## Math Haiku and Limericks

Haiku have 17 syllables, right? Nope. They actually have 17 morae. Don’t know what a mora is? Don’t worry; neither do most linguists.

I find the 5-7-5 structure of haiku too restrictive, and apparently Roger McGough does, too.

The only problem
with haiku is you just get
started and then
~ Roger McGough

And Daniel Mathews thinks the structure is problematic for writing math haiku.

Maths haikus are hard
All the words are much too big
Like homeomorphic.
~ Daniel Mathews

Limericks are a little more forgiving. With five lines in an AABBA pattern, you have a little more time to develop a story. Or not.

There was a young man from Peru
Whose limericks stopped at line two.

If you’re at a cocktail party, and you want to deliver the following one-liner, you better set it up with the two-liner above.

There was a young man from Verdun.

“Then there’s the one about the Emperor Nero,” quipped poets Elliott Moreton and Carl Muckenhoupt.

Personally, I think it’s pretty fun to turn traditional poetry rules on their ear. Here is a tradition-busting limerick for you.

A poet through efforts concerted
Ignored all the rules
He learned in the schools
Tradition he oft times skirted
And wrote all his limericks inverted.

And lest haiku feel neglected as a poetic form, here’s an abomination of that type, too.

The last line goes here.
It’s still 5-7-5, but…
Haiku inverted.

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• 1. xhenderson  |  November 14, 2012 at 6:34 pm

A couple of years ago, I tried to write a couple of proofs in sonnet form. These are the (likely terrible) results:

The Square Root of 2 is Irrational

Suppose, for a while, that there are in Z
An a and b without a factor shared,
With square root 2 equals a over b.
Thus 2 times b squared must equal a squared.
Then, it seems, as the squares of odds are odd,
And squares of evens even, we can say
That a must be of an even facade.
There is in Z: k, where a is 2k.
Thus 2 times b squared is a squared times 4,
Or b squared and 2 a squared are the same.
Then b is even! And, for there is more,
A common factor: 2 shall be its name.
And here, a contradiction we must see.
Square root 2: irrational. Q.E.D.

There Are Infinitely Many Primes

If primes be but finite, list them, bar none!
Take the product of the primes on the list;
Call it N, and add 1. Then N plus 1,
From the set of primes, may be dismissed.
As N plus 1 is certainly not prime,
Among all the many primes there must be
A prime factor. In order to save time,
We shall call this prime by the letter p.
p is prime and divides N, you’ll agree,
Thus it divides N plus one, minus N,
So p divides one quite naturally.
Nonsensical! This is beyond all ken!
Therefore, by contradiction we have shown:
The greatest of primes can never be known.

• 2. venneblock  |  November 15, 2012 at 10:17 am

Proofs in sonnet form?
No tests to grade? Kid asleep?
Too much time on hands.

• 3. xhenderson  |  November 17, 2012 at 9:06 pm

In my defense I wrote those before the kid was born. 😛

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• 5. Mo’ Math Limericks | Math Jokes 4 Mathy Folks  |  August 4, 2017 at 11:33 pm

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