Question

Show that the set S defined by 1$\mathbf{\in }$S and s + t $\mathbf{\in }$S whenever $\mathbf{s}\mathbf{\in }\mathbf{S}$and $\mathbf{t}\mathbf{\in }\mathbf{S}$is the set of positive integers.

Short answer

S is the set of positive integers.

Step-by-step solution

The given data.

The set S is defined by

$\begin{array}{r}1\in \mathrm{S}\\ \mathrm{s}+\mathrm{t}\in \mathrm{S}\text{whenever}\mathrm{s}\in \mathrm{S}\text{and}\mathrm{t}\in \mathrm{S}\end{array}$

To prove n∈S for every positive integer n. 

The proof is given by INDUCTION:

Basis step: $1\in \mathrm{S}$, therefore P(k) is true.

Inductive step: Assume that P(k) is true.

Therefore, $k\in \mathrm{S}$

To prove that P(K+1) is also true.

Since, $1\in \mathrm{S}$and $k\in \mathrm{S}$.

By the definition of S.

k +$1\in \mathrm{S}$

Therefore, P(K+1) is true.

Thus, it is proved by the principle of mathematical induction that P(n) is true where n is a positive integer.