Let be a finite set, with say. Suppose we have a sequence of elements of , with . As only has elements, it is clearly impossible for all the elements to be different. Thus, we can find with .
Let be a finite set, with say. Suppose we have a list of subsets (called pigeonholes) such that every element of lies in one of these subsets. Suppose that (so the number of elements is greater than the number of pigeonholes). Then there exists such that .
List the elements of as . Each element lies in some pigeonhole, so we can choose such that . We now have a sequence in . As , these numbers cannot all be different, so we can choose such that say. This means that and , so . ∎
Show that there are two people in the world with the same number of hairs.
Let be the number of people in the world, so . Let be the maximum number of hairs that anyone has. A typical person has about hairs, so it would be safe to assume that , and certainly is very much smaller than . Let be the number of hairs on the ’th person, so . The sequence is much longer than the size of the set , so there must exist such that as claimed.
As another way to say essentially the same thing, we can divide all people into pigeonholes , where is the set of people with hairs. The number of pigeonholes is smaller than the number of people, so there must be some pigeonhole that contains two people. If contains two people, then those two people have the same number of hairs.
Suppose that people have a meeting, and some of them shake each others hand. (No one shakes their own hand, and no pair of people shake hands more than once.) Then there are two people who shake the same number of hands.
Let be the number of hands shaken by person . There are people whose hands could shake, so . In other words, if we put , then . Note also that because we start at zero, we have . Now suppose, for a contradiction, that the numbers are all different. We now have different numbers all living in the set of size , so the numbers must fill the whole set . In particular, we must have for some , and for some . Because we see that person shakes everyone else’s hand. Because , we see that shakes no-one else’s hand. This is a contradiction, because shakes ’s hand. Thus, the numbers cannot all be different, after all. Thus, we have for some , so and shake the same number of hands. ∎
Let be a set of integers, with all the members of between and (inclusive). Show that there are two distinct disjoint subsets of with the same sum.
As , the number of subsets of is . Let these subsets be , and let be the sum of the elements of . Note that is the sum of distinct terms, each of which is between and . The largest possible sum of this form is . (We can just do this addition by hand, or use the arithmetic progression rule: we have equally spaced terms between and , so the average is and the total is .) We thus have numbers lying in the set . As , there is not enough space for the numbers to all be different. We can therefore find indices such that . In other words, the sets and are different, but they have the same sum. However, we have not yet solved the problem, because we were supposed to find disjoint subsets. However, this is easily fixed. We put and and , so we have a Venn diagram as follows:
It is then clear that and are distinct, disjoint subsets with the same sum. (In a bit more detail, we can let be the sum of all elements of , and similarly for and . We then have and . We chose and such that , and it follows that as required.)
The above discussion is carefully phrased to work around the following finicky point: it is possible for two sets and to be disjoint but not distinct, if they both happen to be empty. However, in the proof we have so at least one of and must be nonempty, and they have the same sum, so both must be nonempty.
Consider a sequence of integers. Show that there is a consecutive subsequence (for some ) such that is divisible by .
For we put
so and and and so on. This gives numbers , all lying in the set , which has size . This means that the numbers cannot all be different, so we can find indices such that . Now note that
We chose and so that , and it follows that . In other words, if we take , we find that is divisible by . ∎
Suppose that I deposit money in my piggybank every day for days. On of the days I deposit , and on each of the other days I deposit . Given an integer with , show that there is a sequence of consecutive days during which I deposit precisely pounds.
For small values of , we can give a direct argument.
For : There are days on which I deposit , and any one of those days counts as a sequence of length one during which I deposit precisely .
For : There are days on which I deposit , and any one of those days counts as a sequence of length one during which I deposit precisely .
For : At some point I must swap over between depositing and . Thus, for some , I deposit on day , and on day , or vice versa. Either way, is a sequence of consecutive days over which I deposit precisely .
For : As there are days on which I deposit , I can certainly find a day when I deposit that is not the beginning or end of the month, so . If I deposit on day , then is a sequence of consecutive days over which I deposit precisely . If I deposit on day , then is a sequence as required. If neither of these possibilities occur, then I must have deposited on days and , so is a sequence as required.
This is clearly becoming increasingly unwieldy, so it is better to take a less direct approach.
Let be the amount deposited on day , so for . Let be the total deposited up to and including day , so . We also put . Note that is the total amount deposited over the whole month, which is . Because some money is deposited every day, we have
In particular, the numbers are all different, so the set has size . Now suppose we are given with . We put , and note that this also has size . Because and we have . We now see that and are both subsets of the set , which has size . If and were disjoint, we would have , which is impossible, as is a subset of . It follows that and are not disjoint, so we can choose . As , we have for some . As , we have for some , so . This means that we must have , and
Thus, over the sequence of days , I deposit precisely pounds.