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In What is mathematics? by Courant, Robbins, and Stewart, "5. An important inequality", the authors change $n$ in this example:

$(1+p)^n\geq1+np$

to $r$ in this example:

$(1+p)^r\geq1+rp$

In other examples given by the book's author, he also switches the variable. I also recall seeing something similar in some other book. Why would one do that?

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    Quite often a sum will be from $r$= 0 or 1 to $n$, and the terms being summed will be expressed in terms of $r$. (It isn't so much a convention as what lots of people do.)2012-08-15

2 Answers 2

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Courant & Robbins' style in that book was to reserve the letter $n$ in an inductive argument for the general case, like this:


We want to prove that some statement $P(n)$ is true for all natural numbers $n$.

First we prove that it is true for $P(0)$.

Now we assume it is true for some natural number $r$, and we will prove it must also be true for $r+1$. That is, we will show $P(r)\implies P(r+1)$.


Here the authors are trying to make explicit the idea that $n$ represents any natural number, and in their induction step they are choosing one particular natural number $r$. This is simply the way that induction is introduced in the book at the top of page 11.

The essential idea in the preceding arguments is to establish a general theorem $A$ for all values of $n$ by successively proving a sequence of special cases, $A_1, A_2, \dots$ . The possibility of doing this depends on two things: a) There is a general method for showing that if any statement $A_r$ is true then the next statement, $A_{r+1}$, will also be true. b) The first statement $A_1$ is known to be true.

Once you are comfortable with this idea, many people simply use $n$ in the second step as well, remembering that in that second step $n$ represents a particular natural number.

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    @GustavoBandeira Congrats! What is Mathematics is a wonderful book - one of my favorites.2012-08-15
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The authors are proving this by induction. They switch from $n$ to $r$ to help readers who would be confused if they assumed something for $n$ when they're trying to prove it in the first place. So they assume the result for $r$ and show how to extend it to $r+1$, thus proving it for all $n$. No big deal.