03-Mathematical_Induction

I Introduction¶

In short, mathematical induction just likes domino, one pushes another.

three simple steps:

Base Case: Prove that P(0) is true.
Induction Hypothesis: For arbitrary k ≥ 0, assume that P(k) is true.
Inductive Step: With the assumption of the Induction Hypothesis in hand, show that P(k + 1) is true

Theorem 3.1

For all \(n\in N, n^{3}-n\) is divisible by 3.

proof proceed by induction (In fact, we can prove it by cases since \(n^{3}-n=(n-1)n(n+1)\) , and one of the factors on the right side of the equation must be a multiple of 3, then we get it.)

We now consider a more advanced proof by induction, which establishes a simplified version of the famous four color theorem. But it is too difficult for us to prove. Let's change it easier:

[!THEOREM 3.2]

(Two Color Theorem) our “map” is given by a rectangle which is divided into regions by drawing straight lines, such that each line divides the rectangle into two regions, then:using no more than two colors (say, red and blue) such that no two bordering regions have the same color(below is an example case)

[!PROOF]

proceed by induction

we set n as the number of lines

Base Case (n = 0): Clearly P(0) holds, since if we have n = 0 lines we can color the entire map using a single color.

Induction Hypothesis: For some arbitrary n = k ≥ 0, assume P(k).

Inductive Step: Whenever we add a line, that is, P(k+1), we can always prove that P(k+1) is true by swapping parts of red and blue (as shown below)

II Strengthening the Induction Hypothesis¶

Sometimes, our Induction Hypothesis is too “weak”; it does not give us enough structure to say anything meaningful, for example:

[!THEOREM 3.3]

For all n ≥ 1, the sum of the first n odd numbers is a perfect square.

In fact, we can not prove it directly. The reason is that this claim did not capture the true structure of the underlying fact we were trying to prove — it was too vague. As a result, our Induction Hypothesis wasn’t strong enough to prove our desired result.

Let us try to show the following stronger claim.

[!THEOREM 3.3']

For all n ≥ 1, the sum of the first n odd numbers is \(n^{2}\) . (It is easy to prove by induction)

III Strong Induction¶

Sometimes we can solve the question difficultly by using P(k) solely, that's why strong induction appears.

strong induction: we assume the stronger statement that P(0), P(1), . . . , and P(k) are all true (i.e. that \(\land_{i=0}^{k} P(i)\) is true)

[!attention]

Is there a difference between the power of strong and weak induction, i.e., can strong induction prove statements which weak induction cannot?

No! Intuitively, this can be seen by returning to our dominoanalogy.

Then, next one:

[!THEOREM 3.4]

Every natural number n > 1 can be written as a product of one or more primes.

proof proceed by induction and cases

Let P(n) be the proposition that n can be written as a product of primes. We will prove that P(n) is true for all n ≥ 2.

Base Case (n = 2): We start at n = 2. Clearly P(2) holds, since 2 is a prime number.

Induction Hypothesis: Assume P(n) is true for all 2 ≤ n ≤ k.

Inductive Step: Prove that n = k +1 can be written as a product of primes. We have two cases: either k +1 is a prime number, or it is not.

For the first case, if k +1 is a prime number, then we are done since k +1 is trivially the product of one prime (itself).

For the second case, if k + 1 is not a prime number, then by definition k + 1 = xy for some x,y ∈ Z + satisfying 1 < x, y < k + 1. By the Induction Hypothesis, x and y can each be written as a product of primes (since x, y ≤ k). But this implies that k +1 can also be written as a product of primes. Then, we get it.

(Recursion, programming and induction are also mentioned here, but these will be covered in FDS, so we'll skip them)

IV False proofs¶

Let's look at one of history's most famous false proofs which makes us laugh:

False theorem

All horses are the same color

proof proceed by induction

Base Case (n = 1): P(1) is certainly true, since if you have a set containing just one horse, all horses in the set have the same color.

Induction Hypothesis: Assume P(n) holds for some arbitrary n ≥ 1.

Inductive Step: Given a set of n + 1 horses {h1,h2,...,hn+1}, we can exclude the last horse in the set and apply the induction hypothesis just to the first n horses {h1,...,hn}, deducing that they all have the same color. Similarly, we can conclude that the last n horses {h2,...,hn+1} all have the same color. But now the “middle” horses {h2,...,hn} (i.e., all but the first and the last) belong to both of these sets, so they have the same color as horse h1 and horse hn+1. It follows, therefore, that all n+1 horses have the same color. We conclude, by the principle of induction, that all horses have the same color.

V Practice¶

Q 1 Airports

简而言之，奇数个机场两两相距不同，一定有一个机场 A 到任意机场 B 的距离比剩余机场其一 C 到 B 的距离要远，因而没有飞机降落在 A。

我们使用数学归纳法证明：

\(n=1\) 时，即三个机场，不妨设 a,b,c 且 ab 之间的距离最近，则没有飞机去 c
\(n=k\) 时，假设成立
当 \(n=k+1\) 时我们先去掉相距最近的两个机场 (a,b)，又回到了 \(n=k\) 的情况，不妨设没有飞机去的是 z，现在我们把 a,b 加回来，他俩的飞机不可能去 z，那么 \(n = k+1\) 命题依旧成立，命题得证。

Q 2 Coins game

简而言之，一堆硬币每次分成两堆，个数相乘为小得分；继续分直到所有堆都只有一个硬币，所得总分为小得分相加，但是总得分终会是 \(\frac{n(n-1)}{2}\)

不难发现其实这与我们证明 Theorem 4 是类似的，将未证明情况转变为已证明情况，下面是参考解答：

Q 3 Grid Induction

Pacman is walking on an infinite 2 D grid. He starts at some location (i, j) ∈ N 2 in the first quadrant,and is constrained to stay in the first quadrant (say, by walls along the x and y axes).Every second he does one of the following (if possible):

(i) Walk one step down, to (i, j − 1). (ii) Walk one step left, to (i − 1, j).

Prove by induction that no matter how he walks, he will always reach (0, 0) in finite time.

这道题很简单，我们甚至可以加强命题：P 由 \((i, j)\) 一定经过 i+j 步后到达 \((0,0)\) （证明略）