7

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 $A$ 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.

M Carl
  • 171
  • 2

2 Answers2

4

This has been asked many times before in MathOverflow

JRN
  • 10,796
  • 2
  • 37
  • 80
  • The links are certainly helpful, but contain very few elementary cases that are suitable as exercises. – M Carl Nov 05 '14 at 13:24
  • 2
    @MCarl There is also this question. – dtldarek Nov 05 '14 at 17:53
  • 2
    @MCarl Consider this one: http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form – user26486 Nov 05 '14 at 18:58
  • Just posted an answer to that math SE question, but considering it's about eigenvalues I'm not sure it's elementary enough for the question here on Maths Ed SE. – DavidButlerUofA Nov 05 '14 at 19:52
  • Thanks to Benjamin Dickman's comment on the question, I found the answer I was referring to in my comment above: http://matheducators.stackexchange.com/a/2171/77 (Note that the answer in the link is quite different from what I said. I have deleted my comment above, as I don't think it's a particularly good answer.) – JRN Nov 06 '14 at 02:45
  • Just to clarify my comment above. I meant that my answer (the comment I deleted) is not particularly good. Neil Strickland's answer (which I linked to) is good. – JRN Nov 06 '14 at 04:28
3

Proving Cauchy-Schwarz for a general inner product is easier than proving it for any particular inner product. For example, it is quite hard to prove for the dot product on $\mathbb{R}^n$.

The reason it is easier to prove it in general is that the definition of an inner product high light only the essential properties needed for the proof, so the "space" of possible proofs is constrained.

I think this is on of the general reasons why it is easier to prove something harder: there is a more limited number of possible approaches to the proof. What is harder is to discover a hard theorem.

Steven Gubkin
  • 25,127
  • 4
  • 61
  • 110