Precalculus (6th Edition) Blitzer

Let us assume ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{1}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $k$. Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $n$. Firstly, show that ${{S}_{1}}$ is true. And write ${{S}_{1}}$ by taking the its term on left and substituting n with $1$ on the right. ${{S}_{1}}:n+2>n$ \begin{align} & 1+2>1 \\ & 3>1 \end{align} This is true. Therefore, the statement ${{S}_{1}}$ is true. Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction. ${{S}_{k}}:k+2>k$ Now, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}:k+1+2>k+1$ $k+3>k+1$ Since, one assume that ${{S}_{k}}$ is true, $k+2>k$ By adding 1 on both sides, one get \begin{align} & k+2+1>k+1 \\ & k+3>k+1 \end{align} So, the final statement is ${{S}_{k+1}}$. Thus, by the principal method of of mathematical induction, the statement $n+2>n$ is true for every positive integer $n$.