1

Show that if we take n+1 numbers from the set [2n] then at least one number will divide the other. Prove using induction.

Now I know a elegant proof exists using php. But our prof specifically asks us to write using mathematical induction. So here's my try.

  • Base Case: n=1 S={1,2}, so we choose 2 numbers, they are 1,2 hence P(1) is true.
  • Induction Hypothesis: Let P(k) be true. We have to prove P(k+1) is true. Consider [2k] numbers.

Case 1: We take >=(k+1) numbers from [2k] set and add either 2k+1 or 2k+2 to it. We are done by induction hypothesis.

Case 2: We take k numbers from [2k] and add both 2k+1 and 2k+2 to it. Now I use PHP. Create k+1 holes by the principle $2^r$b should be in it r>=0 and b is an odd number. So we place k+2 numbers in k+1 holes so we have to place at least 2 numbers in one hole and so we are done.

I know it sounds wrong but strictly using only induction hypothesis in case 2, I cannot proceed. Please help.

RobPratt
  • 45,619
Alex
  • 557
  • I suppose a natural start would be to note that, if your $n+1$ numbers were all $≤2(n-1)$ we'd be done. Indeed, if just $n$ of them are $≤2(n-1)$ we are done. Thus we are down to the case where only $n-1$ are $≤2(n-1)$ and you must have drawn both $2n-1$ and $2n$. – lulu Aug 21 '21 at 14:26
  • @lulu Yes I have done that only. In the last case you mentioned, I am stuck – Alex Aug 21 '21 at 14:31
  • oh sorry this post answers my doubt. I vote to close it sorry for the inconvenience. – Alex Aug 21 '21 at 18:07

0 Answers0