Significant Proofs Understandable to Middle School Students with Pre-Algebra Background

Question

I'm interested in introducing my middle school children to proofs using significant examples (i.e., not just basic geometry proofs in an intro to trig textbook). They have a pre-algebra background (so, polynomials, powers, greatest common factor, etc.). I can think of several proofs off the top of my head:

Proof of the irrationality of $\sqrt{2}$

Cantor's diagonalization proof

Pythagorean's rearrangement proof (and other proofs of the Pythagorean theorem).

What other significant proofs are there that would be explainable to someone with a pre-algebra background?

Answer 1

Here a couple of examples I can think of:

• The sum of the counting numbers from $1$ to $100$ is $5050$. The ambitious student might be able to find the general formula for the sum of an arithmetic series, provided they are given some guidance.
• If $a$ and $b$ are two irrational numbers, then it is possible for $a^b$ to be rational.
• If $n^2$ is even then $n$ is even. This is a nice introduction to proof by contraposition. You could also use contraposition to argue that at least one digit in the decimal representation of $\pi$ occurs infinitely often.*
• There are infinitely many prime numbers.**
• Geometric demonstration of completing the square to solve $x^2+10x=39$. Again, you can generalise this method to prove the quadratic formula, although you might want to limit yourself to the case where the leading coefficient is $1$ so that the algebra doesn't get too messy. You might also be interested in The Simpler Quadratic Formula by 3Blue1Brown.
• A geometric proof that the area of circle is $\pi r^2$. Virtually all of them involve chopping the circle into many tiny pieces. Here is one I quite like.

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*Actually, at least two digits occur infinitely often. That's because if only $1$ digit appeared infinitely often, then there would be a point in the decimal representation of $\pi$ where the digits would only consist of a single number. This would make $\pi$ rational.

**Don't prove this by contradiction. See here.

Answer 2

If you mention Cantor's diagonal argument then you should also consider Julius König's proof of the Schröder–Bernstein theorem – the one that makes chains of function applications between two sets to prove a bijection. This is significant because it ultimately allows the notion of cardinality by establishing a bijection to some standard set.

In graph theory there is the Seven Bridges problem and the proof of its lack of solution, which only relies on following a path in and out of a vertex (and also indicates that proofs don't have to show something positively). In number theory there is Euclid's proof of the infinitude of primes, as mentioned in comments.

Combinatorics and its "twin city" of probability are full of simple but profound results too. The Fibonacci numbers appear everywhere – the proof that they enumerate tilings of a $1×n$ rectangle by single squares and dominoes is but one example in the former topic; gambler's ruin and the Monty Hall problem are examples in the latter (these shed a little light on the psychology of betting and why people lose when betting in the long run).

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There has never been something called "pre-algebra" or "middle school" in Singapore's educational system, and furthermore I had gifted education (Integrated Programme, NUS High School). Some of the examples above may therefore be too high-level.

Answer 3

Little Fermat? Pick a prime $p$ and a number $n$. Then $n^p-n$ is a multiple of $p$.
If $n$ is a multiple of $p$ we are done. So suppose it isn't.
Multiplying the numbers from $1$ to $p-1$ by $n$, then taking the remainder, shuffles the numbers. (The difference between any two is not a multiple of $p$ so they are all different.) So $1×2×3×...×(p-1)$ and $n×2n×3n×...(p-1)n$ leave the same remainder when you divide by $p$.
So $(n^{p-1}-1)(p-1)!$ is a multiple of $p$. $(p-1)!$ is not, so $n^{p-1}-1$ must be.

Answer 4

In Ontario I have met many high-school graduates who are vague or confused about $\Bbb R,$ especially on the issue of infinitesimals, that is, whether a number can be positive but less than every member of $\Bbb Q+.$ I think this is because they were not taught properly. I'm sure they never heard of a Dedekind cut. I suggest you give an axiomatic def'n of $\Bbb R,$ and show from the def'n that (i) the Archimedean property, (ii) the non-existence of infinitesimals in $\Bbb R$, (iii) the order-density (in $\Bbb R$) of $\Bbb Q$ & of its complement, (iv) every non-empty subset of $\Bbb R$ with an upper (lower) bound has a lub (glb) and from this deduce that $\sqrt 2$ exists in $\Bbb R.$

I think it would be helpful to emphasize that these follow from the def'n of a certain structure ($\Bbb R$) but that we can also define larger structures (e.g. the hyper-reals), so we cannot derive properties of the "real" numbers without a definition. Without a def'n it's like trying to prove that widgets are green.