"Fixed $k$" in Mathematical Induction

Question

On page 34, in his Calculus book, Apostol gives the following description of proof by induction:

Method of proof by induction. Let $A(n)$ by an assertion involving an integer $n$. We conclude that $A(n)$ is true for every $n \ge n_1$ if we can perform the following two steps:

a. Prove that $A(n_1)$ is true.
b. Let $k$ be an arbitrary but fixed integer $ \ge n_1 $. Assume that $A(k)$ is true and prove that $A(k+1)$ is also true.

I'm not new to proof by induction, but I don't understand the intention behind $k$ being "an arbitrary but fixed integer": isn't it necessary to prove, in the inductive step, that it is the case for all integers $k \ge n_1$ that if $A(k)$ is true, then $A(k+1)$ is true as well?

Answer 1

I think you're the statement of the author more complicated than need be, or reading "too much" into "arbitrary but fixed." [I'd have preferred to see the word "but" omitted, because "but" makes it seem that "arbitrary and fixed" are at odds in what they mean, and they really aren't! It is just a way of saying: "for any".]

The key word is arbitrary: we consider, in the inductive hypothesis, any (arbitrary) integer $ \geq n_1$ and call it $k$. Then we make an assumption (the inductive hypothesis) about $k$, which we then hold as fixed, so that in the inductive step, when considering $P(k + 1)$, "$k+1$" makes sense: our hypothesis is assumed true for the arbitrarily chosen $n = k$, which we "fix" so that we can consider the hypothesis about $n = k+1$.

The point of a proof by induction is to be able to say something, if the proof succeeds, about every $n\geq n_1, n\in \mathbb N$. The proof of $P(n)$ itself needs only to consider a base case $P(n_1)$, some (any) one arbitrary integer $k\geq n_1$ for which we assume $P(k)$, and then the what this means for $P(k+1)$, where $k+1$ is the immediate successor of that arbitrarily chosen, then fixed $k$.

Answer 2

I'd like to suggest the following explanation of the phrase "arbitrary but fixed" and along the way I'll also try to categorize some of our uses of variables.

Consider the following sentences:

Case 1

1. Let $\mathbb{N}$ be the domain of the variable $x$.
2. It is the case that $x+2=7$.

Notice that we never assigned a value to $x$. Thus, in this case, $x$ is not really a 'variable value' rather it can be thought of as a blank or a placeholder for some future value. The only restriction we do have (given bySentence 1) is that whenever we do assign a value to $x$, it be some element of $\mathbb{N}$. Since it has this blank or 'hole', logically, Sentence 2 is known as an open sentence, that is, it is not actually asserting anything since $x$ is 'empty'.

How do we 'fill the hole' in Sentence 2 so that it actually makes some sort of claim? By means of quantification; particularly, by using the $\forall$ (for all) and $\exists$ (exists) quantifiers. Applying these to Sentence 2, we can revise the sentence and say

Case 2

For all $x$, it is the case that $x+2=7$.

Here, we are assigning a value to $x$. In this case, we are actually ('iteratively') assigning every value of $\mathbb{N}$ to $x$, and it is as if we are claiming that "${0+2=7}$ and $1+2=7$ and $2+2=7 \ldots$ An obviously false statement, but an assertion nonetheless. Using the variable $x$ rather than the latter explicit form, clearly saves us a lot of writing. This is an example of using a variable as shorthand.

Likewise, we can state Sentence 2 as

Case 3

There exists some $x$ such that $x+2=7$.

Again, we are 'filling the blank', but in this case, we are only claiming that at least a single value of $\mathbb{N}$ which $x$ can take on will satisfy the formula. Why use a variable instead of explicitly using a single value? Well, we may know it to be the case that $x+2=7$ from some prior axiom or theorem, but we may not know which value of $\mathbb{N}$ $x$ takes on; hence, the use of a variable. This is an example of a variable being used for an unknown.

Now, let us consider our case, in which we would like to prove for all $k$ that if $A(k)$ is true, then $A(k+1)$ is true as well.

Well, as Qiaochu Yuan points out, using $k$ with universal quantification to refer to all of $\mathbb{N}$ collectively, would force us repeatedly say things such as "for all $k$ ..." throughout our proof; pretty painful to say the least. We may also, fallaciously, attempt to formulate our proof as follows:

Let $k$ be a variable over the domain $\mathbb{N}$.

"Proceed to prove that if A(k) is true, then A(k+1) is also true."

This approach obviously fails since we are leaving $k$ 'open'.

How can we construct a logically valid but clear proof? We choose a single element of $\mathbb{N}$ and carry out our proof on this single element showing that if the assertion $A$ is true for this element, then it is true for the 'next' element as well. To complete our proof for all $\mathbb{N}$, we (implicitly) extrapolate our proof to the rest of the natural numbers. To this end, when selecting the number on which to execute our proof, we make an arbitrary choice and prefer to leave the number anonymous. This emphasizes the fact that nothing about our proof depends on the particulars of a specific number, rather all that is assumed about our number is the fact that it is a member of $\mathbb{N}$. This makes it easy for us to see that our proof can be extended to all other natural numbers. How do we 'anonymize' our choice in $\mathbb{N}$? By using a variable as in Apostol's

Case 4

Let $k$ be an arbitrary but fixed integer $\ge n_1$.

This is an example of a variable being used for 'arbitration' and 'anonymization'.

Now, in this last case, where we make an arbitrary choice of $k$ from $\mathbb{N}$ for our proof, one may misunderstand our approach as follows: since we believe our proof will hold for any element of $\mathbb{N}$, (rather than lay out the proof for a single number and then generalize for all natural numbers), we use as the object of our proof a single 'abstract' natural number, $k$, which suffices as proof for the entire set $\mathbb{N}$.

Such an approach would definitely suffer from being 'open', and is not logically valid. I believe the phrase "arbitrary but fixed" was coined to steer us away from this misconception: yes, our choice of number is "arbitrary" and in fact we are not even naming it, yet do not be misled to think we are using an 'abstract' $k$, rather "$k$ is fixed", and it is from this fixed $k$ that we will subsequently extend our proof to all of $\mathbb{N}$.

Answer 3

$k$ is a sort of index function. You may want to prove your statement for $n_1+1, n_1+2,n_1+3,...$. The point is that if the statement works for a number and the following numbers, then it works for every number above that one. (simply by rescaling $\tilde{k}=k+1$)

Answer 4

The principle of induction says that if $P(n)$ is a proposition about a natural number $n$, and you can prove both of these:

• $P(0)$ holds

• For every natural number $k$, if $P(k)$ holds then $P(k+1)$ holds

then that is enough to show that for all $n$, $P(n)$ holds. So, yes, it is necessary to prove the second bullet for all natural numbers $k$ (although the thing that has to be proved is only "if $P(k)$ then $P(k+1)$", which is often simpler than trying to directly prove $P(k)$).

There is a standard way to prove a statement of the form "for every natural number $k$, ...". You assume that you have been given a fixed but unknown natural number $k$, and you prove the "..." part for that (unknown) number. That is the method that Apostol is alluding to when he says

Let k be an arbitrary but fixed integer ≥n1. Assume that A(k) is true and prove that A(k+1) is also true.

So this general method is not something particular to induction, it is just the general method for proving a universal statement. It just happens that the second bullet that you have to prove to do a proof by induction is a universal statement.