I reading about of proof of the claim "If $a \ge 0$ and $b > 0$, then $a \le ab$. (Here $a$ and $b$ are integers.) The proof the author is employing is inductive. I understand the basis case; however, I do not understand the proof the conditional relationship. The author assumes that it is true for the case for some arbitrary positive number $a = n$, and then wants to show that this implies that the statement is true for $n=1$. Here are the steps he takes to do just this:
"Suppose now that the proposition is true for $a = n \ge 0$. Then for all $b >0$, we have $n \le nb$, so $n+1 \le nb + 1 \le \underbrace{nb + b} = (n+1)b$."
I am not sure I understand the underbraced portion of the proof. He is the only way I can account for it:
Now, because we know that $b$ is greater than $0$, then it is possible for $b$ to assume the value of $1$ or any number greater than $1$. Thus, by replacing $1$ with $b$, we are either replacing $1$ with the same thing, or something greater. Such a replacement, then, keeps the inequality the $n+1 \le nb + 1$ the same, or it makes the left hand side even greater.
Could someone help me with this proof?
It is not true that for all $b>0$ and given $n\ge0$ that $n\le nb$. Take $b=\frac{1}{2}$ and $n\ge0$. I suspect this is for $b\in\mathbb{N}$. Then it is the case that $b\ge1$. This makes this inequality valid and it explains the portion you put a brace under.