# Best basic algebra examples to show students that proof by example is not sufficient [duplicate]

Often, students will try to 'prove' a propositon by checking some examples and 'concluding' that it will be true for all $n \in N$.

I'm looking for some good, non-trivial examples from highschool and lower college mathematics showing that checking a finite number of cases is generally not sufficient. There is already a similar thread with examples from higher level math (Examples of apparent patterns that eventually fail).

The best non-trivial example I currently have to show this fallacy is:

3 is odd and prime
5 is odd and prime
7 is odd and prime
11 is odd and prime


therefore, all odd numbers are prime. :/

## marked as duplicate by Michael Albanese, Grigory M, mdp, Hakim, Asaf Karagila♦Jun 3 '14 at 11:00

One well-known example is the statement that $n^2 - n + 41$ is prime for $n \in \mathbb{Z}^{+}$.

Further historical remarks can be found here.

Note, in particular, that this is prime for $1 \leq n \leq 40$, but since $n^2 - n + 41 = n(n-1) + 41$, it is clear that the number is composite when $n = 41$ (and, incidentally, $42$).

• Very nice, thank you! – Floyd Jun 3 '14 at 10:10

Fermat primes. First five candidates in the form $2^{(2^n)}+1$ are all primes (increasingly more difficult to check), but then it all falls apart! If you do it in the classroom where you can't really afford to factor billion-scale numbers, it's pretty convincing to thing they are all primes.

• Very nice as well, thank you! – Floyd Jun 3 '14 at 10:10

The product of the first primes $+1$ is prime... $$2\ 3 + 1 = 7,$$ $$2\ 3\ 5 + 1 = 31,$$ $$2\ 3\ 5\ 7 + 1 = 211,$$ $$2\ 3\ 5\ 7 \ 11 + 1 = 2311,$$ until $$2\ 3\ 5\ 7 \ 11\ 13 + 1 = 30031 =59\ 509.$$

• beautiful, and at first sight the elegant pattern makes you want it to be correct! – Floyd Jun 3 '14 at 10:12
• And is related with the Euclid's proof of of the infinity of primes. – Martín-Blas Pérez Pinilla Jun 3 '14 at 10:14

A geometric example: Let $\Omega \in \mathbb{R}^2$ be any bounded connected and open domain. A Neumann Cheeger cut $\gamma$ of this domain is defined to be the curve which separates this domain into two connected disjoints open subdomains $\Omega_1, \Omega_2 \subset \Omega$ (i.e. $\overline{\Omega_1} \cup \overline{\Omega_2} = \overline{\Omega},$ $\overline{\Omega_1} \cap \Omega = \overline{\Omega_2} \cap \Omega = \gamma,$ $\Omega_1 \cap \Omega_2 = \emptyset$) such that the follow ratio is minimized:

$R_\Omega^N(\gamma):=\frac{\text{length}(\gamma)}{\min\{\text{surface}(\Omega_1),\text{surface}(\Omega_2)\}}$

Not easy to formulate rigorously but very easy to explain intuitively (even more with a drawing on the black board). Lots of example can be shown quite easily, e.g. if $\Omega$ is a circle then any diameter is optimal, if it is a square then any symmetry axis, etc.. People thoughts a very long time that the optimal curve is attained when $\text{surface}(\Omega_1)=\text{surface}(\Omega_2)$ (I think that also papers were published under this assumption) but this is wrong. A counter example (provided by Klaus Dieter Semmler) can be easily constructed taking two disjoints balls of different radius $r_1,r_2$ and a rectangle joining them, just make the width of the rectangle small enough so that the cut has to be this width.

An image may help :): Examples: circle ($\infty$ solutions), rectangle isosceles triangle ($2$ solutions), rectangle ($1$ solution)) Counter example