NUMB3R5

Relatively prime numbers and Pi

2006-12-04T04:48:00.000-08:00

A prime number is a number whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11, 13, and so on. Two numbers are relatively prime if they have no common factors aside from 1. For example, 12 and 35 are relatively prime since 12 = 2*2*3 and 35 = 5*7 have no common factors.

The chance that two randomly chosen numbers are relatively prime is 6/Pi^2 or approximately 60%.

To test this, I let 10 students write two numbers. The result is close. Out of 10 pairs, 7 pairs are relatively prime. That is 70%. If there are 100 random pairs, the result would be closer.

Game 23

2006-12-03T18:19:00.000-08:00

Here's a trick that will surely amaze students who have not yet discovered the real power of algebra:

Face the students with the blackboard behind you. Ask for one volunteer to write a number between 50 and 100 on the board. Ask him/her to add 76 to the number. Next, ask him/her to add the first digit of the result to the last two-digit number. Finally, subtract the result from the original number he wrote on the board. Surprise the entire class by "guessing" that the final number on the board is 23.

Example:
original number = 53
add 76 to 53 = 129
add 1 to 29 = 30
subtract 30 from 53 = 23

* I discovered recently that there is a coming movie with the title "Game 23".

Goldbach's counterexample?

2006-11-01T04:26:00.000-08:00

Find a composite even number that cannot be expressed as a sum of two prime numbers.

Mystical ball

2006-05-24T21:49:00.000-07:00

--- NATS wrote:

> try this magic:
>
> http://www.mysticalball.com/

THIS IS NOT MAGIC. of course!

any two digit number can be written as

a = 10*n + m

where n and m are any two numbers from 0 to 9

the first step involves taking the sum of n and m

first number = n + m = b

the second step involves subtracting b from a

second number = a - b
= (10*n + m) - (n + m)
= 9*n

which means that no matter what two-digit number you chose, you will always get a number which is a multiple of 9 since n=0,1,2,...,9

so the only possible final numbers are
9*0 = 0
9*1 = 9
9*2 = 18
9*3 = 27
...
9*9 = 81

now, look at the symbols for these numbers and you will see what i mean

that's the power of algebra!

Think of a number

2006-03-18T07:21:00.000-08:00

This very old trick still works. I first know of this during my 2nd year in high school. It was an oral exercise in our algebra class. It goes like this:

1. Think of a number.
2. Multiply the number with 6.
3. Add 10 to the result in #2.
4. Divide the result in #3 by 2.
5. Add 4 to the result in #4.
6. Divide the result in #5 by 3.
7. Subtract your original number from the result in #6.
8. What you will get is 3 in the end.

For people who have a good grasp of algebra, this is not a trick at all. But for those people who are ignorant of algebra (many people can solve algebraic equations even without really having an idea of what algebra really is all about), this trick excites them and makes them wonder how you guessed at the final result without even knowing the number that they are thinking.

I will not explain the trick here. It's for you to discover. For your questions, you can email me at kgargar at yahoo.

Mathematical Constants

2006-02-14T19:09:00.000-08:00

I am so lucky to have stumbled upon a cheap hardbound copy of Mathematical Constants by Steven Finch. The book is part of the Encyclopedia of Mathematics and Its Applications. What an addition to my collection of mathematical books for P350 only where else but at Booksale!

I have a hard time finding a constant named after me. Hahaha.

Infinite series for my sister

2005-11-03T19:07:00.000-08:00

To my sister who is an education student major in mathematics in MSU-IIT, solve for the following infinite series:

1. 3/10 + 3/100 + 3/1000 + 3/10000 + ... ad infinitum
2. 9/10 + 9/100 + 9/1000 + 9/10000 + ... ad infinitum
3. 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... ad infinitum

Kana lang sa. Inform me when you get them!

Continued fractions in the heavens

2005-03-18T16:04:00.000-08:00

7

Having nothing to write at the moment, I surfed the internet and discovered the following article dated October 13, 1997 by popular math and science writer Ivar Peterson in his regular Science News column Math Trek. It mentioned some applications of continued fractions. Read on.

Fractions, Cycles, and Time

In our efforts to measure and understand the universe in which we live, we often find ourselves dealing with "messy" numbers. Our tendency is to replace those inconveniently long strings of digits by rough approximations. In the everyday world, we say that a inch is about two and a half centimeters, light travels at a speed of nearly 300,000 kilometers per second, and the number pi is close to 22/7 or 3.14.

In ancient times, people had to confront awkward numbers in astronomical contexts when they compared the motions of the sun and moon. The unfailing, daily passages of the sun across the sky and the corresponding movements of the stars at night represented one highly predictable cycle. The periodic changes in the moon's appearance and position represented another cycle.

Closer observations over months and years revealed subtle shifts in these patterns. The sun, for instance, doesn't rise at precisely the same point on the horizon every day. The location of sunrise drifts back and forth along the horizon. These recurring excursions define a longer cycle tied to the changing seasons. Similarly, particular stars rise at different locations along the horizon and, at certain times, disappear from the sky for lengthy periods. These movements also have a definite rhythm attuned to the seasons.

In modern terms, taking the day as the standard unit of measurement, the seasons recur every 365.242199 days (a year), while the period of the moon's phases is 29.530588 days (a month). And to be really precise, we must also note that these numbers decrease each century by 1 or 2 in the last decimal place because tidal friction is slowing Earth's rotation and making the day longer. Indeed, official timekeepers add a second every year or so to keep their clocks in sync with Earth's rotation rate.

The ancients didn't use decimals, but they could represent these cycles with remarkable precision by considering ratios. The Athenian astronomer Meton (5th century B.C.), for example, noted that 235 months very nearly equals 19 years. The so-called Metonic cycle is still used to determine the Jewish calendar and set the date of Easter.

The following table gives the error when various numbers of months are compared with the corresponding numbers of years. The listed entries represent successive improvements in the accuracy of the ratio of months to years used to approximate a cycle. Each line is obtained by adding a certain multiple of its predecessor to the one before that. For example, to get 99 months in line 5, you add two times 37 (fourth line) to 25 (third line).

No. of months	No. of years	Error	Multiplier
1 month	0 year	+29.530588	12
12 months	1 year	-10.875143 days	2
25 months	2 years	+7.780302 days	1
37 months	3 years	-3.094841 days	2
99 months	8 years	+1.590620 days	1
136 months	11 years	-1.504221 days	1
235 months	19 years	+0.086399 days	17
4131 months	334 years	-0.035438 days

Meton's approximation is off by just 2 hours 4.4 minutes! And it's bettered only by comparing 4,131 months with 334 years.

Now consider the successive fractions (number of months divided by the corresponding number of years): 12/1, 25/2, 37/3, 99/8, 136/11, 235/19, 4131/334. Those fractions, in turn, can be written in the following manner:

Such expressions are known as continued fractions. They can be used in designing gear trains, including those that might be found in a planetarium to simulate the relative motion of the sun and moon around Earth.

The so-called Antikythera mechanism, apparently constructed in the first century B.C., recovered in 1900 from a Mediterranean shipwreck, and analyzed just a few decades ago, is one of the most striking examples of such engineering in the ancient world. It contained a system of gears whose gear ratios corresponded to well-known astronomical cycles involving the moon, including the Metonic cycle. The mechanism was clearly a type of analog computer, using fixed gear ratios to make calculations displayed as pointer readings on a dial.

The Antikythera mechanism -- the sole survivor of what was undoubtedly a long tradition of astronomical automata -- served primarily as an elegant simulation of the heavens. It was a tabletop monument to Greek and Alexandrian astronomy. Such ingenious devices also illuminated the intimate link between mathematics and astronomy, especially the role of number in astronomical prediction.

By demonstrating an ability to predict the movements of the moon, the rising and setting times of stars, and the changes of the seasons, astronomers could please their rulers while contemplating the subtleties of the evident mathematical order displayed in the heavens.

Source: http://www.maa.org/mathland/mathtrek_10_13.html

Another counterexample

2005-03-09T06:31:00.000-08:00

00000110

Here's another counterexample to the (wrong) conjecture. This time it's from Marty Golubitsky from the University of Houston.

Date: Tue, 8 Mar 2005 23:31:51 -0600 (CST)
From: "Marty Golubitsky"

I believe that the conjecture is false.

A colleague mentioned this counterexample.

Let e(n) be the sequence

1, 1/2, -1/3, -1/4, 1/5, 1/6, -1/7, -1/8 , ...

Define c(n) by c(1) = 0 and c(n) = c(n-1) + e(n-1)

Then c(2) = c(1) + e(1) = 1
c(3) = c(2) + e(2) = 1 + 1/2
c(4) = c(3) + e(3) = 1 + 1/2 - 1/3
etc

Claim: c(n) is convergent

c(n) = 1 + (1/2-1/3) - (1/4-1/5) + (1/6-1/7) + ...
=1 + 1/6 - 1/20 + 1/42 - 1/72
=1 + 14/120 + ...

In particular, note that

1/k - 1/(k+1) - 1/(k+2) + 1/(k+3) = (4k+6)/(k(k+1)(k+2)(k+3))
< 4/(k(k+1)(k+2))
< 4/k^3

whose sum converges.

It follows that

d(n) = e(n-1)/e(n)

Then e(n-1)/e(n) = n/(n-1) times 1, -1, 1, -1, 1, -1, ...

which does not converge.

Approximation by continued fractions

2005-03-06T07:17:00.000-08:00

00000101

The idea of approximating irrational numbers with rational numbers has been brought to my attention upon seriously reading the book Continued Fractions by C.D. Olds. What I am referring to is illustrated in the following example.

The square root of 2 is very simply written as the infinite continued fraction [1,2,2,2,2,2,...]. See previous post for the meaning of this notation. This means that the successive convergents for the square root of 2 are
c₀ = [1] = 1
c₁ = [1,2] = 3/2 = 1.5
c₂ = [1,2,2] = 7/5 = 1.4
c₃ = [1,2,2,2] = 17/12 = 1.41666...
c₄ = [1,2,2,2] = 41/29 = 1.41379...
c₅ = [1,2,2,2,2] = 99/70 = 1.41428...
and so on

It can be shown that the convergents are given by the fraction c_n = p_n/q_n where the numerators and denominators are given in terms of a recursion formula:
p_n = 2p_n-1 + p_n-2
q_n = 2q_n-1 + q_n-2
with the following initial values
p_-1 = 1
q_-1 = 0
Since c₀ = 1 = p₀/q₀, the zeroth convergent, we deduce the values of p₀ and q₀ to be both 1. This gives us an easy method of finding the next closer approximation of the square root of 2 in terms of the two preceding fractional approximations instead of the more cumbersome brute force simplification of the continued fraction expansion. (This statement would be understood if someone simplifies the continued fraction [1,2,2,2,2] and compare it with finding the successive convergents c₀ up to c₅ using the recursion formula above.)

There are theorems on continued fraction approximations of irrational numbers that provide a way of determining the earliest convergent that gives an approximation up to a desired precision. Moreover, it can be shown that all even convergents are less than the irrational number while all odd are greater.

Finally, for the fractional approximations of the square root of 2, here are some facts: (Find some pattern.)

Latest convergent with numerators and denominators less than...

10	c₂	7/5
100	c₅	99/70
1,000	c₇	577/408
10,000	c₁₀	8119/5741
100,000	c₁₂	47321/33461
1,000,000	c₁₅	665857/470832

Appendix:
square root of 2 = 1.4142135623730950488016887242097...
Search the internet for "continued fractions"

The original conjecture

2005-03-03T17:47:00.000-08:00

00000100

This is the observation that prompted me to generalize the (wrong) conjecture. Continued fractions are used to get an optimum rational number approximation to irrational numbers. To approximate a number x, we find positive integers a_n such that

the successive convergents to x are the rational numbers

The original conjecture
In the language of the first (wrong) conjecture, let {c_n} be the sequence of successive convergents of an infinite continued fraction. Then, the derived sequence {d_n} is convergent.

References:
http://mathworld.wolfram.com/ContinuedFraction.html
Olds, C. D., Continued Fractions, New York: Random House, 1963

The counterexamples

2005-02-28T06:02:00.000-08:00

00000011

Here are the three counterexamples to the conjecture below. I don't want to write anything about them so I just post the original reply. Again, my sincerest thanks to the three for making my life simpler.

From: "James Yorke"
Date: Tue, 22 Feb 2005 14:17:02 -0500
Try 10 to the power - n squared for c(n).

From: "Jose Perico Esguerra"
Date: Wed, 23 Feb 2005 13:11:15 +0000 (GMT)
Counterexample: c(n)=0.9^(n^3)

From: "Richard Taylor"
Date: Wed, 23 Feb 2005 15:13:52 -0500
I think this is not true. For example try c(n)=e(n)/2^n where e(n) is 1 if n has remainder 0 or 1 when divided by 3 and where e(n) is -1 if n has remainder 2 when divided by 3. If I didn't make a mistake this gives d(n)=-6 if 3 divides n, d(n)=2/3 if n has remainder 1 when divided by n and d(n)=-1 if n has remainder 2 when divided by n. This sequence does not converge to anything.

Best wishes,
Richard Taylor

A conjecture

2005-02-27T04:46:00.000-08:00

2

I was too hasty to generalize an observation. Here's my conjecture (which was easily rejected by a counterexample by no less than James Yorke of University of Maryland):

Let {c_n} be a convergent infinite sequence that does not converge to a constant sequence. Derive another infinite sequence {d_n} as follows:

d_n=(c_n-c_n-1)/(c_n+1-c_n)

Then, {d_n} is also convergent.

Unable to prove or disprove the conjecture, I emailed several mathematicians around the world to contaminate the challenge. Within few hours, James Yorke replied. Days later, Richard Taylor of Harvard University provided another counterexample. Actually, I am not sure if Taylor's sequence was really a counterexample since I did not check it out. The third counterexample, similar to Yorke's, is from my masteral thesis adviser and theoretical physics buddy Perry Esguerra of University of the Philippines.

I thank them all.

Who is Ed Matthew?

2005-02-27T04:01:00.000-08:00

1

He is Quennie's eldest son*. I forgot his birthday but I remember giving him his first birthday anniversary gift a book entitled The Lore of Large Numbers by Philip J. Davis. I also remember giving him a poem along with the book but I forgot the text of that poem.

So this site is all about NUMBERS. Obviously, I'm very fond of them.

* Quennie is a cousin.