1

I have been looking at this problem for a while, but I cannot find an exact solution to it, although the statement and the conclusion are obviously wrong. Here is the proof:

Claim: All tables are the same height

To prove this by induction, we let P(n) be the statement "For any set of n tables, all n tables are the same height." If we prove this true for all n, it will certainly be true for n = the number of tables that exist

Now we proceed by induction on the number of tables. The base case is the case in which there is one table. Since this table is the same height as itself, the base case is true. Now assume that the statement holds for any set of n tables, and consider a set of n + 1 tables.

Put the tables in a line. If we remove the first table, we are left with a set of n tables. Then by the inductive hypothesis, these n tables must be all be the same height. If, instead, we had removed the last table, we would again have n tables, which would now include the first one, and again by inductive hypothesis all n tables would be the same height. Therefore, all of the tables must be the same height as, for instance, the second table from the front, and consequently must be the same height as one another. The result then follows from induction.

I do think the base case is true. Although I hate to say that I am getting convinced by this proof, I do have a strong feeling that the flaw is in the inductive step. It's so obvious that I'm confusing myself, but what would be wrong with this proof??

Isaac Seo
  • 125

1 Answers1

2

Your argument fails for $ n=2$

When you set one table aside there is no middle table to make the comparison.

It works for $n\ge 3$ but not for $n=2$

You can not start with $n=2$ either because not every two tables have the same height.

Mohammad Riazi-Kermani
  • 69,891