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MATH 135 SOS Review Package

Induction

Mathematical induction is a proof technique used to prove statements that depend on a natural number n,i.e. statements of the form ∀n ∈ P, P (n) where P (n) is some statement about n. We review the statementsof the Principle of Mathematical Induction and the Principle of Strong Induction below.

Theorem (Principle of Mathematical Induction). Let P (n) be a statement about n ∈ P. If1. P (1) is true2. and if for each k ∈ P, P (k) is true implies P (k + 1) is truethen P (n) is true for all n ∈ P.

Theorem (Principle of Strong Induction). Let P (n) be a statement about n ∈ P. If1. P (1) is true2. and if for each k ∈ P, P (1), . . . , P (k) are true implies P (k + 1) is truethen P (n) is true for all n ∈ P.

The main difference between induction and strong induction is in the second hypothesis. In induction, weassume only that P (k) is true and we use this to prove P (k + 1) is true, whereas in strong induction, weassume P (1), . . . , P (k) are true and we use this to prove P (k + 1).

We separate a proof by induction into three steps. First we verify P (1) or some other base case(s). Thenwe make an induction hypothesis. If we are using induction, we assume P (k) is true. If we are using stronginduction, we assume P (1), · · · , P (k) are true. Finally, in the induction conclusion, we use our inductionhypothesis to prove P (k + 1).

We give three examples to illustrate typical uses of induction.

Example. Prove that for each n ≥ 2, n ∈ P

n∏r=2

(1 +

1

r2 − 1

)=

2n

n+ 1.

Solution. Let P (n) be the statement∏nr=2

(1 + 1

r2−1

)= 2n

n+1 . We prove the result by induction on n.

Base Case: We verify P (2) is true. P (2) is the statement∏2r=2

(1 + 1

r2−1

)= 4

3 . We have

2∏r=2

(1 +

1

r2 − 1

)= 1 +

1

3=

4

3

which verifies P (2).

Induction Hypothesis: Assume P (k) is true for some k ≥ 2, k ∈ P. That is

k∏r=2

(1 +

1

r2 − 1

)=

2k

k + 1.

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Induction Conclusion: We verify P (k + 1) is true. P (k + 1) is the statement

k+1∏r=2

(1 +

1

r2 − 1

)=

2(k + 1)

(k + 1) + 1.

We have

k+1∏r=2

(1 +

1

r2 − 1

)=

k∏r=2

(1 +

1

r2 − 1

)·(

1 +1

(k + 1)2 − 1

)=

2k

k + 1·(

1 +1

(k + 1)2 − 1

)using the induction hypothesis

=2k

k + 1· k

2 + 2k + 1

k(k + 2)

=2k

k + 1· (k + 1)2

k(k + 2)

=2(k + 1)

(k + 1) + 1

which verifies P (k + 1). Therefore the result follows by the Principle of Mathematical Induction.

Example. A sequence {xn} is defined by

x1 = 0

x2 = 30

xn = xn−1 + 6xn−2 for n ≥ 3

Prove that for each n ∈ P xn = 2 · 3n + 3 · (−2)n.

Solution. Let P (n) be the statement xn = 2 · 3n + 3 · (−2)n. We prove the result by strong induction on n.Base Case: We verify P (1) and P (2). P (1) is the statement x1 = 2 · 31 + 3 · (−2)1. We have x1 = 0 and2 · 31 + 3 · (−2)1 = 0 so P (1) is true. P (2) is the statement x2 = 2 · 32 + 3 · (−2)2. We have x2 = 30 and2 · 32 + 3 · (−2)2 = 18 + 12 = 30 so P (2) is true.Induction Hypothesis: Assume P (1), · · ·P (k) are true for some k ≥ 2, k ∈ P. That is xi = 2 · 3i + 3 · (−2)i

for 1 ≤ i ≤ k.Induction Conclusion: We verify P (k+ 1) is true. P (k+ 1) is the statement xk+1 = 2 · 3k+1 + 3 · (−2)k+1.We have

xk+1 = xk + 6xk−1 by the recursive definition of {xn}= (2 · 3k + 3 · (−2)k) + 6(2 · 3k−1 + 3 · (−2)k−1) by the induction hypothesis

= 2 · 3k + 3 · (−2)k + 4 · 3k + 32 · 2 · (−2)k−1

= (2 + 4) · 3k + (−6 + 18)(−2)k−1

= 2 · 3k+1 + 3 · (−2)2 · (−2)k−1

= 2 · 3k+1 + 3 · (−2)k+1

which verifies P (k + 1). Therefore the result follows by the Principle of Strong Induction.

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Example. Prove that a set with n elements has 2n subsets.

Proof. Solution Let P (n) be the statement: “A set with n elements has 2n subsets.” We prove the resultby induction.Base Case: We verify P (1). P (1) is the statement “A set with 1 element has 21 subsets.” Let S be a setwith 1 element. The subsets of S are ∅ and S. Thus S has 2 = 21 subsets which verifies P (1).Induction Hypothesis: Suppose P (k) is true for some k ∈ P. That is suppose a set with k elements has 2k

subsets.Induction Conclusion: We verify P (k + 1). P (k + 1) is the statement “A set with (k + 1) elements has2k+1 subsets.” Let S be a set with (k + 1) elements and let x ∈ S. Then S − {x} is a set with k elementsso by induction hypothesis it has 2k subsets. Let T be a subset of S. Either x ∈ T or x /∈ T . If x ∈ T thenT − {x} is a subset of S − {x}. So we get 2k such subsets T by the induction hypothesis. If x /∈ S, thenT is a subset of S − {x}. So we get 2k such subsets T by the induction hypothesis. Thus in total we get2k + 2k = 2 · 2k = 2k+1 subsets.

Exercise. Let S be a non-empty subset of P. Show that S has a least element. (that is, show that thereexists x ∈ S such that for all y ∈ S we have x ≤ y.)

Solution. We prove the result by contradiction. Suppose there exists a non-empty subset S of P whichdoes not have a least element. We consider the set T defined by

T = P \ S = {x ∈ P | x /∈ S}.

We show that T = P by strong induction. Let P (n) be the statement n ∈ T .Base Case: We verify P (1). P (1) is the statement 1 ∈ T . Suppose for a contradiction that 1 /∈ T . Then1 ∈ S. But then since for each x ∈ P we have 1 ≤ x so we have found a least element for S. This is acontradiction and so we must have 1 ∈ T .Induction Hypothesis: Suppose that P (1), . . . , P (k) are true for some k ∈ P.Induction Conclusion. We verify P (k + 1) which is the statement k + 1 ∈ T . Suppose for a contradictionthat k + 1 /∈ T . Then k + 1 ∈ S. But since 1, 2, · · · , k /∈ S then k + 1 is a least element for S which is acontradiction. Thus we must have k + 1 ∈ T .Therefore by strong induction we have n ∈ T for each n ∈ P, i.e. T = P. This implies S = ∅ which is acontradiction. Thus S must have a least element.

Divisibility (Again...)

We quickly review the material covered before the midterm.

Definition. Let a and b be integers. We say a divides b and we write a | b if there exists an integer ksuch that b = ka.

Definition. Let a and b be integers, both not zero and let d be a positive integer. We say that d is thegreatest common divisor of a and b and we write gcd(a, b) = d if:1. d | a and d | b2. if c is an integer such that c | a and c | b, then c ≤ d.

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We now recall several propositions and theorems covered before the midterm.

1. Transitivity of Divisibility. If a | b and b | c, then a | c.

2. Divisibility of Integer Combinations. If a | b and a | c, then a | bx+ cy for any x, y ∈ Z.

3. Bounds by Divisibility. a | b and b 6= 0, then |a| ≤ |b|.

4. GCD with Remainder. Let a and b be integers, both not zero. If a = qb + r for some integers qand r, then gcd(a, b) = gcd(b, r).

5. GCD Certificate Theorem. Let d be a positive integer and let a and b be integers. If d | a, d | b,and there exist integers x and y such that ax+ by = d, then d = gcd(a, b).

6. Extended Euclidean Algorithm. Let a and b be integers and let d be a positive integer. Ifd = gcd(a, b), then there exist integers x and y such that ax+ by = d.

7. Linear Diophantine Equation Theorem 1. The linear Diophantine equation

ax+ by = c

has solution if and only if gcd(a, b) | c.

8. Linear Diophantine Equation Theorem 2. If gcd(a, b) = d 6= 0 and x = x0, y = y0 is a particularsolution , then the complete integer solution is

x = x0 + nb

d, y = y0 − n

a

d

for all n ∈ Z.

We give two examples to review some of the ideas.

Example (Practice Problem 5). Let a, b, c ∈ Z. Prove that if a and c are coprime, then gcd(ab, c) =gcd(b, c).

Solution. Let d = gcd(b, c). We show d = gcd(ab, c). Since gcd(a, c) = 1, then there exists x, y ∈ Z suchthat ax+cy = 1 by the Extended Euclidean Algorithm. Multiplying this by ab we get b(a2x)+c(abx) = ab.Since d | b and d | c we get d | b(a2x) + c(aby) = ab by Divisibility of Integer Combinations. By definitionwe have d | c. Thus d is common divisor of ab and c. Suppose e | ab and e | c. We multiply ax+ cy = 1 byb to obtain ab(x) + c(by) = b. Since e | ab and e | c then e | ab(x) + c(by) = b. Thus e is a common divisorof b and c. Therefore we conclude e ≤ d as gcd(b, c). This shows d = gcd(ab, c).

Example. Let k ∈ P. Show that that there exists k consecutive composite integers.

Solution. Let n = k + 1 and consider the k consecutive integers n! + 2, n! + 3, · · · , n! + (k + 1). For each2 ≤ i ≤ k + 1 = n, we know that i | n! because i ≤ n and i | i so i | n! + i by Divisibility of IntegerCominbations. Since i 6= 1 and i 6= n! + i this shows n! + i is composite. Thus we have found k consecutivecomposite integers.

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Another way to think about divisibility

We saw another way to think about divisibility in terms of prime factorizations. First, we recall that werecall that we can always factor a positive integer into primes.

Theorem. Let x ∈ P. Then x can be written as a product of primes which is unique up to re-ordering.That is, we can write x = pr11 · · · p

rkk where p1, · · · , pk are distinct primes and r1, · · · , rk ∈ P. Moreover if

x = pr11 · · · prkk = qs11 · · · q

sll

are two prime factorizations, then k = l and after re-ordering pi = qi for each 1 ≤ i ≤ k.

The next proposition connects divisibility and prime factorizations.

Proposition. If x = pr11 · · · prkk is a prime factorization of x, then d is a positive divisor of x if and only

d = ps11 · · · pskk where for each 1 ≤ i ≤ k we have 0 ≤ si ≤ ri.

We also get a formula for the greatest common divisor of x and y as follows: suppose x = pr11 · · · prkk and

y = ps11 · · · pskk where p1, · · · , pk are distinct primes and r1, · · · , rk, s1, · · · , sk ≥ 0. Then

gcd(x, y) = pe11 · · · pekk

where ei = min(ri, si).

Example. How many positive divisors does 6696 have?

Solution. We have 6696 = 23 · 33 · 31. By our proposition, d is a positive divisor of 6696 if and only ifd = 2d1 · 3d2 · 31d3 where 0 ≤ d1 ≤ 3, 0 ≤ d2 ≤ 3, and 0 ≤ d3 ≤ 1. Thus there are 4× 4× 2 = 32 positivedivisors.

Example. Find gcd(12!, 2113457).

Solution. It is enough to just count the number of factors of 2,3, and 5 in

12! = 12× 11× 10× 9× 8× 7× 6× 5× 4× 3× 2× 1.

We get 10 factors of 2, 5 factors of 3, and 2 factors of 5. Thus gcd(12!, 2113457) = 2103452.

Congruences

Definition. Let m be a fixed positive integer. Let a, b ∈ Z. We say a is congruent to b modulo mand we write a ≡ b mod m if m | a− b.

We have a ≡ a + m ≡ a + 2m ≡ a + 3m ≡ · · · ≡ a + km mod m so we can intuitively think of mod mas treating integer multiples of m as zero.

We have the following basic properties of the congruence relation.

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Proposition (Congruence is an Equivalence Relation). Let m be a fixed positive integer. Let a, b, c ∈ Z.Then1. a ≡ a mod m [identity]2. if a ≡ b mod m then b ≡ a mod m [reflexive]3. if a ≡ b mod m and b ≡ c mod m then a ≡ c mod m [transitive].

Proposition (Properties of Congruence). Let m be a fixed positive integer. Let a, b, a′, b′ ∈ Z and letk ∈ P. If a ≡ a′ mod m and b ≡ b′ mod m, then1. a+ b ≡ a′ + b′ mod m2. a− b ≡ a′ − b′ mod m3. ab ≡ a′b′ mod m4. ak ≡ (a′)k mod m

What about division? In R, we know if ac = bc and c 6= 0 then a = b. (Another way to say this is thatevery non-zero real number has a multiplicative inverse.) But when we consider the integers modulo 4, weknow 5× 2 ≡ 7× 2 mod 4 since 10 ≡ 14 mod 4 but 5 6≡ 7 mod 4. When can we “cancel out” or “divide”common factors?

Proposition (Congruence and Division). Let m be a fixed positive integer. Let a, b, c ∈ Z. If ac ≡ bcmod m and gcd(c,m) = 1 then a ≡ b mod m.

We have now proven several propositions about the congruence relation. However we still do not fullyunderstand the relation. The following proposition tells us that the congruence relation is the same as the“remainder when divided by m” relation.

Proposition (Congruent Iff Same Remainder). Let m be a fixed positive integer. Let a, b ∈ Z. Thena ≡ b if and only if a and b have the same remainder when divided by m.

Proof. By the Division Algorithm we have

a = q1m+ r1 where q1, r1 ∈ Z and 0 ≤ r1 < m

b = q2m+ r2 where q2, r2 ∈ Z and 0 ≤ r2 < m

(=⇒). Suppose a ≡ b mod m. We must show r1 = r2. Since a ≡ b mod m then m | a − b. We see thata− b = (q1 − q2)m+ (r1 − r2). Thus we have m | (a− b)− (q1 − q2)m = r1 − r2 by Divisibility of IntegerCombinations.On the other hand we have −m < r1 − r2 < m. Since the only integer dividing m strictly between −mand m is 0, we must have r1 − r2 = 0 or r1 = r2.(⇐=). Suppose r1 = r2. Then a− b = (q1 − q2)m and thus m | a− b so a ≡ b mod m.

We can use this proposition to find remainders (without actually performing the division!) and to provetests for divisibility.

Example. What is the last digit of 333 × 77?

Solution. The last digit is the same the remainder when we divide by 10. Instead of performing the division,we can simply compute 333 × 77 mod 10. We notice 32 = 9 ≡ −1 mod 10 so that

333 = (32)163 ≡ (−1)163 ≡ 3 mod 10.

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Similarly we have 72 = 49 ≡ −1 mod 10 so that

77 = (72)37 ≡ (−1)37 ≡ 3 mod 10.

Therefore333 × 77 ≡ 3× 3 ≡ 9 mod 10

so the last digit of 333 × 77 is 9.

We have one last extremely useful result about congruences.

Theorem (Fermat’s Little Theorem). Let p ∈ Z be a prime. If a is an integer such that p does not dividea then

ap−1 ≡ 1 mod p.

This theorem is very handy when dealing with prime moduli. We give a typical example.

Example. Suppose that n ∈ Z. Prove that n91 ≡ n7 mod 91.

Solution. We want to apply Fermat’s Little Theorem so we must turn the problem into one about primes.We observe 91 = 7× 13 so instead we try and show

n91 ≡ n7 mod 7 (1)

n7 ≡ n7 mod 13. (2)

First we prove a claim which says that we can do this. (We could alternatively appeal to the ChineseRemainder Theorem which we review below.)

Claim. Let x ∈ Z and let p, q ∈ Z be distinct primes. If p | x and q | x then pq | x.

Verification. Since p and q are distinct, gcd(p, q) = 1 so by the Extended Euclidean Algorithm there existu, v ∈ Z such that pu+ qv = 1. Multiplying this by x we obtain pux+ qvx = x. Now p | x and q | so thereexists integers m and n such that x = mp and x = nq. Thus we have

x = pu(nq) + qv(mp) = pq(un+ vm)

which implies pq | x as required.

Thus if n91 ≡ n7 mod 7 and n91 ≡ n7 mod 13 then 7 | n91 − n7 and 13 | n91 − n7 so by our claim91 | n91 − n7 which is equivalent to n91 ≡ n7 mod 91.

To verify the first equation, we consider two cases. If n ≡ 0 mod 7 then we have n91 ≡ n7 ≡ 0 mod 7.Otherwise, we suppose n 6≡ 0 mod 7. Then we may apply Fermat’s Little Theorem to obtain n7 ≡ nmod 7 which implies

n91 = (n7)13 ≡ n13 ≡ n6n7 ≡ n7 mod 7.

Similarly we have n91 ≡ n7 mod 13. Thus n91 ≡ n7 mod 91.

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The integers modulo m: Zm

Definition. Let m be a fixed positive integer. Let a ∈ Z. The congruence class of a modulo m is theset

[a]m = {x ∈ Z | x ≡ a mod m}

For example if m = 6, we have

[5]6 = {· · · ,−7,−1, 5, 11, 17, · · · } = [11]6.

Intuitively we have [5]6 is all the integers which have remainder 5 when we divide by 6.

We define Zm = {[0]m, · · · , [m− 1]m}. We define addition and multiplication in Zm as follows

[a]m + [b]m = [a+ b]m

[a]m · [b]m = [ab]m.

We saw Zm does not have the same algebraic properties as Z. For example, in Z6, 2 · 3 = 0 (i.e. 2 · 3 ≡ 0mod 6), but neither 2 nor 3 is 0 in Z6! Also, in Zp, for p prime, we saw that every non-zero element has amultiplicative inverse. But in Z only 1 and −1 have multiplicative inverses.

Congruences

Definition. Let m be a positive integer. Let a, c ∈ Z. A linear congruence in the variable x is arelation of the form

ax ≡ c mod m.

We notice

ax ≡ c mod m ⇐⇒ m | ax− c⇐⇒ ∃y ∈ Z s.t. ax− c = my

⇐⇒ ∃y ∈ Z s.t. ax+my = c

Similar to our theorem about linear diophantine equations we have the following theorem about linearcongruences.

Theorem (Linear Congruence Theorem). Let gcd(a,m) = 1. The linear congruence

ax ≡ c mod m

has a solution if and only if d | c. Moreover, if x = x0 is a particular solution then the complete solution is

x ≡ x0 modm

d

orx ≡ x0, x0 +

m

d, · · · , x0 + (d− 1)

m

d.

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Example. Solve the congruence 21x ≡ 12 mod 15

Solution. Since gcd(21, 15) = 3 and 3 | 12 the congruence has 3 solutions. To find the solutions we solvethe linear Diophantine equation 21x+ 15y = 12. We get

21 = 1 · 15 + 6

15 = 2 · 6 + 3

6 = 2 · 3 + 0

so q3 = 1 and q4 = q5 = 2.

Next we use the Extended Euclidean Algorithm to get

x y ri qi1 0 21 0

0 1 15 0

1 -1 6 1

-2 3 3 2

5 -7 0 2

Thus 21(−2) + 15(3) = 3. Multiplying by 4, we get 21(−8) + 15(12) = 12 so a particular solution isx = −8, y = 12. Thus a particular solution to the congruence is

x ≡ −8 ≡ 7 mod 15.

The complete solution is

x ≡ 7, 7 +15

3, 7 + 2

15

3≡ 2, 7, 12 mod 15.

We also have the following result for solving systems of linear congruences.

Theorem (Chinese Remainder Theorem). Let a1, a2,m1,m2 ∈ Z with gcd(m1,m2) = d. Consider thesystem S

x ≡ a1 mod m1

x ≡ a2 mod m2

of linear congruences. If d does not divide a2− a1 then S has no solutions. If d | a2− a1 then the completesolution is given by

x ≡ x0 modm1m2

d

where x0 is a particular solution to S.

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Example. Solve the system of congruences

x ≡ 2 mod 7

x ≡ 5 mod 8.

Solution. Since x satisfies 1 we may write x = 2+7y for some y ∈ Z and we substitute this into 2 to obtain

7y ≡ 3 mod 8.

We could solve this linear congruence as above, however, we present an alternative solution. We notice72 ≡ 1 mod 8 so that

y ≡ 21 ≡ 5 mod 8.

Thus we may write y = 5 + 8z for some z ∈ Z. Substituting this back into x = 2 + 7y, we get

x = 37 + 56z

orx ≡ 37 mod 56.

Example. Solve the congruencex5 + x3 ≡ 2 mod 5.

Solution. By Fermat’s Little Theorem we know that x5 ≡ x mod 5 so we can instead solve the polynomialcongruence x3 + x ≡ 2 mod 5. We make a table.

x 0 1 2 3 4

x3 0 1 3 2 4

x3 + x 0 2 0 0 3

Thus the solution is x ≡ 2 mod 5.

Complex Numbers

Definition. We define C to be the set {x + iy | x, y ∈ R} where i2 = −1. Let z = x + iy ∈ C. The realpart of z is Re(z) = x. The imaginary part of z is Im(z) = y. The conjugate of z is z = x− iy. Thenorm of z is |z| =

√x2 + y2.

We can define addition, subtraction, multiplication, and division of complex numbers in the natural wayand they have the same properties as addition, subtraction, multiplication, and division of real numbers.(see your notes!)

We can also represent z = x+ iy as z = r(cos θ + i sin θ) which is called the polar representation of z.

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Example. Write z = 1 + i in polar form.

Solution. Sketching the point (1, 1) we see that r =√

1 + 1 =√

2 and θ = π4 . Thus

z =√

2(

cosπ

4+ i sin

π

4

).

The polar representation of z gives us a nice way of performing exponentiation.

Theorem. Let z = r(cos θ + i sin θ) ∈ C and let n ∈ Z. Then

zn = rn(cosnθ + i sinnθ).

Example. Provecos 3θ = 4 cos3 θ − 3 cos θ

sin 3θ = 3 sin θ − 4 sin3 θ.

Solution. We consider z = cos θ + i sin θ and compute z3 in two ways. On the one hand,

z3 = (cos θ + i sin θ)3

=

3∑i=0

(3

i

)(cos θ)3−i(i sin θ)i

= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ

= (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ).

Whereas, using the theorem above, we have z3 = cos 3θ + i sin 3θ. Equating real and imaginary parts, weget

cos 3θ = cos3 θ − 3 cos θ sin2 θ

= cos3 θ − 3 cos θ(1− cos2 θ)

= 4 cos3 θ − 3 cos θ

sin 3θ = 3 cos2 θ sin θ − sin3 θ

= 3(1− sin2 θ) sin θ − sin3 θ

= 3 sin θ − 4 sin3 θ.

Example. Find a real cubic polynomial with roots 1 and i.

Solution. We know that if f is a real polynomial and z is a complex root of f then z is also a root of f .Thus we see

f(x) = (x− i)(x+ i)(x− 1) = (x2 + 1)(x− 1) = x3 − x2 + x− 1

is a real cubic polynomial with roots 1 and i.

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Example. Prove that if |z| = 1 or |w| = 1 then |z − w| = |1− zw|.

Solution. We have

|1− zw|2 = (1− zw)(1− zw)

= (1− zw)(1− zw)

= 1− zw − zw + zzww

Suppose |z| = 1. Then zz = 1 and so we get

|1− zw|2 = 1− zw − zw + zzww

= zz − zw − zw + ww

= |z − w|2

and thus |z − w| = |1− zw|. The case |w| = 1 is similar.