[Math] Examples where it is easier to prove more than less

Question

Especially (but not only) in the case of induction proofs, it happens that a stronger claim $B$ is easier to prove than the intended claim $A$ (e.g. since the induction hypothesis gives you more information). I am trying to come up with exercises for beginner students that help to demonstrate this point (and also interested in the general phenomenon). Do you know any good examples (preferably elementary ones) where strengthening a claim makes the proof easier?

Here is an example of what I mean (Problem 16 from chapter 7 of Engel's `Problem solving strategies'):

Show that $\frac{1}{2}\frac{3}{4}…\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n}}$ for $n\geq 1$. This is much harder than proving the stronger statement that $\frac{1}{2}\frac{3}{4}…\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$ for $n\geq 1$, which is a straightforward induction.

Answer 1

Here's my favorite proof - for the reasons that it's very easy and elementary enough that it takes almost no mathematical knowledge at all - for an example of the general case that is easier than the special case:

Problem: For every $2^n\times2^n$ grid where $n$ is a natural number, there is a covering with $L$ shaped pieces, all have an area of $3$ squares, that leaves one central square (one of four central squares) empty.

Here is an example for $n=2$:

General case is:

Problem: For every $2^n\times2^n$ grid where $n$ is a natural number, there is a covering with $L$ shaped pieces, all have an area of $3$ squares, that leaves only one arbitrarily specified square empty.

General case claims that we can choose any square (not necessarily center as opposed to special case) we want to be empty, before covering. Proof is very easy and uses induction on $n$:

Let $P(n)$ be the proposition that problem states.

Base case: $P(0)$ is obvious.

Inductive step: Assume $P(n)$ for some $n\in\mathbb{N}$. Take any $2^{n+1}\times2^{n+1}$ board and arbitrarily choose one square which will be empty. Divide the board four $2^n\times2^n$ quadrants. From hypothesis; there is a covering of the quadrant which has chosen square on it, and covering of all other quadrants except three center squares of the big square like this:

Here, red is chosen square and blues are empty squares for other quadrants. Now, put one piece on blue squares and you have $P(n+1)$. $\blacksquare$

For me; this argument is also very effective way to make people, who have little or no mathematical experience, to comprehend that sometimes why and how proving general case may be easier than special one. Therefore make an excellent example to demonstrate this point for beginner students.