Mathematical Rebus II

Mathematical Rebus III

$$ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\\ (-\infty,...,-1,0,1,...,\infty)\times(-\infty,...,-1,0,1,...,\infty)\\ \forall\begin{bmatrix}{-1}&{0}\\ {0}&{-1} \end{bmatrix} $$

  • 1
    $\begingroup$ This is beautiful! I think I've half-realized the solution, though there's a piece that's giving me a hard time! $\endgroup$
    – leoll2
    Jun 2 '15 at 11:48
  • $\begingroup$ nice one. i think i've got everything except for the matrix. that's the part giving me a headache $\endgroup$
    – Paul
    Jun 2 '15 at 12:15
  • $\begingroup$ This definitely involves some circular reasoning, but I also haven't figured out the right way to read the matrix. $\endgroup$
    – Glen O
    Jun 2 '15 at 12:19
  • 2
    $\begingroup$ Really , nice one! $\endgroup$ Jun 2 '15 at 12:48
  • 3
    $\begingroup$ Perfect puzzle! +1 $\endgroup$
    – BmyGuest
    Jun 2 '15 at 20:01

The first line is

equal to Pi.

The second line is

the integers, Z, multiplied by itself, making Z2.

The third line is

multiplying some matrix ∀ by the negative identity matrix, negating its value. The letter ∀ negated is A.

All together,

Pi + Z2 + A = Pizza.

  • 9
    $\begingroup$ Wow, the last line was very lateral-thinking! I really couldn't realize the meaning of "for every point reflection" $\endgroup$
    – leoll2
    Jun 2 '15 at 12:27
  • 2
    $\begingroup$ When I thought about using the Matrix for making $\forall$ become A, I thought about the 180º rotation Matrix. $\endgroup$
    – Masclins
    Jun 2 '15 at 12:28
  • 1
    $\begingroup$ Ah! Rotation. I figured there was a more specific term than "negating". I should have been thinking more geometrically. $\endgroup$
    – Kevin
    Jun 2 '15 at 12:43
  • 1
    $\begingroup$ Surely if you're applying a rotation matrix to $\forall$, the matrix ought to be placed before the $\forall$. $\endgroup$ Jun 2 '15 at 22:07
  • 4
    $\begingroup$ @DavidZhang - $\forall$ must be a row vector, I guess. $\endgroup$
    – Glen O
    Jun 3 '15 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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