# Suppose the numbers $1,2,3,…,10$ are spilt into two disjoint collections

Suppose the numbers $$1,2,3,...,10$$ are spilt into two disjoint collections $$a_1,a_2,..a_5$$ and $$b_1,b_2,...,b_5$$ such that

$$a_1 $$b_1>b_2>b_3>b_4>b_5$$

Show that the larger number in any pair $${a_j,b_j}, 1≤j≤5,$$ is at least $$6$$.

It's a question from a Regional math contest. I don't understand how to approach it. The proposed solution (which I do not understand) is:

Suppose $$j=3$$ then $$a_3,b_3$$ are two distinct numbers.

$$a_1$$ and $$a_2$$ are both less that $$a_3$$ so $$a_3$$ must be at least $$3$$

$$b_4$$ and $$b_5$$ are both less than $$b_3$$ so $$b_3$$ must be at least $$3$$

but we can't have $$a_3=b_3$$ so one of them must be at least $$4$$. That's what we get from the first attempt. But we can do better.

Now suppose $$a_3$$ and $$b_3$$ are both less than $$6$$ then $$a_1, a_2, b_4, b_5$$ (the four numbers referred to in the proof you have quoted) must all be less than $$6$$ too - that is six distinct positive integers less than $$6$$, which is impossible, so the largest, which will be $$a_3$$ or $$b_3$$, will be at least $$6$$.

The proof you have been given takes this observation and generalises it to any $$j$$ rather than just $$3$$. Always try to work out a specific example of a general statement if you don't quite understand what it is saying.

Another way of approaching this is to look at the way the numbers $$6,7,8,9,10$$ are distributed between the $$a_i$$ and $$b_i$$. If three are $$b$$s then two must be $$a$$s, for example.

• Thank you Mark! I understand your solution very clearly. – Ice Inkberry Oct 6 '18 at 12:36