), for every n ∈ N.
3. 16 | n4 + 4n2 + 11, for every odd n ∈ Z+ .
Proof.
1. The proof is by induction on n. For the induction basis we consider n = 0. Here we have that 6 | 0
holds. We proceed to the induction step. Assume the claim holds for n ≥ 0, that is we assume that
6 | (2n3 + 3n2 + n). We prove that the claim holds for n + 1.
2(n + 1)3 + 3(n + 1)2 + (n + 1) =
2n3 + 6n2 + 6n + 2 + 3n2 + 6n + 3 + n + 1 =
2n3 + 3n2 + n + 6(n2 + 2n + 1).
By the induction assumption 6 | 2n3 + 3n2 + n, and since 6 | 6(n2 + 2n + 1) it also holds that
6 | 2(n + 1)3 + 3(n + 1)2 + (n + 1).
2. The proof is by induction on n. For the induction basis we consider n = 0. Here we have that 6 | 0
holds. We proceed to the induction step. Assume the claim holds for n ≥ 0, that is we assume that
6 | (n3 + 5n). We prove that the claim holds for n + 1.
(n + 1)3 + 5(n + 1) =
n3 + 3n2 + 3n + 1 + 5n + 5 = n3 + 5n + 3n2 + 3n + 6 =
n3 + 5n + 3(n2 + n + 2) = n3 + 5n + 3(2 + n(n + 1)).
By the induction assumption 6 | n3 + 5n. 2 | n(n + 1), therefore 2 | (2 + n(n + 1)), hence, by
Proposition 2.12 of the lecture notes 6 | 3(2 + n(n + 1)). We conclude that 6 | (n + 1)3 + 5(n + 1).
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Number Theory ArielU 2016
Practical Session 2
3. The proof is by induction on n. For the induction basis we consider n = 1. Here we have that
16 | 1 + 4 + 11 holds. We proceed to the induction step. Assume the claim holds for n ≥ 1, that is
we assume that 16 | n4 + 4n2 + 11. We prove that the claim holds for next odd integer i.e. n + 2.
Since n is odd there exists m ∈ Z such that n = 2m − 1. Therefore:
(n + 2)4 + 4(n + 2)2 + 11 =
(2m + 1)4 + 4(2m + 1)2 + 11 = (2m − 1 + 2)4 + 4(2m − 1 + 2)2 + 11 =
(2m − 1)4 + 8(2m − 1)3 + 24(2m − 1)2 + 32(2m − 1) + 16 + 4(2m − 1)2 + 16(2m − 1) + 16 + 11 =
(2m − 1)4 + 4(2m − 1)2 + 11 + 8(2m − 1)3 + 24(2m − 1)2 + 48(2m − 1) + 32 =
(2m − 1)4 + 4(2m − 1)2 + 11 + (2m − 1)2 (8(2m − 1) + 24) + 48(2m − 1) + 32 =
(2m − 1)4 + 4(2m − 1)2 + 11 + 16(2m − 1)2 (m + 1) + 48(2m − 1) + 32.
By the induction assumption 16 | (2m − 1)4 + 4(2m − 1)2 + 11 and since 16 | 16(2m − 1)2 (m + 1) +
48(2m − 1) + 32, we conclude that 16 | (n + 2)4 + 4(n + 2)2 + 11.
Exercise 2.8.
1. Define the sequence a1 = a2 = 2, an+2 = an + 1, for every n ∈ Z+ . Prove that an = 12 (n + 1) +
1
n
+
4 (1 + (−1) ), for every n ∈ Z .
2. Define the sequence b0 = 1, b1 = 2, b2 = 3, bn+3 = bn+2 + bn+1 + bn , for every n ∈ Z+ . Prove that
bn ≤ 2n , for every n ∈ N.
3. Define the Fibonacci sequence f0 = 0, f1 = 1, fn = fn−1 + fn−2 , for every n ∈ Z+ , n ≥ 2. Prove
n−2
that fn ≥ 32
for every n ∈ Z+ .
Proof.
1. The proof is by induction on n. For the induction basis we consider n = 1, 2. Here we have that
a1 = 2 = 21 (1 + 1) + 41 (1 − 1) and a2 = 2 = 12 (2 + 1) + 14 (1 + 1) holds. We proceed to the
induction step. Assume the claim
holds for every integer k ≤ n, for n ≥ 1, that is we assume that
1
1
k
ak = 2 (k + 1) + 4 1 + (−1) . We prove that the claim holds for n + 1.
1
1
an+2 = an + 1 = (n + 1) + (1 + (−1)n ) + 1 =
2
4
1
1
1
1
(n + 3 − 2) +
1 + (−1)n+2 + 1 = (n + 3) +
1 + (−1)n+2 .
2
4
2
4
2. The proof is by induction on n. For the induction basis we consider n = 0, 1, 2. Here we have that
b0 = 1 ≤ 20 , b1 = 2 ≤ 21 and b2 = 3 ≤ 22 holds. We proceed to the induction step. Assume the
claim holds for every integer k ≤ n, for n ≥ 1, that is we assume that bk ≤ 2k . We prove that the
claim holds for n + 1.
bn+1 = bn + bn−1 + bn−2 ≤ 2n + 2n−1 + 2n−2 = 4 · 2n−2 + 2 · 2n−2 + 2n−2 = 2n+1 .
3. The proof is by induction on n. For the induction basis we consider n = 1, 2. Here we have that
f1 = f2 = 1 ≥ 32 holds. We proceed to the induction step. Assume the claim holds for every integer
k−2
k ≤ n, for n ≥ 1, that is we assume that fk ≥ 32
. We prove that the claim holds for n + 1.
n−3 n−4
n−2
n−2
3
3
2 3
4 3 n−2 10 3 n−2
3
fn = fn−1 + fn−2 ≥
+
=
+
=
≥
.
2
2
3 2
9 2
9 2
2
6