Much of what passed for proof with the pioneers
would not now be tolerated in a college textbook.
E.T. Bell, Mathematics: Queen & Servant of Science
Dover, 1987 (oiriginal 1951)

A Formal Framework

A charming proof by Ian Stewart, being a proof without a theorem, couldn't but raise some important issues. Following is a sample of the correspondence with a response and an attempt to salvage the proof by putting it into a formal framework.

Patrick Morris-Suzuki sent me the following letter

Dear Alex,

Very nice explanation... One problem, though. Rythm is a valid word with no vowels... and so are sly, spy... You should say every valid word contains either a vowel or y... But this may be problematic for your assertion that one of the intermediate words contains 2 vowels. Better to say that every word contains 2 letters which are either vowels or y (I guess you could try to count y as a vowel...)


to which I replied as follows

Dear Patrick:

You are of course right. Both Stewart and I have overlooked the fact that there are vowelless words.

What's, however, interesting is that, as far as the deduction is concerned, it's still not only valid but is edifying. If you have no objections I am going to place your remark and my response on the page.

Alex Bogomolny

Larry Davidson picked up the gauntlet

As a math teacher, I am enjoying your Web site and I am convinced that you have a lot of material there that will be of great interest to students and teachers of mathematics. However, both as a math educator and as a former student of linguistics, I was rather taken aback-- perhaps horrified would not be too strong a word--by your reply to Patrick Morris-Suzuki. You actually agreed with his incorrect claims that words like "rhythm" and "sly" have no vowels. But the letter y serves two functions in English! It can represent a consonant sound, as in "yellow" and "young," or it can represent a vowel sound, as in "rhythm" and "sly"; only in Wheel of Fortune is y always a consonant. The requirement that every English word--in fact, every English syllable--must contain a vowel is a fact of pronunciation much more than spelling, and it's reflected in English spelling by the presence of a, e, i, o, u, or sometimes y in every natural English word. The rare exception is an artificial word like the mathematical "nth," which is of course pronounced "enth."

Note that the confusion between vowel sounds and their representation in spelling sometimes leads to severe misunderstanding about when to use "an" rather than "a" as the indefinite article. Most of us learn that "an" is used before a vowel, but it then can become mysterious why we say and write "a union" and "an heir." The reason, of course, is that "union" begins with a consonant sound but "heir" begins with a vowel sound, despite their spellings.


Jacob Mandelson found another exception. Wrote he

There's also the word "cwm". It more clearly qualifies as a word than nth, but isn't in as common usage, and the "w" may count as a (nonorthodox) vowel.

As The American Heritage Dictionary reveals, "cwm" is pronounced "koom" and stands for College of William and Mary. Micah Stetson adds

Alex, I ran across your site today and I love it. However, the discussion of 'cwm' as an example of a "word without a vowel" is more interesting considering that it is not only an acronym for College of William and Mary. Webster's third international dictionary says that it is from a Welsh word meaning "valley" ('w' is apparently a common vowel in Welsh) and that it is synonymous with 'cirque'. 'Cirque' is defined as "a deep steep-walled basin high on a mountain, usu. shaped like half a bowl and often containing a small lake, caused esp. by glacial erosion and usu. forming the blunt head of a valley." 'Cwm' is commonly used among climbers and mountaineers; one may see a photo at

There is another omission detected by Sara Farmer:

There's one more caveat to put on the proof that there's always a word with two vowels in it: the vowels in the start and end words (SHIP, DOCK) must be in different positions. It's a pretty pedantic and not very helpful point, but someone had to say it :-).

It's not pedantic and very helpful. It's right to the point.

At this point it might be the right time to put on a blue formal hat and see what the discussion comes to. There is a couple of points to make:

  1. That the problem was stated in English is clearly unimportant. Dealing with English in its every day manifestation has probably broadened the perceptive audience but resulted in conflicting opinions as to the notion of vowel in the English language.
  2. Either way, what ever is the English usage or its rules, one can't help but feel there is something right, some truth in Ian Stewart's proof. Perhaps with some omission, he has really proven something.

Following is an attempt to distill the proof to an unambiguous framework.

Let there be an alphabet - a collection of symbols which will come in two varieties: truant and trowel. Out of these symbols, collectively known as tatters, we are allowed to form strings. As usual, not all strings are equal. Some strings are better than others. The bad ones have no special name attached to them (but sometimes are referred to as tards.) The good ones are collectively called tords with a strange sounding word tord reserved for individual good strings. Only one rule is imposed on construction of tords:

(*) No string may be a tord unless it contains at least one trowel.

A sequence of tords is called a truss iff any two consecutive tords differ by a single tatter. A truss is said to traject its extreme tords. Now we are in a position to state a general statement:


Let there be two tords with the same number of tatters. Assume each of the given tords has a single trowel and those trowels occur in different positions. Prove that every truss that trajects the two tords contains at least one tord with a minimum of 2 trowels.


|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny