# Examples of problems where proving something more general leads to a much simpler solution

The following is written in my textbook:-

"The approach of solving a problem by proving something more general can sometimes be very useful."

The textbook says this in the context of giving a solution to a question involving proving a result of matrices by induction. However, I solved the problem using a simpler, non-general approach.

But, it got me thinking whether there were certain problems that can be solved much more easily by using the above mentioned approach.

Could anyone give some examples(limited to advanced high school math)?

• feynmans trick.
– Anon
Apr 12, 2023 at 17:20
• "Show that $100!$ is divisible by $18!19!20!21!22!$" I would hate to try to do this directly without generalizing. Instead, we can learn that $\dfrac{N!}{n_1!n_2!\cdots n_k!}$ where $n_1+n_2+\dots+n_k=N$ is the number of ways to split $N$ objects into labeled groups of size $n_1,n_2,\dots,n_k$ respectively. Since $\frac{100!}{18!19!20!21!22!}$ is the answer to a counting question it must be a whole number, thus proving the result. Apr 12, 2023 at 17:35
• Apr 12, 2023 at 17:50

Here's a question I like to give to high schoolers:

Show that if we remove any square from a $$8 \times 8$$ chessboard, then the rest can be perfectly tiled with $$21$$ L-shaped triminos.

It can be approached by tedious case checking. Using symmetry we still have to check 10 cases, and that can be frustrating.

Or it could be done by induction. What do you induct on?

Here's a not-so-good example

Prove that in an acute triangle, we can inscribe a square with a base on one side of the triangle, and vertices on the other 2 sides of the triangle.

There are numerous ways to approach the problem directly like using similarity (which is why this is a not-so-good example).

The reason it's relevant: One approach that I like is to relax the "square" condition to "rectangle" (the general case), with the similar constraints on the vertices/edges. It is obvious that we can construct such a rectangle for any height.
Then, we use IVT to show that "length = height" and hence we have a square. In fact, the length and height varies linearly, and we could even graphically determine the height of the square.