nature of mathematics

In praise of the Queen of all Sciences !

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  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

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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, MoMath‘s 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.  mathplay.momath.org

“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.” — 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.  studentsessions.momath.org

“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!” — 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 mathgift.momath.org.

For more mathematical fun, be sure to check out MoMath’s complete lineup of upcoming adult events and children’s programs at events.momath.org.Regards,
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 🙂

Wisdom of V. I. Arnold, immortal Russian mathematician

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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

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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.

QED.

Cheers,

Nalin Pithwa

Number theory: let’s learn it the Nash way !

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Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

A fifth primer: plane geometry tutorial for preRMO and RMO: core stuff

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  1. Show that three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.
  2. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
  3. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: prove that BX and DY trisect the diagonal AC.
  4. If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. Prove this.
  5. Show that the straight lines which join the middle points of opposite sides of a quadrilateral bisect one another.
  6. From two points A and B, and from O the mid-point between them, perpendiculars AP, and BQ, OX are drawn to a straight line CD. If AP, BQ measure respectively 4.2 cm, and 5.8 cm, deduce the length of OX. Prove that OX is one half the sum of AP and BQ. or \frac{1}{2}(AP-BQ) or \frac{1}{2}(BQ-AP) according as A and B are on the same side or on opposite sides of CD.
  7. When three parallels cut off equal intercepts from two transversals, prove that of three parallel lengths between the two transversals the middle one is the Arithmetic Mean of the other two.
  8. The parallel sides of a trapezium are a cm and b cm respectively. Prove that the line joining the middle points of the oblique sides is parallel to the parallel sides, and that its length is \frac{1}{2}(a+b) cm.
  9. OX and OY are two straight lines, and along OX five points 1,2,3,4,5 are marked at equal distances. Through these points parallels are drawn in any direction to meet OY. Measure the lengths of these parallels : take their average and compare it with the lengths of the third parallel. Prove geometrically that the third parallel is the mean of all five.
  10. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is outside the parallelogram : prove that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair.  (Draw the diagonals,and from their point of intersection suppose a perpendicular drawn to the given straight line.)
  11. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side. It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken).
  12. The sum of the perpendiculars drawn from any point within the an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore, constant. Prove this.
  13. Equal and parallel lines have equal projections on any other straight line. Prove this.

More later,

Cheers,

Nalin Pithwa.