Example Proofs: Induction
Proofs Workshop
Note: the proofs in this handout are not necessarily in the same form as they were presented at the workshop. In particular, any errors you spot here are entirely accidental, not deliberate. Theorem Let { f n } be the sequence of integers satisfying: Prove that 1 + f1 + f3 + … + f2n 1 = f2n −
f1 = 1, f2 = 2, and fr = fr-1 + fr-2 for all r ≥ 3 .
for all positive integers n.
Proof: Let P(n) be the proposition 1 + f1 + f3 + … + f2n-1 = f2n .
Reminder: P(k+1) is the statement:
Initial step:
1 + f1 + f3 + … + f2(k+1) 1 = f2(k+1) i.e. 1 + f1 + f3 + … + f2k+1 = f2k+2
P(1) is true, since 1 + f1 = 1 + 1 = 2 = f2 (= f2n).
−
Inductive step: Suppose P(k) is true for some integer k ≥ 1, so 1 + f1 + f3 + … + f2k 1
= f2k
−
Then
1 + f1 + f3 + … + f2k-1 + f2k+1 = f2k + f2k+1 (adding f2k+1 to both sides) = f2k+2
(by definition, since 2k + 2 ≥ 3)
and this is the statement P(k + 1). So we have shown that for all integers k ≥ 1, whenever P(k) is true, then P(k+1) is also true. So P(n) is true for all positive integers n by the principle of mathematical induction.
Theorem Prove that
d n x = nx n −1 using the product rule, where n is a positive integer. dx
( )
Proof: Let P(n) be the proposition
d n x = nx n−1 dx
( )
Initial step: Let n = 1 , then
d n d x = (x ) = 1 = 1⋅ x 0 = 1⋅ x1−1 = nx n −1 . Therefore P(1) is true. dx dx
( )
Inductive step: Assume the proposition is true when n = k , where k ≥ 1, i.e.
d k +1 d x = x ⋅ xk dx dx d (x ) ⋅ x k + x ⋅ d x k = dx dx k k −1 = 1 ⋅ x + x ⋅ kx
( )
(
)
( )
k
= x + kx
(by the product rule)
d k x = kx k −1. Then for n = k+1, dx
( )
Reminder: P(k+1) is the statement:
(by assumption)
k
= (k + 1)x k This shows that if the proposition is true for n = k , k ≥ 1 , then it is true for n = k + 1. We have shown that P(1) is true, and that whenever P(k) is true, where k ≥ 1, then P(k+1) is true. Hence by the principle of mathematical induction, P(n) is true for all n ≥ 1. Therefore
d n x = nx n −1 for all positive integers n. dx
Shirleen Stibbe
( )
http://www.shirleenstibbe.co.uk1
Theorem
Deliberate error on presentation: We left out 'convex' from the statement, to see if anyone would question the insertion of the line segment.
Prove that the sum of the internal angles of an n-sided convex polygon is 180(n − 2)° where n ≥ 3. ("Convex" means that any point on a line segment joining two vertices lies inside the polygon or on its border.) Proof: Let P(n) be the statement: "The sum of the internal angles of an n-sided convex polygon is 180(n − 2)° ". Initial step: The sum of the internal angles of a triangle is 180° = 180(3 − 2)°, so P(3) is true. Inductive step: Assume that P(k) is true for some k ≥ 3; that is, the sum of the internal angles of a k-sided convex polygon is 180(k − 2)°, k ≥ 3. Consider a (k+1)-sided convex polygon: Reminder: P(k+1) is the statement: The sum of the internal angles of a k+1-sided convex polygon is 180(k − 1)°
We insert a line segment to form a triangle consisting of any two adjacent edges of the polygon and the line segment. (The line segment lies inside the polygon, since it is convex.) There are now k −1 edges on one side of the segment, and 2 edges on the other side.
We have thus split the original (k+1)-sided polygon into a triangle and a polygon with k sides. The sum of the internal angles of the triangle is 180°, and, by our assumption, the sum of the internal angles of the k-sided polygon is 180(k − 2)°. Adding the sum of the internal angles of the two smaller polygons gives the sum of the internal angles of the original (k+1)-sided polygon. Therefore, the sum of the internal angles of the (k+1)-sided polygon is 180 + 180(k − 2) = 180(k − 1)°. Therefore P(k+1) is true, and this completes the inductive step. Hence, by the principle of mathematical induction, the sum of the internal angles of an n-sided convex polygon is 180(n − 2)°, where n ≥ 3.
Theorem Prove that any two consecutive Fibonacci numbers are coprime, i.e. fn and fn+1 have no common factor greater than 1. Proof by induction Let P(n) be the statement “ fn and fn+1 are coprime” f1 = f2 = 1, so P(1) is true. Assume P(k) is true for some positive integer k ≥ 1, i.e. fk and fk+1 are coprime. (We wish to prove that P(k+1) is true, i.e. fk+1 and fk+2 are coprime) If fk+1 and fk+2 are not coprime, then they have a common factor, d. But this cannot happen, since d would also divide fk+2 − fk+1 = fk , (by the definition), and so fk and fk+1 would not be coprime. Hence whenever fk and fk+1 are coprime, fk+1 and fk+2 must be coprime, and P(k+1) is true. We have shown that P(1) is true, and that whenever P(k) is true for some positive integer k ≥ 1, P(k+1) is true. Hence, by the principle of mathematical induction P(n) is true for all n ∈ ℕ.
Shirleen Stibbe
http://www.shirleenstibbe.co.uk
Theorem Let a ≠ 1 be a real number. For all positive integers n, an − 1 . 1 + a + a 2 + ... + a n −1 = a −1 Proof
Note: we also proved this by inspiration (see Miscellaneous methods)
The proof is by induction on n. Let P(n) be the statement: 1 + a + a 2 + ... + a n −1 = When n = 1, LHS = 1 and RHS =
a −1 = 1 . Therefore P(1) is true. a −1
Assume that, for some integer k ≥ 1, P(k) is true; that is, 1 + a + a 2 + ... + a k −1 = Then
1+ a + a 2 + ... + a k!1 + a k =
a k !1 k +a a !1
=
a k !1+ a k (a !1) a !1
=
a k !1+ a k+1 ! a k a !1
=
a k+1 !1 . a !1
an − 1 . a −1
ak − 1 . a −1
[by the assumption]
Hence P(k + 1) is true, which completes the inductive step: P(k) ⇒ P(k + 1). Therefore P(n) is true for all positive integers n, by induction.
Shirleen Stibbe
http://www.shirleenstibbe.co.uk3