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CSE 21: Notes 3
October 5, 2009
Starter
DON’T DO THAT!
Proof by Induction
A method for proving that something is true for infinitely many cases.
Basic Format:
Suppose that S(n) denotes some statement that depends on the integer n. For example, S(n) could the the statement:
S(n) : Every positive integer is the sum of exactly 4 perfect squares.
There are two steps:
- Show that S(n 0 ) is true for some value n 0 (typically n 0 = 1);
- Show that if S(n) is true then S(n + 1) is also true.
From this you can conclude that S(n) is true for all values of n greater that or equal to n 0.
Some Examples:
- Use induction to prove: ∑n
k=
k = n(n + 1) 2
Sketch of proof: (a) Step 1: ∑^1 k=
k = 1 =
(b) Step 2: n∑+
k=
k =
∑^ n
k=
k + (n + 1) = n(n + 1) 2
(n + 1)(n + 2) 2
- Use induction to prove: ∑n
k=
k(k + 1)
n + 1
Sketch of proof:
(a) Step 1: ∑^1 k=
(b) Step 2: n∑+
k=
k(k + 1)
∑^ n k=
k(k + 1)
(n + 1)(n + 2)
= 1 −
n + 1
(n + 1)(n + 2) = 1 − n + 2 (n + 1)(n + 2) +^
(n + 1)(n + 2) = 1 − n + 1 (n + 1)(n + 2) = 1 −
n + 2. §
- Use induction to prove: ∑n k=
k^2 = n(n + 1)(2n + 1) 6
Sketch of proof: (a) Step 1: ∑^1 k=
12 = 1 =^1 ·^ (1 + 1)^ ·^ (2 + 1)
(b) Step 2: n∑+
k=
k^2 =
∑^ n k=
k^2 + (n + 1)^2
= n(n^ + 1)(2n^ + 1) 6
= n^ + 1 6 {n(2n + 1) + 6(n + 1)}
= n^ + 1 6 { 2 n^2 + n + 6n + 6}
= n^ + 1 6 (2n + 3)(n + 2).
§
Bijective proofs (proof technique)
Bijective proofs proceed by counting some quantity in two different ways, thereby deriving an equality.
Some Examples:
- Try to prove that (^) ( n k
n k + 1
n + 1 k + 1
Proof by definition: Recall that (^) ( n k
n! k!(n − k)!
Thus, we have ( n k
n k + 1
n! k!(n − k)! +^
n! (k + 1)!(n − k − 1)! = n! k!(n − k − 1)!
n − k
k + 1
n! k!(n − k − 1)!
n + 1 (n − k)(k + 1) = (n + 1)! (k + 1)!(n − k − 1)!
Proof by bijection: Consider the problem of counting the number of (k + 1)-subsets of an (n + 1)-set S = {s 0 , s 1 , s 2 ,... , sn}. We know that there are
(n+ k+
such subsets in total. We also notice that the number of (k +1)-subsets that don’t contain s 0 is
( (^) n k+
, and the number of (k +1)-subsets that contains s 0 is
(n k
. Thus, in total there are (^) ( n k
n k + 1
(k + 1)-subsets, which implies that ( n k
n k + 1
n + 1 k + 1
- Try to prove that ∑n
k=
n k
2 n n
Proof by bijection: Consider a box that contains n red balls and n blue balls. Clearly, the number of ways to choose n balls from the box is
( 2 n n
. One method for choosing n balls from the box is that you first choose k red balls, and then choose (n − k) blue balls. Alternatively, there are
(n k
ways to pick k red balls from n red balls and there are there are
( (^) n n−k
(n k
ways to pick n − k blue balls from n blue balls. Thus, there are in total ∑n
k=
n k
ways to pick n balls, which implies that ∑^ n
k=
n k
2 n n
