19
$\begingroup$

Explaining math is hard, but when 'squaring the circle' came up in class, I could not resist to explain the connection between circles and squares.

I started easy by introducing Pythagoras' theorem

Then I introduced pi as the is the ratio of a circle's circumference to its diameter, and continued with explaining more geometry, esp. $sin^2+cos^2 = 1$;

enter image description here

Having established the connection, I could see the class was already very confused by this 'useless' information, so I continued quickly with

"Pi is a perfect square!" enter image description here

Now I completely lost my class, who thought they knew better.
One said: "Pi is not a perfect square, e.g. Four is."
After this it was clear that even the brightest of my students did not get it. After all, four clearly is not a square.

So I switched to some number theory

"An integer number is 9142057518"
"Yes", said the class.
"A real number is 2571.72"
"Yes" said the class.
"A version number is 2.3.5569.6739151"
The class looked confused but agreed.
"An odd number is 4.193.2"
One started smiling. I thought she finally understood it, but then she said jokingly: "That is an odd number!"
"Of course not! That is an even number", I said.

As last one I mentioned:
"An imaginary number is 3300647498274i/300"

No matter how hard I tried to explain it; they did not understand it.

Can you explain the class why 3300647498274i/300 is imaginary? (and remove any remaining confusion)

A minor hint for if you still want to solve this, but avoid the big spoiler when reading the answer:

Apply Pythagoras on the bottom picture - why is it correct?
or
Look at the statement about integers, this exact integer is chosen for a reason.

$\endgroup$
2
  • $\begingroup$ Pi is a square of what? geometrically or 3.14 algebraically? So what's the square root of PI?1.77... $\endgroup$ Sep 16, 2022 at 6:39
  • 1
    $\begingroup$ @LifeInTheTrees I find it hard to say something useful without spoiling the puzzle. I added some hints that hopefully do not confuse you even more. $\endgroup$
    – Retudin
    Sep 16, 2022 at 7:15

1 Answer 1

18
$\begingroup$

Your students should probably brush up on their

A1Z26/alphanumeric substitution knowledge. Under this substitution, the letter A becomes the number 1, B becomes 2, and so on until Z, which becomes 26. For example, the word NUMBER under the A1Z26 substitution is 1421132518 - the N becomes 14, the U becomes 21, etc.

So let's clear things up, shall we?

Why is pi a perfect square?

Because under A1Z26, PI becomes 169, which is 13 squared. The diagram is correct because according to the modified Pythagorean Theorem, 42 + 32 = 16 + 9 = P + I (under A1Z26) = PI (concatenation) = 169 = 132 (thanks @T. Dirks for spotting that!).

Why is four not a perfect square?

Because FOUR is 6142118, whose square root is not an integer as can be checked on a standard calculator.

What's going on with the number theory?

Each number encodes the word before "number" with an additional twist: a decimal point in the numbers represents the act of multiplying. Hence, we have

INTEGER <--> 9142057518
REAL <--> 185112 = 2571 × 72
VERSION <--> 225181991514 = 2 × 3 × 5569 × 6739151
ODD <--> 1544 = 4 × 193 × 2

"That is an even number" - why?

Because the last digit of the A1Z26 encoding of THAT is 0, so it is an even number (the mathematical kind), not odd.

Why is that weird expression an imaginary number?

This one requires putting everything we've learned together. First, convert the i into the digit 9 to get the expression 33006474982749/300. Evaluate it to get 110021583275.83. Finally, treat the decimal point as a multiplication sign and evaluate to get 9131791411825, which is IMAGINARY encoded in A1Z26. Phew!

$\endgroup$
3
  • $\begingroup$ I'm not sure what.. You have to apply Pythagoras and then it should be clear. $\endgroup$
    – Retudin
    Sep 16, 2022 at 6:10
  • 2
    $\begingroup$ I believe for the Pythagoras part rot13(vs jr svyy va 4 sbe n naq 3 sbe o va n²+o²=p² jr trg 16+9=p². Pbairegvat gurfr gb yrggref lbh trg C+V=p² fb gur fdhnerq qvntbany rdhnyf CV) $\endgroup$
    – T. Dirks
    Sep 16, 2022 at 9:44
  • 1
    $\begingroup$ @T.Dirks ohh how in the world did I miss that? Thanks, the answer should be complete now! $\endgroup$
    – HTM
    Sep 16, 2022 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.