The previous chapter makes frequent use of the word number, of which there are various kinds.

The counting numbers are symbolized by \(\N\), a group of numbers for which not all of the aforementioned properties of numbers hold, such as \(P2, P3\).

• \(P2\) contains zero, which isn't a counting number, and \(P3\) contains a negative number, and negatives also aren't counting numbers.

The most basic property of this group of numbers is the induction principle, which says that if \(P(x)\) is true for some \(x\), the property should hold for all numbers \(x\), and whenever \(P(k)\) is true \(P(k+1)\) should be as well.

The second condition, is asserted to be true under the assumption that the first is, and in the event that \(P(1)\) is true, we can show the property for the case \(P(2)\), using the second condition with the case \(k=1\), and so on.

An example of induction is a formula that provides the sum of the first \(n\) numbers:

\(1 + \ldots + 1 = \frac{n(n+1)}{2}\).

In order to prove this is true, we first show that it holds for \(n=1\).

• \(\frac{1(1+1)}{2} = 1\).

Now, we assume for some \(k, k \in N\), we have: \(1 + \ldots + k = \frac{k(k+1)}{2}\).

Then \(\\1 + \ldots + k + (k+1) =\\[3pt] \frac{k(k+1)}{2} + (k+1) =\\[7pt] \frac{k(k+1) + 2k + 2}{2} =\\[5pt] \frac{k^2 + 3k+2}{2} =\\[5pt] \frac{(k+1)(k+2)}{2}\)

The inductive principle can be shown without reference to numerical properties, more precisely, if \(A\) is a set and \(1 \in A\), as well as \(k+1 \in A\), if \(k \in A\), \(A = \N\).