0

I'd like to clarify the definition of mathematical induction, with the context of functions:

To define a function f on all nonnegative integers $x$ it suffices to define:

  • $f(0)$, the function on $0$, and
  • $f(n + 1)$ in terms of $f(n)$, for all nonnegative integers $n$.

So for example, with a function $f(n) = n + 2$, is it accurate to say:

  • $f(0) = 0 + 2 = 2$
  • $f(n) = n + 2$
  • $f(n+1) = f(n) + 1 = (n + 2) + 1$, since the function is a linear function and can be expressed in terms of $f(n)$.
  • Therefore, all numbers in $n$ have the property, $f(n)$.
JMP
  • 21,771
fncischen
  • 31
  • You problem is related to induction, but induction I suspect, is not enough to show that the function exists (it should suffice to show it is unique): https://math.stackexchange.com/questions/1421626/halmos-naive-set-theory-recursion-theorem-proof-why-must-he-do-it-that-way?rq=1 – William M. May 16 '17 at 19:17
  • What do you mean when you say that "all numbers in $n$ have the property $f(n)$"? First, do you mean "all numbers $n$" or "all numbers in $\Bbb N$", perhaps? Also, what does it mean for a number to have the property $f(n)$, when $f(n)$ is the function $n+2$? To me, your last line is like saying, "all numbers in 17 have the property 19", which makes little sense. – Théophile May 16 '17 at 19:18
  • There is confusion between recursive definition (The function is not really a rule so to speak) to to the mothod of induction that can sometimes be used to extract a rule from the recursive definition – user76568 May 16 '17 at 19:30
  • @Théophile I originally should have stated all numbers in ℕ. – fncischen May 16 '17 at 22:54

2 Answers2

2

Your $f$ is a function, not a property. You are confusing inductive proofs with inductive definitions. To define $f(n)$ inductively, don't start with $f(n) = n +2$. That's already a definition, so you don't have to re-define it inductively. Instead, define $f(n)$ by

$f(0) = 2$, $f(n+1) = f(n) + 1$ for $n > 0$.

Then you can prove that $f(n) = n+2$ using an inductive proof.

Jair Taylor
  • 16,852
  • Don't you mean "recursive definitions" ? – user76568 May 16 '17 at 19:36
  • I think either usage is okay. e.g., Wikipedia says "A recursive definition (or inductive definition) in mathematical logic and computer science is used to define the elements in a set in terms of other elements in the set." – Jair Taylor May 16 '17 at 20:37
  • Thank you, @JairTaylor. In other words, could you say that you are using the inductive definition of the basis f(0) and f(n+1) (which can be expressed in terms of f(n)) to prove that the property f(n) = n+2 is true? – fncischen May 16 '17 at 22:37
  • Yes, you can use the inductive definition of $f(n)$ to prove that $f(n) = n+2$, if that is what you mean. – Jair Taylor May 16 '17 at 22:44
1

I am worried by the private and non-standard use of terminology here.

The word property has a precise meaning in mathematics; a property is a predicate that can be true or false. An utterance such as

"Therefore, all numbers in $n$ have the property, $f(n)$."

is quite simply nonsense.

What would it mean to say that e.g. "all numbers in $4$ have the property $6$"?

The kind of recursive definition alluded to in the question is a simple form of what is known as primitive recursion: We define $f(0)$ directly and we define $f(n+1)$ as a simple function of $n$ and $f(n)$.

It is not the case that every function $f: \mathbb{N} \rightarrow \mathbb{N}$ can be expressed using primitive recursion. A simple way to see this is to remember that every definition of this form is a description of finite length over some finite alphabet. There are only countably infinitely many such descriptions, but there are uncountably many functions over the natural numbers.

Hans Hüttel
  • 4,271
  • Thank you for the clarification, Hans, the clarification regarding definitions really is something I'm going to look out for! – fncischen May 16 '17 at 20:44
  • To clarify the last paragraph, @Hans-Hüttel, what you are saying is f:ℕ→ℕ cannot be expressed in terms of primitive recursion since you can have an infinite amount of functions for all n in ℕ? For example, you can say f(n) = 2*n + 5 or f(n) = n/5 + 8. These two functions have different sets of x and y values. Thus, you can express individual functions with a unique set of numbers in terms of primitive recursion, but not f:ℕ→ℕ. – fncischen May 16 '17 at 23:06
  • It makes no sense to say things like "You can have an infinite amount of functions for all $n \in \mathbb{N}$" or "these two functions have different sets of x and y values". Please use the same language for mathematics as everyone else. – Hans Hüttel May 17 '17 at 01:09
  • @fncischen it'll take some work on our behalf to "clarify that last paragraph". Worth it though. – user76568 May 17 '17 at 17:41
  • Moreover, the cardinality of the set of all functions $\mathbb{N} \to \mathbb{N}$ is exactly the cardinality of all infinite sequences of natural numbers. Intuition says this cardinality is at least that of $(0,1)$ which is $C$. – user76568 May 17 '17 at 17:50
  • @HansHüttel, after reviewing countable and uncountable sets, and the last paragraph, essentially, while addition/division/multipication/subtraction can be expressed primitively recursively (https://en.wikipedia.org/wiki/Primitive_recursive_function), there are functions that exist that cannot be expressed primitively recursively (the first thing that pops up is the Ackermann function). Therefore, it would be inappropriate to state all numbers in ℕ can be mapped f:ℕ→ℕ & expressed by primitive recursion due to the existence of Ackermann function? – fncischen May 17 '17 at 21:00
  • @Dror, which two sets' cardinalities are you referring to, in regards to (0,1)? – fncischen May 17 '17 at 21:00
  • What does "all numbers in ℕ can be mapped f:ℕ→ℕ & expressed by primitive recursion " mean? Sorry, but you prefer to use mathematical terminology in a private and impenetrable way. Primitive recursion is a definitional mechanism that can be used to define a function, not a mechanism that can be used to define a number. – Hans Hüttel May 17 '17 at 21:11
  • $(0,1) \subset \mathbb{R}$ , i.e all real numbers between $0$ and $1$. It has cardinality $2^{\aleph_{0}} = C $ (Continuum). The other set is the set of all sequences of natural numbers. – user76568 May 18 '17 at 03:52