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I know mathematical induction can be used to prove that a statements is true for all natural numbers (or those belonging to a certain subset of N). However, it is pretty obvious, unless I'm terribly mistaken, that this method can be expanded for any integer, as any integer can be expressed as a natural number or its opposite. However, since rational numbers are defined as a fraction of two integers, induction should also be expandable into rational numbers as well. If x is a rational number, it can be expressed as a/b where a,b are integers and b does not equal 0. So, if I want to prove a statement P(x)=P(a/b) is true for every x, is it sufficient to prove it for (a+1)/b and a/(b+1)? If not, would it be sufficient to also prove it for (a+1)/(b+1), or should a different method altogether be used?

Gepapado
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    What would the initial case be if you expanded induction to integers? – Michael Albanese Jul 23 '13 at 22:15
  • You could prove a prove that a statement is true of all rationals in $[0,1)$, then prove that if some statement is true of $r$ then it is also true of $r\pm 1$. You could also try an induction of the form where you show that some thing holds for $\frac{1}{n}$ then show that if something holds for $\frac{k}{n}$ then the property also holds for $\frac{k+1}{n}$. I do not know of any applications of this technique off the top of my head. – Baby Dragon Jul 23 '13 at 22:20
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    One should place a well ordering on the rational numbers (possible as there is a bijection with the natural numbers, although the ordering is not canonical). You can then replace the rational number $r\in\mathbb{Q}$ with its representative under this ordering $r=f(n)$. Now prove the statement for $f(n)$ in the usual manner. – Dan Rust Jul 23 '13 at 22:21
  • @DanRust Such a proof would be correct, but you need to prove the "f(n) implies f(n+1)" part of the induction proof, and in practice this is only possible if f has some sort of recurrence relation or recurrence property. If you know of such a bijection f, I'd be very interested. – Stef Aug 21 '22 at 10:12

5 Answers5

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You need to do a little more work, because as you noted in the case for integers you must take into account negatives. Thus it suffices to show the following:

  1. $P(0)$
  2. $P(x)\implies P(-x)$
  3. $P(a/b)\implies P((a+1)/b)$
  4. $P(a/b)\implies P(a/(b+1))$
Alex Becker
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  • Concerning integers, any x that is an integer can be equal to a y that is natural if x>=0, and x=-y if x<0. So if the statement can also be proven for x=-y for y and y+1, it is proven for every x, since when x>=0 it is the same as x being a natural. Concerning rationals, however, are (3) and (4) sufficient, or is it required to also prove that P(a/b) true => P((a+1)/(b+1)) true? It seems to me that the latter is redundant, but I'm not really certain. – Gepapado Jul 23 '13 at 22:37
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    The condition you have written is redundant. For rationals, the 4 conditions I have given are all necessary and together sufficient. – Alex Becker Jul 23 '13 at 22:41
  • Great! Thanks a lot! – Gepapado Jul 23 '13 at 22:47
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    Instead of $P(x)\implies P(-x)$, you could equivalently show $P(x)\implies P(x-1)$, or something else like that. (You just need some way to be able to reach any $P(x)$.) – AJMansfield Jan 29 '14 at 12:02
  • Sorry for being almost 6 years late to the party, but is rule 4 really necessary? For any b, we can assume P(0/b) holds after all, so by repeatedly applying rule 3 we can then get to an arbitrary P(a/b) right? – MrHug Dec 11 '19 at 11:35
  • @MrHug It is not necessary, check my answer. – Aleksandr Kalinin Apr 22 '23 at 06:43
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You can use induction on every infinite set, a way to do it is to endow the set with what is called a "well order". A well order is an order which has the property that every non-empty subset has a minimal element, and assuming the axiom of choice every set has a well-order.

So let us assume that $(X, <)$ is a well ordered set. Every element $x$ of $X$ has a successor $x^+$, by the properties of $<$ (indeed, the set $\{y \in X : x < y\}$ is non empty, hence it has a minimal element). Now if you can show that a property $P$ holds for an initial segment of $<$ (i.e a part $Y\subseteq X$ such that if $x\in Y$ and $y < x$ then $y\in Y$) and moreover that $P(x) \Rightarrow P(x^+)$ then you have $\forall x\in X, P(x)$.

For $\mathbb Z$ for example, a well-order could be $0 < 1 < -1 < -2 < 2 < 3 <-3 < \ldots$. And for the rationals, just take the well-order image of $\mathbb N$ by your favorite bijection between $\mathbb N$ and $\mathbb Q$ and it will work just as well.

zarathustra
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    I think by "every infinite set", you mean "every countable set"? – Josie Thompson Oct 08 '21 at 15:09
  • I think there are two errors in this answer. First: If I read your answer correctly, you are saying that even $\mathbb{R}$, the set of all reals, has an order such that every non-empty subset has a minimal element? – Stef Aug 21 '22 at 10:32
  • Second: Consider the set $X = { 1 - \frac{1}{n+1}, n \in \mathbb{N} } \cup { 2 - \frac{1}{n+1}, n \in \mathbb{N} }$, with the usual order $<$. It is easy to see that $(X, <)$ is a well-ordered set, and the successor function is the function that maps $1-1/n$ to $1-1/(n+1)$ and $2-1/n$ to $2-1/(n+1)$. And yet it is easy to come up with a property $P$ such that $P(0)$ and $\forall x,, P(x) \Rightarrow P(x^{+})$, but $P$ is false for the second "half" of $X$. – Stef Aug 21 '22 at 10:38
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There are many ways.

  1. Perhaps the problem allows for an enumeration $a_n$ of the rationals. Then you use induction on $n$.
  2. Another way could be as a double induction in increasing numerators and denominators. You would prove the initial case for $0/1$ and then assuming the statement for $a/b$ prove it for $(a+1)/b$ and $a/(b+1)$. Or increasing factors of numerator and denominator.

It depends on the problem. But in a way what you are using is induction on the naturals.

OR.
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If you are only interested in $\mathbb Q_+$ then it is enough to have this:

  1. $P(1)$
  2. $P(x) \implies P(x + 1)$
  3. $P(x) \implies P({1 \over x})$

This "induction over $\mathbb Q_+$" principle can be justified using an interesting enumeration of $\mathbb Q_+$ based on continues fraction decomposition [1]. The enumeration has some nice properties:

  1. It justifies "induction over $\mathbb Q_+$" principle.

  2. It gives fast (logarithmic) and explicit way to get assign a natural number to any rational one and vice versa.

  3. It has some smoothness properties in some precise sense.

Clearly, if you want to use that enumeration for in $\mathbb Q$ then you can change #1 to

1'. $P(0)$

and add

  1. $P(x) \implies P(-x)$

[1] Д.Н.Андреев Об одной замечательной нумерации положительных рациональных чисел. Сборник "Математическое Просвещение" Третья серия Выпуск 1

Aleksandr Kalinin
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I think your statement is a bit imprecise. Let's say your hypothesis is $p:\mathbb{Q}\rightarrow\left\{0,1\right\}$ ($0$ denotes falsehood, $1$ denotes truth). If you can find a bijection $f\colon \mathbb{N} \rightarrow \mathbb{Q}$ between $\mathbb{N}$ and $\mathbb{Q}$ with:

  1. $p\left(f\left(0\right)\right)=1$
  2. $p\left(f\left(n\right)\right)=1 \implies p\left(f\left(n + 1\right)\right)=1$

Then you are done.

parsiad
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