Are proofs by induction limited to cases where there is an explicit dependance on a integer, like sums? I cannot grasp the idea of induction being a proof in less explicit cases. What if you have a function that suddenly changes behavior? If a function is positive up to some limit couldn't I prove by induction that it is always positive by starting my proof on the left of the limit?

I feel like I am not getting a grasp on something fundamental and I hope you can help me with this. Thank you.

  • $\begingroup$ No there's such a thing as transfinite induction, where you can extend to well-ordered sets. $\endgroup$ Oct 17, 2012 at 19:28
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    $\begingroup$ Do you know a little bit about graph theory? I can provide some nice examples of induction on graphs. $\endgroup$ Oct 17, 2012 at 19:31
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    $\begingroup$ Sure you can show the function is positive to the left of your limit, but you will have trouble showing the inductive step precisely because of the sudden behavior change. $\endgroup$
    – EuYu
    Oct 17, 2012 at 19:32
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    $\begingroup$ It is useful to think of induction proofs as an "outline" for an infinite length proof. In particular, what you a providing is a way to write a proof for any particular $n$. For example, say you've proven $1+2+...+n=n(n+1)/2$ by induction. We can think of this as giving me a 'program' to write a proof for, say, $n=6$ or $n=100000$ or $n=10^100$. $\endgroup$ Oct 17, 2012 at 19:40
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    $\begingroup$ @user948652 I don't want to put words in Thomas' mouth, but I doubt he intended to say that induction is a "weak" form of proof. The heart of the proof is showing that the pattern you perceive does indeed hold everywhere. It is a perfectly valid and useful form of proof. $\endgroup$ Oct 17, 2012 at 19:56

6 Answers 6


The title suggests puzzlement about the very idea of a proof by induction (maybe I am misconstruing the question). Anyway, let's take the simplest case, where we want to show that all natural numbers have some property $P.$ We obviously can't give separate proofs, one for each $n$, that $n$ has $P$, because that would be a never-ending task. So how can we proceed?

Suppose we can show that (i) $0$ has some property $P$, and also that (ii) if any given number has the property $P$ then so does the next: then we can infer that all numbers have property $P$. Using $\varphi$ for an expression attributing some property to numbers, we can put the principle like this:

Given (i) $\varphi(0)$, and (ii) $\forall n(\varphi(n) \to \varphi(n + 1))$, we can infer $\forall n\varphi(n)$.

And the headline question seems to be: Why are arguments which appeal to this sort of principle good arguments?

Well, suppose we establish both the base case (i) and the induction step (ii). By (i) we have $\varphi(0)$. By (ii), $\varphi(0) \to \varphi(1)$. Hence we can infer $\varphi(1)$. By (ii) again, $\varphi(1) \to \varphi(2)$. Hence we can now infer $\varphi(2)$. Likewise, we can use another instance of (ii) to infer $\varphi(3)$. And so on and so forth, running as far as we like through the successors of $0$ (i.e. through the numbers that can be reached by starting from zero and repeatedly adding one). But the successors of $0$ are the only natural numbers. So for every natural number $n$, $\varphi(n)$.

The arithmetical induction principle is underwritten, then, by the basic structure of the number sequence, and in particular by the absence of 'stray' numbers that you can't get to step-by-step from zero by applying and reapplying the successor function.

So much for the simplest case. But the principles underlying fancier inductive arguments (not over numbers but e.g structural inductions or transfinite inductions) are similar enough to become clear if you are really clear about this simple case. The crucial thing, in each case, will be to find an appropriate induction step which tells that a certain property gets passed down from 'predecessors' to 'successors', and so ripples through the whole of the relevant domain.

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    $\begingroup$ I think the very last sub-clause is the most relevant statement here. For the induction principle, the order in which you deal with the elements does not matter, as long as you cover the whole domain in the limit. $\endgroup$
    – Raphael
    Oct 17, 2012 at 20:26
  • $\begingroup$ All of the answers here have cleared my confusion in one way or another but accepting yours as it was most clearly written. Thank you all! $\endgroup$
    – arynaq
    Oct 17, 2012 at 20:38

The Principle of Mathematical Induction is equivalent to the Well-Ordering Principle, which states that every non-empty set of positive integers has a least element. You either assume (as an axiom of your number system) PMI and prove WOP from this assumption, or vice versa.


All forms of mathematical induction $-$ ordinary induction, so-called strong induction, structural induction, transfinite induction, etc. $-$ can usefully be thought of in the following way, in terms of the non-existence of a minimal counterexample to the theorem being proved. This view makes it easier to avoid the error of thinking of induction as a sequential process of some kind.

I have some collection $A$ of objects and some kind of ordering of the objects, which I’ll denote by $\preceq$, that has the following property:

$(*)\quad$ if $C$ is any non-empty subset of $A$, there is a $c_0\in C$ that is minimal with respect to $\preceq$, meaning that if $c\in C$ and $c\preceq c_0$, then $c=c_0$. In other words, no element of $C$ is strictly less than $c_0$ with respect to the ordering $\preceq$. Equivalently, if $a\in A$ and $a\prec c_0$, then $a\notin C$. (Here $a\prec c_0$ is short for $a\preceq c_0\text{ and }a\ne c_0$.)

There is some property that objects in $A$ might possess; call it $P$. I want to prove that every $a\in A$ has property $P$, so I let $B=\{a\in A:a\text{ does not have }P\}$ and assume that $B\ne\varnothing$; my goal is then to get a contradiction, because that will imply that $B$ must actually be empty and hence that every $a\in A$ has $P$.

I first apply $(*)$ to $B$ to conclude that there is a $b_0\in B$ that is $\preceq$-minimal: if $a\in A$ and $a\prec b_0$, then $a\notin B$. If you think of $B$ as the set of ‘bad’ objects, meaning ones that don’t have the property $P$, then $b_0$ is a minimal bad object: any objects earlier in the order are ‘good’, meaning that they have the property $P$. I then show that

$(\dagger)\qquad\qquad\quad$ if $a\in A$, and every $a'\prec a$ has property $P$, then $a$ has $P$ as well.

My minimal ‘bad’ element $b_0$ satisfies the hypothesis of $(\dagger)$: every $a\prec b_0$ has property $P$. Thus, $(\dagger)$ implies that $b_0$ has property $P$, and I have my contradiction.

The argument can be summarized as follows. The nature of $A$ and $\preceq$ means that if my theorem (that every $a\in A$ has $P$) is false, there must be a minimal counterexample, $b_0$. But I’m able to prove $(\dagger)$, which implies that there can be no minimal counterexample. Thus, there cannot be any counterexample at all, and the theorem must therefore be true. Many proofs by induction in more advanced settings are actually worded in just this way: Suppose that the result is false, and let $\text{thing}$ be a minimal counterexample. Then ....

In ordinary induction $A$ is typically $\{n\in\Bbb Z:n\ge m\}$ for some fixed integer $m$. Often $m=1$ and $A=\Bbb Z^+$, but examples are generally presented very early to show that the induction doesn’t have to start at $1$. The ordering $\preceq$ is simply ordinary $\le$ on the integers: if $C$ is any nonempty subset of $\{n\in\Bbb Z:n\ge m\}$, then $C$ has a minimal element, so $(*)$ is satisfied. (In fact $C$ even has a minimum element.) The induction step is proving $(\dagger)$ for those $a\in A$ that actually have a predecessor in $A$, and the basis step is proving it for the one $a\in A$ that has no predecessor. (If $A=\{n\in\Bbb Z:n\ge m\}$, the unique element of $A$ with no predecessor is $m$.)

Technical Note: $(*)$ is a somewhat informal statement of the requirement that the partial order $\langle A,\preceq\rangle$ be well-founded.


Induction is normally defined for the case when you have an explicit dependence on an integer (or an ordinal for transfinite induction, but I'm not going to talk about that). I'm not sure what you're talking about with the function, but since induction requires you to prove that $P(k)\Longrightarrow P(k+1)$ for all $n$, you wouldn't be able to 'prove' that the function was everywhere positive, because the statement $P\Longrightarrow P(n+1)$ would fail at the point where we cross the 'limit point'.

You can prove that proof by induction is a proof as follows:

Suppose we have that $P(1)$ is true, and $P(k)\Longrightarrow P(k+1)$ for all $n\ge 1$. Then suppose for a contradiction that there exists some $m$ such that $P(m)$ is false. Let $S=\{n\in \mathbb{N} : P(k) \text{ is false}\}$. $S$ is non-empty (since it contains $m$), so it has a least element $s$ (The statement that every non-empty set of natural numbers has a least element is known as the Well Ordering Principle).

Now, since $P(1)$ is true, $s\neq 1$. So $s-1$ is a natural number. Now, if $s-1$ were true, then $P(k)\Longrightarrow P(k+1)$ would imply that $s$ were true (setting $k=s-1$). Since $s$ is not true, $s-1$ must not be true as well. But $s-1<s$, and we had supposed $s$ to be the smallest natural number $n$ such that $P(n)$ was not true. So we have a contradiction.

Therefore, $P(n)$ is true for all $n\ge 1$.

It turns out there actually is an analogue of induction which you can use if you want to prove that a statement is true for all real numbers. You can't apply the above argument to the real numbers, as the real numbers don't satisfy the Well Ordering Theorem. For example, the set $\{1,0.1,0.01,0.001,\dots\}$ doesn't have a smallest element. Neither does $\{x\in \mathbb{R} : x>3\}$. However, the real numbers do satisfy what is known as the Greatest Lower Bound property. This means that every set of real numbers which is bounded above has a greatest lower bound. For example, the set of real numbers that are greater than $3$ has no least element: if you take some number greater than $3$ - $3.001$, say - I can tell you another number that is greater than $4$ but still less than your number (like $3.0001$). But it does have a greatest lower bound, which is $3$. The set $\{x\in\mathbb{R} : x^2 > 2\}$ has no least element, but it has a greatest lower bound, which is $\sqrt{2}$.

Exercise: what is the greatest lower bound for the set $\{1,0.1,0.01,0.001,\dots\}$?

In general, a lower bound for a set is a number $b$ that is smaller than every element in the set. The Greatest Lower Bound Property is the statement that whenever you've got a lower bound, there is always a greatest lowers bound. We denote the greatest lower bound of a set $S$ by $\inf S$ ($\inf$ is short for infimum, another word for greatest lower bound).

How can we use the Greatest Lower Bound Property to form an analogue of induction? I'll give you the statement of continuous induction first, and then prove it.

Theorem Let $P(x)$ be a statement about an arbitrary real number $x$. Suppose we know the following two facts:

  • $P(x)$ is always true if $x<y$, where $y$ is a fixed real number.
  • If $P(x)$ is always true up to some number $c$ (whenever $x<c$) then, for some $\varepsilon > 0$, $P(x)$ is always true up to $c+\varepsilon$ (whenever $x<c+\varepsilon$).

Then $P(x)$ is true for all real numbers $x$.

The proof is similar to the proof for natural-number induction:

Proof: Suppose for a contradiction that, for some real number $z$, $P(z)$ is false. Let $A=\{x\in \mathbb{R} : P(x) \text{ is false}\}$. Since $P(x)$ is true for all $x<y$, $y-1$ is a lower bound for $A$. Since $z\in A$, $A$ is non-empty. So $A$ has a greatest lower bound $a=\inf A$. Now, if $x<a$, then $P(x)$ is true (as $a$ is a lower bound for $A$, the set of real numbers $x$ such that $P(x)$ is false). Therefore, for some $\varepsilon > 0$, $P(x)$ is true for all $x<a+\varepsilon$. Therefore, $a+\varepsilon$ is a lower bound for $A$. But $a+\varepsilon>a$, and we had supposed $a$ to be the greatest lower bound for $A$. So we have a contradiction.

So $P(x)$ is true for all real numbers $x$.

This is really the same thing as proof by induction for integers, but using the Greatest Lower Bound Property instead of the Well Ordering Principle. It is worth mentioning that I have never had occasion to use continuous induction over real numbers, and it's not a standard technique: mathematicians tend to prove statements about the real numbers straight from the Greatest Lower Bound Principle, and not use any fancy tricks (just as you can turn any proof by induction into a proof where you try to find a 'minimal counterexample', as I did when I proved that induction works above). But it's quite fun, and it shows that you can generalize things like the induction principle on to more general sets, as long as you have some statement similar to the Well Ordering Principle.

  • $\begingroup$ In your theorem about the greatest lower bound, I think you need to reverse some quantifiers in the second bullet point. It's crucial that the $\epsilon$ not depend on $c$, so I think you should rephrase as "There is an $\epsilon$ such that if $P(x)$ is always true up to some number $c$....". $\endgroup$ Oct 17, 2012 at 22:05
  • $\begingroup$ @Jason - it is not crucial that the $\varepsilon$ not depend on $c$ - that is the power of this method and is the reason we need to use the Greatest Lower Bound property. If $\varepsilon$ were just a constant independent of $c$, then it is trivial that $P(x)$ is true for all $x\in\mathbb{R}$ - just partition the real numbers up into countably many subintervals of size $\varepsilon$ and use induction to show that $P(x)$ is true over each subinterval. If you look at the proof of continuous induction, then you will see that the only fact I used about $\varepsilon$ was that it was positive. $\endgroup$ Oct 18, 2012 at 11:41
  • $\begingroup$ I see - I had misunderstood what you were saying. I was thinking "True up to $c$" meant true for $x\leq c$ (despite the fact that you wrote after that $(x<c)$ ). Sorry about that! I can delete my previous comment if you prefer. $\endgroup$ Oct 18, 2012 at 12:20

This math.SE question has a lot of great answers as far as induction over the reals goes. And as Austin mentioned, there are many cases in graph theory where you can use induction on the vertices or edges of a graph to prove a result. an example:

If every two nodes of $G$ are joined by a unique path, then $G$ is connected and $n = e + 1$ Where $n$ is the number of nodes and $e$ is the number of edges.

$G$ is connected since any two nodes are joined by a path. To show $n = e + 1$, we use induction. Assume it’s true for less than $n$ points (this means we're using strong induction). Removing any edge from $G$ breaks $G$ into two components, since paths are unique. Suppose the sizes are $n_1$ and $n_2$, with $n_1 + n_2 = n$. By the induction hypothesis, $n_1 = e_1 + 1$ and $n_2 = e_2 + 1$; but then$$n = n_1 + n_2 = (e_1 + 1) + (e_2 + 1) = (e_1 + e_2) + 2 = e − 1 + 2 = e + 1$$

So our statement is true by strong induction

  • $\begingroup$ Yes you are right. I will edit it. $\endgroup$
    – Jemmy
    Oct 17, 2012 at 19:59

Here is another example from graph theory.

Claim: There is a $4$-regular graph on $n$ vertices for all $n \geq 5$.

Base Case: The complete graph on $5$ vertices is four regular.

Inductive Step: Assume there is a $4$-regular graph $G$ on $k$ vertices. We want to show there is a $4$-regular graph on $k+1$ vertices. Remove any two disjoint edges $e$ and $f$ from $G$. The resulting graph is almost $4$-regular, except it has four vertices of degree $3$ (the vertices involved with the edges $e$ and $f$). Introduce a new vertex to $G - \{e,f\}$ (so now $G$ has $k+1$ vertices) and connect it to each of the aforementioned degree $3$ vertices. The result graph is now $4$-regular, as desired.

How does this proof work? We showed explicitly there is a $4$-regular graph on $5$ vertices. The inductive step tells us how to take this graph and turn it into a $4$-regular graph on $6$ vertices. Applying the inductive step again would take the $4$-regular graph on $6$ vertices and turn it into a $4$-regular graph on $7$ vertices, and so on for any number of vertices you like.

The reason this all works is because my induction step does not depend on any properties of particular values of $k$. It goes through no matter what integer I set to $k$. This assures me that there won't be a sudden change in behavior of the graphs.


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