Sometimes, ordinary induction doesn't give us quite enough to prove what we want to prove. In this lecture, we'll introduce a reformulation of induction called strong induction.
One example of what strong induction can help us prove is the Fundamental Theorem of Arithmetic:Every integer $n \gt 1$ can be written as a product of primes. This product is unique up to reordering.
McDonald’s is currently selling chicken nuggets in packages of 3 or 5 nuggets. Let $P(n)$ denote the proposition that you can buy exactly n nuggets. For example, $P(6)$ is true because you can buy two packs of three, and P(8) is true, but P(7) is false. Show that for all n ≥ 8, $P(n)$ is true.
Let's think about how we'd prove this chicken nuggets probelm. We want to show that, for every natural number $n \geq 8$, we can buy exactly $n$ nuggets (served in packs of 3 or 5).
Assuming that we set everything up properly, our inductive step would involve showing that if we can buy exactly $k$ nuggets, then we can buy exactly $k+1$ nuggets. But knowing that we can buy exactly $k$ nuggets doesn't quite help us, since we can't buy a single nugget to achieve $k+1$ nuggets. What we really want to know is whether we can buy $k + 1 - 3$ or $k + 1 - 5$ nuggets, and then add in a 3-pack or a 5-pack. However, our inductive hypothesis doesn't help us with that!
Strong induction is a reformulation of induction that gives us a "stronger" hypothesis to say these kinds of things.
Let $P(n)$ be a predicate with domain $\mathbb N$. If
are both true, then the following is true:
This is called the principle of strong induction.
What is the difference between ordinary mathematical induction and strong induction? Or, in other words, what makes strong induction strong? The hypothesis that we make in part 2 is stronger. Recall that ordinary induction was stated as a chain of implications $P(k) \rightarrow P(k+1)$. Strong induction is a chain of implications of the form $$P(0) \wedge P(1) \wedge P(2) \wedge \cdots \wedge P(k) \rightarrow P(k+1).$$ Then what we're really proving is all of the following statements, \begin{align*} &P(0), \\ P(0) \rightarrow &P(1), \\ P(0) \wedge P(1) \rightarrow &P(2), \\ P(0) \wedge P(1) \wedge P(2) \rightarrow &P(3), \\ & \vdots \end{align*}
It's important to note that strong induction is not stronger in the sense that it allows us to prove things that we wouldn't be able to with mathematical induction. This is why I described strong induction as an alternate form of mathematical induction. It is possible to use mathematical induction to prove anything we would prove with strong induction, but this involves slightly altering the statement that you want to prove to incorporate your stronger assumptions.
A proof by strong induction looks similar to a proof by mathematical induction.
We proceed by strong induction on $n$ to show that for all $n \geq 8$, we can buy $n$ nuggets. Define $P(n)$ to mean that we can buy $n$ nuggets.
Base Case: We’ll need three base cases here: 8, 9, 10. Note that
Thus, P(8), P(9), and P(10) are all true.
Assume that $P(i)$ is true for some arbitrary $k \in \mathbb{N}$, $k \geq 10$, where $8 \leq i \leq k$. We aim to show that this implies that $P(k + 1)$ is true — we can buy exactly $k + 1$ nuggets.
Note that $k + 1$ nuggets can be bought as a combination of $k − 2$ nuggets plus 3 nuggets. We can of course buy 3 nuggets, and by the inductive hypothesis we can buy $k − 2$ nuggets. Thus, we can buy $k + 1$ nuggets. Thus, we have shown that $P(8) ∧ P(9) ∧ . . . ∧ P(k)$ implies $P(k + 1)$, so $∀n P(n)$.
For the remainder of the lecture, we are going to develop some more number theory and apply strong induction to prove some famous theorems.
The study of primes is a central topic of number theory and has fascinated people for thousands of years.
An integer $p$ greater than 1 is called prime if the only positive divisors of $p$ are 1 and $p$. Otherwise, $p$ is called composite.
Primes are obviously relatively prime to any positive number less than them, since they don't have any divisors except 1 and themselves. However, non-prime numbers (called composite numbers) can be relatively prime, even to each other. The numbers 10 and 21 are not prime, since $10 = 2 \cdot 5$ and $21 = 3 \cdot 7$. However, they are relatively prime, since the divisors of $10$ are $\{\pm 1, \pm 2, \pm 5, \pm 10\}$ and the divisors of $21$ are $\{\pm 1, \pm 3, \pm 7, \pm 21\}.$
Simple questions about primes immediately get into deep mathematical territory. However we can establish several important results in this lecture.
Let's prove Theorem 4.1, The Fundamental Theorem of Arithmetic.
We will first prove that for $n \gt 1$, $n$ can be written as a product of primes. (That is, we'll put off proving the uniqueness up to reordering for now.) We will proceed by strong induction on $n$.
Base case. Let $n = 2$. 2 is a product of primes because 2 itself is a prime number.
Inductive Hypothesis. Let $k \in \mathbb N$ be arbitrary. Assume that for $2 \leq m \leq k$, $m$ can be written as a product of primes.
Inductive step. We want to show that $k+1$ can be written as a product of primes. There are two cases to consider.
This completes the proof that any $n$ can be written as a product of primes.
If $p$ is prime, and $a,b$ are integers such that $p \mid a \cdot b$, then $p \mid a$ or $p \mid b$.
There are two cases to consider: $p \mid a$ and $p \not \mid a$. If $p \mid a$, then we're done. Suppose $p \not \mid a$. We will show that $p \mid b$. If $p \not \mid a,$ then $\gcd(p,a) = 1$ and there exist integers $x$ and $y$ such that $xa + yp = 1$. Then multiplying both sides by $b$, we have $$b = xab + ypb.$$ Since $p \mid ab$, let $n$ be an integer such that $ab = pn$. Then we have $$b = xab + ypb = xpn + ypb = p \cdot (xn+yb).$$ Therefore, by definition of divisibility, $p \mid b$.
Since $3\mid 10\cdot 15 = 150$, the theorem says that $3$ must divide $10$ or $15$ and indeed $3\mid 15$. This isn't true if $p$ is not prime. For instance, $6\mid 10\cdot 15$ but $6$ does not divide either $10$ or $15$ (it gets the factor of $2$ from $10$ and the factor of $3$ from $15$).
We will prove this by (non-strong) induction on $k$, the number of primes.
Base case. In this case, $p \mid p_1$, so $p$ is a positive divisor of $p_1$. Since $p_1$ is prime, its only positive divisors are $1$ and $p_1$. But $p\neq 1$, so it must be the case that $p = p_1$.
Inductive hypothesis.Suppose our lemma is true for $k$.
Inductive step. Consider $p \mid p_1 \cdots p_k \cdot p_{k+1}$. Let $a = p_1 \cdots p_k$ and $p_{k+1} = b$. Since $p \mid ab$, by Theorem 7.7, $p \mid a$ or $p \mid b$. If $p \mid a$, then $p = p_i$ for some $i$ with $1 \leq i \leq k$ by our inductive hypothesis and our result holds. If $p \mid b$, then an argument similar to the base case shows that $p = p_{k+1}$, and our result holds. This completes the induction.
With this corollary in tow, we can finally prove that factorization is unique.
We'll prove this by strong induction on $n$ again.
Base case. Our base case is the claim with $n=2$. This is easy to establish (any other product of primes will be larger than $2$).
Inductive step. Now we suppose the theorem is true for all $1,\ldots,n$ and show it is true for $n+1$. Suppose that $$ n+1 = p_1p_2\dots p_s = q_1q_2 \dots q_t. $$ Then $p_1$ divides $q_1 q_2 \dots q_t$. By Corollary 7.10, $p_1 = q_j$ for some $j$. Dividing both sides by $q_j$, we have $$ p_2\dots p_s = q_1\cdots q_{j-1} q_{j+1} \dots q_t. $$ Call this value $m$. Then $1 \leq m \leq n$, so the strong inductive hypothesis applies, we have that $q_1\cdots q_{j-1} q_{j+1} \dots q_t$ is a reordering of $p_2\dots p_s$. Therefore $p_1\dots p_s$ is a reordering of $q_1\dots q_t$.
We move on to two of the most famous proofs ever. Both use the results about prime numbers and factorization that we proved last time. They also both wield contradiction, further demonstrating the surprising power and potentially broad nature of proofs in the space.
There exist infinitely many prime numbers.
Assume for contradiction that there are a finite number of primes. Let $p_1,\ldots,p_k$ be the $k$ primes. To find a contradiction, it is enough to show that some prime is not included on this list. Consider $x=p_1 p_2\cdots p_k+1$. Observe that $x$ is not divisible by any of $p_1,\ldots, p_k$ (as this would imply that $p_i\mid 1$, which is impossible). On the other hand, by the Fundamental Theorem of Arithmetic, $x$ can be factored into primes, and in particular some prime $p$ divides $x$. Since this prime can't be any of $p_1,\ldots,p_k$, we have found an additional prime beyond $p_1,\ldots,p_k$.
The rest of the notes, presenting some more results related to primes and prime factorization, are optional.
Any square root of $2$ is irrational. More explicitly, for all $a,b\in\mathbb{Z}$ with $b\neq 0$, $(a/b)^2 \neq 2$.
Let's continue avoiding division, and prove the equivalent statement that $a^2 \neq 2b^2$ for all $a,b\in\mathbb{Z}$ with $b\neq 0$. (Note that the assumption that $b$ is non-zero is necessary, since otherwise we can take $a=0$ and $b=0$.)
Now suppose for the purpose of contradiction that there exist $a,b\in\mathbb{Z}$, $b\neq 0$, such that $a^2 = 2 b^2$. By negating $b$ if necessary, we may assume that $b$ is positive. Since there exists at least one such pair with positive $b$, we can select a pair that minimizes the value of $b$. We are going to contradict our choice of $a,b$ by showing that we can find another pair with a smaller $b$.
Since $a^2=2b^2$, we have that $2|a^2$. Since $2$ is prime, we have that $2|a$ by Euclid's Lemma from the last lecture. Thus there exists $k\in\mathbb{Z}$ such that $a=2k$.
Next we have that $2b^2 = a^2 = (2k)^2 = 4k^2$. Thus $b^2 = 2k^2$. We can repeat the argument from the previous paragraph to show that $2|b$ as well, so there is $\ell\in\mathbb{Z}$ such that $b=2\ell$.
Since $a^2=2b^2$, we have that $(2k)^2=2(2\ell)^2$, which implies that $k,\ell$ is another pair satisfying $k^2=2\ell^2$. Moreover $\ell$ is positive and $\ell \lt b$, which contradicts our choice of $a$ and $b$. This completes the proof.
Quite a few striking results about primes have been established! Here are a few, though we won't go into their proofs.
Let $\pi(x)$ be the number of primes at most $x$. Then
$$ \lim_{x\rightarrow\infty} \frac{\pi(x)}{x/\ln(x)} = 1. $$Here $\ln(x)$ is the natural logarithm of $x$. So this theorem not only says there are infinitely many primes, but also that, amongst the $x$ numbers $\{1,\ldots,x\}$, approximately a $1/\ln(x)$-fraction of them are prime! (And in the limit the fraction is exactly $1/\ln(x)$, which provides an example of the Euler's number $e$ from calculus showing up in number theory.) This means there are a lot or primes. A concrete analysis shows, for example, that a random $1000$-digit number has about a $1/7000$ chance of being prime. Combined with the fact that we can efficiently recognize prime numbers (even large ones, but this is a story for another day), this means that finding a $1000$-digit prime is a simple matter with a computer, which is important in cryptography.
The Prime Number Theorem is a hard theorem that was famously proved using complex analysis (i.e. calculus with imaginary numbers). We definitely won't be proving it in this course.
Here's an even harder question, which is open. Define twin primes to be numbers $p, p+2$ that are both prime. For example, $3,5$ are twin primes, as are $137,139$.
Open Question. Do there exist infinitely many twin primes?
It's a very old question that remains unsolved. However there has been some spectacular progress on this question in the last decade. This started with the following breakthrough result in 2013:
There exist infinitely many pairs of primes $p,q$ satisfying $|p-q| < 10^7$.
The $10^7$ number is far cry from 2, but at least it's finite, and this was the first theorem showing there infinitely many primes within some constant distance of each other. The number $10^7$ was quickly improved to $246$, which remains the record today. You can read more about these events here.
Rosen Chapter 5.2 and Velleman Chapter 6 both have a plethora of great examples.