And so if we have $f:(m+a+1) o (m+1)$, then it is also $((m+1)+a) o(m+1)$, and its restriction to $(m+1)$ is not injective. By this time we hopefully know that addition is commutative and associative, so $m+a+1=m+1+a$. Now to complete the proof, we just have to handle the case where $a e 0$ in $|m+a+1|>|m+1|$. But then either $f(p)$ or $f(q)$ equals $f(m+1)$, and $f$ is again not injective. Otherwise $f(p) e f(q)$, but this can only be the case if one of them is $m$ and the other is $b$.

If $f(p)=f(q)$ then $f$ is not injective, and we're done. Then by the induction hypothesis $g$ is not injective, so there is $p$ and $q$ with $g(p)=g(q)$. $0$ is the empty set, so there is no function for $1 o 0$ at all, so in the impossible case that we get one, we can safely claim it will be non-injective.įor the induction case, assume that $|m+1|>|m|$ and we need to show that $|m+1+1|>|m+1|$. The actual content of the proof now is to show that these two competing definitions of $>$ agree! Of course we first need to define addition, but once we have done that, it is a fairly simple matter of induction.įirst we prove that $|m+1|>|m|$ (as cardinals) for all $m$. This would be trivial (and pointless) if we're using the cardinality definition of what $m>n$ means - so even though that is the most common choice, let's assume that we want to define $m>n$ to mean 'there is a natural $a$ such that $n+a+1=m$. Then all we have to do is to prove this from a definitions of 'has $m$ elements' and '$mn$ and $f:m o n$, then $f$ is not injective. If $m$ and $n$ are naturals, and $m>n$, and $A$ has $m$ elements and $B$ has $n$ elements, and $f$ is a function $A o B$, then $f$ is not injective. Of course, once we define natural numbers, we might want to prove a 'finite pigeonhole principle':

So if we use that as our definition, the pigeonhole principle is not a matter of proof - instead it is part of the definition of what it means for one set to be larger than the other. There is an injective function $B o A$, but there is no injective function $A o B$. In particular, what does '$A$ has more elements than $B$' mean? In the usual development of set theory we use this phrase to mean the same as '$A$ has larger cardinality than $B$', which again means If $A$ and $B$ are sets, and $A$ has more elements than $B$, and $f$ is a function $A o B$, then $f$ is not injective.Ĭan it be proved? That depends. A fairly precise statement of the pigeonhole principle would be: This can be reworded as: if m pigeons occupy n pigeonholes, where m > n, then there is at least one pigeonhole with two or more pigeons in it.Yes and no.

Then, under any assignment of objects to the boxes, there will always be a box with more than one object in it. Theorem 1 (The Pigeonhole Principle) : Let there be n boxes and (n+1) objects. This is actually an application of the pigeonhole principle, which we now state. Wouldn’t you agree that regardless of the way the objects are placed in the two boxes at least one box will have more than one object in it? On the face of it, this seems obvious. ANSWER – Let us start with considering a situation where we have 10 boxes and 11 objects to be placed in them.

Pigeonhole principle