Most elementray aha! moments
The identity
although simple, is rather surprising. It could be verified by raising the right-hand side to the power of 3:

One way of accidently running into this identity is by means of Cardano's cubic formula applied to the third degree equation
The formula will produce terms that include cubic roots. On the other hand, the equation has an integer root
and, therefore, the remaining two roots satisfy a quadratic equation
with integer coefficients. Professor Tom Osler from Rowan University has observed
It is known that
is not constructible. Yet
which looks more complicated than
is constructible because it can be expressed by the golden section. This is an unexpected result. It is surprises of this type that make it a joy to be a mathematician.
A most engaging remark, indeed.
More recently, I had a similar experience on a smaller scale. Say, consider a problem: Prove that the equation
has no solutions in positive integers. A more or less common way of approaching a problem like that would be to apply Fermat's infinite descent. (The case where at least one of
or
is assumed to be even is easy. Assuming both of them odd leads to no obvious contradiction.) But here's a simpler approach. By adding to and subtracting
from both sides gives
and
Divide one by another and take the square root:
This is of course impossible as the left-hand side is rational, while the right-hand side is not.
Martin Gardner had two collections of surprises in this spirit - aha! Insight and aha! Gotcha - that are also available in a single volume.
Not every child is born with a mindset fit to see and appreciate those instances of - real, albeit "small" - mathematics. Most kids do not naturaly come across such small surprises. However, there are plenty of elegant mathematics even on a most elementary level.
Years ago the Australian mathematician M. L. Urquhart discovered a theorem while working on "some fundamental concepts of the theory of special relativity." He dubbed it the most elementary theorem of Euclidean geometry since it only involves the concepts of straight line and distance. Now known as Urquhart's theorem it reads, If ABB' and AC'C are straight lines with BC and B'C' intersecting in D and if AB + BD = AC' + C'D, then AB' + B'D = AC + CD. This is not what I mean by an "elementary level."
There are problems whose solution requires little or no up-front knowledge. Socrates' doubling (or halving) a square is one example. I'll list a few more.
A pin has two ends: a pointed one and a blunt one. If three pins form a triangle then either all vertices look the same (pointed/blunt) or all look different (pointed/pointed, blunt/blunt, blunt/pointed.)
4-letter word ladders, one of Lewis Carroll's inventions. The task is to convert one 4-letter word into another by changing a letter at a time so as to have meaningful 4-letter words at every step of the process. Claim: some of the words in the chain is bound to have two vowels.
Take two very different 3D bodies, say, a ball and a cube or a potato and a kettle, etc. On each body in the pair draw curves with a marker. Is it always (or never) possible to draw identical curves on any two bodies? (Fro an answer, imagine the two bodies made of some amorphous material so that they can physically intersect, i.e. penetrate each other. Their intersection will be a curve on the surfaces of both.)
Assume you mastered the Toads and Frogs puzzle. Then it must be easy for you to solve its 2D analog. Is it?
I think that what is missing from school curricula and from course offerings intended for potential teachers is this kind of simple, almost unversally accessible mathematical morsels. I actually would like to solicit additional examples of simple mathematics. Not mathematical facts, but statements that require and admit a proof, a proof no longer than a step or two, based more on logic and every day experience than on knowledge of some axioms.
To start with, it is always possible to comment on this post. If the thing picks up, I promise to find a more suitable arrangement. Actually, why not to declare a competition for the Most elementray aha! moments. A winner will be determined monthly and announced at the end of every month. A monthly prize of $1 will be paid to the winner's paypal account.
No related posts.
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[...] like to solicit such examples of “minimal mathematics”. I have put together a post Most elementary aha! moments to which comments can be [...]
July 13th, 2010 at 11:12 amReally liked the idea behind this.
July 13th, 2010 at 5:23 pmMy favourite was the old $latex 0.\dot{9}=1$
If $latex \frac{1}{3}=0.\dot{3}$
multiply by three and you are forced to conclude that $latex 0.\dot{9}=1$
Alternatively $latex x=0.\dot{9}$
$latex 10x=9.\dot{9}$
then, subtracting $latex 9x=9$
and therefore $latex x=0.\dot{9}=1$
That didn't come out quite right, but it was meant to be about 0.9(recurring)=1
July 13th, 2010 at 5:26 pmJames, in this blog LaTex goes between two dollar signs on both sides like this:
Multiplying by 3 gives
I am going to see how it looks like and then later on shall add a comment.
July 13th, 2010 at 6:19 pmThe problem of .999... being 1 is quite involved. I have two discussions on this: http://www.cut-the-knot.org/arithmetic/999999.shtml and http://www.cut-the-knot.org/WhatIs/Infinity/9999.shtml. The first one looks in the traditional framework, the second in the framework of the non-standard analysis. There are two questions that I may ask right away
1. What is 0.333... and why it is equal to 1/3.
July 13th, 2010 at 7:11 pm2. Do we actually know how to calculate with the decimals that we so easily assume the validity of 3×0.333... operation.
Thank you for directing me to a couple of very interesting articles. I hadn't appreciated the problem from a non-standard analysis point of view. When thinking of the questions you asked it does seem an altogether more complicated issue. If we can't properly define these decimals then non of it really makes sense.
July 14th, 2010 at 12:57 pmIncidentally, my other argument for 1=0.999... was
let x=0.999...
then 10x=9.999...
subtracting gives 9x=9
therefore x=1=0.999...
always felt it was a very nice little proof, though I'm not as confident it's satisfactory now.
Here is in fact the same problem. Unless there is a reasonable definition of .999... there is no reason to assume that 10×.999... = 9.999...
Assume x = 1 + Δ. Then
10x = 10 + 10Δ
Subtracting gives
9x = 9 + 9Δ, or
x = 1 + Δ,
which is nothing new.
July 14th, 2010 at 1:18 pm[...] a related post I have shown that Without resorting to this proof, I am going to show [...]
July 19th, 2010 at 11:33 am[...] A beautiful mathematical equality July 22, 2010 — ckrao Here is an equality worth sharing, found at the CTK Insights blog here. [...]
July 22nd, 2010 at 5:31 am