In praise of the Queen of all Sciences !

  1. Computo ergo sum: D H Bailey
  2. The only thing I know is that I don’t know anything. : Socrates.
  3. Also, he made a molten sea of ten cubits from brim to brim, round in compass, and five cubits the height thereof, and a line of thirty cubits did compass it roundabout. : Old Testament I, Kings 7:23
  4. We would be amiss, however, if we did not emphasize that the extended precision calculation of “pi” / \pi has substantial application as a test of the “global integrity” of a supercomputer. : D.H, Bailey, J.M. Borwein, and P.B. Borwein
  5. The purpose of computing is insight, not numbers. Richard Hamming.

A little big announcement from National Museum of Math, NYC

Dear MoMath families,

There’s always more to learn at MoMath!  The Museum’s pre-K through 12 programming reveals a world of engaging mathematics and offers fun, hands-on activities your child can do right at home, led by classroom-seasoned, top-notch educators.

MathPlay, MoMaths program for preschoolers
Whether your preschooler is just learning to count or is gearing up for kindergarten, MathPlay will instill a love of mathematics in each child through educational games, catchy songs, and intriguing problem-solving challenges.

“It was the first time my daughter saw the use of a number line for addition and she really enjoyed the class.  She was very focused for the full 30 minutes thanks to our engaging instructor who walked her through step by step.” β€” a parent writing about MathPlay

Student Sessions for grades pre-K through 12
Join young math fans from around the world for one of MoMath’s 45-minute, interactive Student Sessions, such as the ever-popular “Shape Shifters” or “Breaking Codes,” or the newest session, “Color Grids.”  In “Color Grids,” students generate their own infinite patterns while exploring simple cellular automata.

“My son spent the weekend continuing to play with his shapes and his interest in the content was elevated due to this session.  Zoom was an excellent platform and the educator was very engaging.  Excellent session!” β€” a parent writing about MoMath Online: Student Sessions

Now you can share the gift of math with friends!  Gift registrations for Student Sessions are available at

For more mathematical fun, be sure to check out MoMath’s complete lineup of upcoming adult events and children’s programs atΒ,
National Museum of Mathematics.

PS: I am sharing the email content I received for the benefit of Indian students….Hope the awareness spreads and Indian kids enrich their love and knowledge of math πŸ™‚

Wisdom of V. I. Arnold, immortal Russian mathematician

Development of mathematics resembles a fast revolution of a wheel: sprinkles of water are flying in all directions. Fashion — it is the stream that leaves the main trajectory in the tangential direction. These streams of epigone works attract most attention, and they constitute the main mass, but they inevitably disappear after a while because they parted with the wheel. To remain on the wheel, one must apply the effort in the direction perpendicular to the main stream.

V I Arnold, translated from “Arnold in His Own words,” interview with the mathematician originally published in Kvant Magazine, 1990, and republished in the Notices of the American Mathematical Society, 2012.

Symmetric difference is associative

We want to show that : A \triangle (B \triangle C) = (A \triangle B) \triangle C

Part I: TPT: LHS  \subset RHS

Proof of Part I: let x \in LHS

Then, x \in A \triangle (B \triangle C). By definition of symmetric difference,

x \in \{ A - (B \triangle C)\} \bigcup \{(B \triangle C) -A \}

Hence, x \in A, x \notin (B \triangle C), OR x \in B \triangle C, x \in A

That is, x \notin A \bigcap (B \triangle C).

Hence, x \notin A, and x \notin B \triangle C

Hence, x \notin A and x \in B \bigcap C.

Hence, x \in (B \bigcap C)-A.

Hence, x \in B, x \in C, but x \notin A.

Therefore, x \in B, but x \notin A \bigcap C. —- Call this I.

Consider y \in (A \triangle B) \triangle C.

Therefore, y \notin (A \triangle B) \bigcap C.

Therefore, y \notin C, and y \notin A \triangle B

Therefore, y \notin C, and y \in A \bigcap B.

Therefore, y \in (A \bigcap B)-C.

Hence, y \in A, y \in B, but y \notin C.

That is, y \in B, y \notin A \bigcap C. Call this II.

From I and II, LHS \subset RHS.

Part II: TPT: RHS \subset LHS.

Quite simply, reversing the above steps we prove part II.



Nalin Pithwa