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
    Commented Jun 2, 2015 at 11:48
  • $\begingroup$ nice one. i think i've got everything except for the matrix. that's the part giving me a headache $\endgroup$
    – user12241
    Commented Jun 2, 2015 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
    Commented Jun 2, 2015 at 12:19
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    $\begingroup$ Really , nice one! $\endgroup$ Commented Jun 2, 2015 at 12:48
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    $\begingroup$ Perfect puzzle! +1 $\endgroup$
    – BmyGuest
    Commented Jun 2, 2015 at 20:01

1 Answer 1


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.

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    $\begingroup$ Wow, the last line was very lateral-thinking! I really couldn't realize the meaning of "for every point reflection" $\endgroup$
    – leoll2
    Commented Jun 2, 2015 at 12:27
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    $\begingroup$ When I thought about using the Matrix for making $\forall$ become A, I thought about the 180º rotation Matrix. $\endgroup$
    – Masclins
    Commented Jun 2, 2015 at 12:28
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    $\begingroup$ Ah! Rotation. I figured there was a more specific term than "negating". I should have been thinking more geometrically. $\endgroup$
    – Kevin
    Commented Jun 2, 2015 at 12:43
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    $\begingroup$ Surely if you're applying a rotation matrix to $\forall$, the matrix ought to be placed before the $\forall$. $\endgroup$ Commented Jun 2, 2015 at 22:07
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    $\begingroup$ @DavidZhang - $\forall$ must be a row vector, I guess. $\endgroup$
    – Glen O
    Commented Jun 3, 2015 at 3:06

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