## 305 -Algebra and induction (3)

I won’t be here this Friday. Professor Marion Scheepers will cover the lecture for me.

Definition. $p>1$ is prime iff the only positive divisors of $p$ are $1$ and $p$.

Proposition. A number $p>1$ is prime iff for all $m$, either $p|m$ or else $(p,m)=1$. $\Box$

Proposition. Let $p$ be prime, and let $m,n$ be integers. If $p|mn$ then either $p|m$ or $p|n$. $\Box$

In the more general setting of rings, of which the integers are an example, it is customary to call a number $p$ prime when it satisfies the second proposition, and to call it irreducible when it satisfies the definition above. Both notions coincide for the integers, but there are examples of rings where there are irreducible elements that are not prime. We will introduce rings later on in the course.

Mathematical induction. Let $N\in{\mathbb Z}$. Let $P(\cdot)$ be a statement about integers, so for each integer $m$, $P(m)$ may or may not hold. Assume

1.  $P(N)$ holds, and
2. $\forall k\ge N\,($ if $P(k)$ holds then $P(k+1)$ holds$).$

Then $\forall k\ge N\,(P(k)$ holds$)$.

Strong induction. Let $N\in{\mathbb Z}$. Let $P(\cdot)$ be a statement about integers. Assume

• $\forall k\ge N\,($ if $\forall m$ with $N\le m it is the case that $P(m)$ holds, then also $P(k)$ holds$).$

Then $\forall k\ge N\,(P(k)$ holds$)$.

Both induction and strong induction are consequences of the well-ordering principle. In fact, all three statements are equivalent. Most properties about the natural numbers are established by inductive arguments. Here are two examples:

Example 1. For all $n\ge1$, $\displaystyle \sum_{ i=1}^n i=\frac {n(n+1)}2$.

We also have $\displaystyle \sum_{ i=1}^n i^2=\frac {n(n+1)(2n+1)}6$ and $\displaystyle \sum_{ i=1}^n i^3=\left(\frac {n(n+1)}2\right)^2$. In fact, for all $k\ge 1$, there is a polynomial $q(x)$ with rational coefficients, of degree $k+1$ and leading coefficient $\displaystyle \frac1{k+1}$ such that for all $n\ge1$$\displaystyle \sum_{ i=1}^n i^k=q(n)$

Example 2. For all integers $a,b$, if $(a,b)=1$ then for all $n\ge1$, $(a^n,b)=1$.

The book discusses additional examples and has many more in the exercises. Other examples will be discussed in the first homework set, to be posted later today.