# Mathematical Rebus

Mathematical Rebus II

Mathematical Rebus III

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

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

## 1 Answer

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.

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