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English Vocabulary / Some Words Added to the Dictionary
« on: July 21, 2014, 06:20:08 PM »
Muggle: n.       
In the fiction of J.K. Rowling: a person who possesses no magical powers. Hence in allusive and extended uses: a person who lacks a particular skill or skills, or who is regarded as inferior in some way.

 A method of collectively finding one to blame for a mistake no one is willing to confess to. Often occurs in the form of a meeting of colleagues at work, gathered to decide who is to blame for a screw up.

 A homosexual person's ability to identify another person as homosexual by interpreting subtle signals conveyed by their appearance, interests, etc.

A young woman regarded as independent and strong or aggressive, especially in her attitude to men or in her sexuality.

 The third film, book, event, etc. in a series; a second sequel.

A person closely resembling a smaller or younger version of another.

A person in their teens or twenties who has an aptitude for computers and the Internet.

Spending one's employer's Internet and email facilities for personal activities during working hours.

Prejudice or discrimination on the grounds of appearance.

Derogatory a food that contains genetically modified ingredients.

Guitar riffs.

n. The ability to recover from near-defeat in a competition; the ability to recover from a setback.

n. A rebuttal for an accusation before it is made.

v. Searching the Internet for instances of one's own name or links to one's own website.


n. The physical world, as opposed to virtual.

Departments / Nanofluid and smart fluid
« on: July 17, 2014, 02:16:28 PM »
A Nanofluid is a fluid containing nanometer-sized particles, called nanoparticles. These fluids are engineered colloidal suspensions of nanoparticles in a base fluid.The nanoparticles used in nanofluids are typically made of metals, oxides, carbides, or carbon nanotubes. Common base fluids include water, ethylene glycol and oil.

Nanofluids have new properties that make them potentially useful in many applications in heat transfer, including microelectronics, fuel cells, pharmaceutical processes, and hybrid-powered engines, engine cooling thermal management, inland refrigerator, hair-raiser, heat exchanger,in grinding, machining and in boiler flue gas temperature reduction. They exhibit enhanced thermal conductivity and the convective heat transfer coefficient compared to the base fluid. Knowledge of the rheological behaviour of nanofluids is found to be very critical in deciding their suitability for convective heat transfer applications.

In this new age of energy awareness, our lack of abundant sources of clean energy and the widespread dissemination of battery operated devices, such as cell-phones and laptops, have accented the necessity for a smart technological handling of energetic resources. Nanofluids have been demonstrated to be able to handle this role in some instances as a smart fluid.

Departments / Re: math problem of the day(another)
« on: September 22, 2013, 01:23:57 PM »
When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y?
ans: 35

Nutrition and Food Engineering / John Venn's biography
« on: August 04, 2013, 03:54:44 PM »
please see the The following link to learn about John Venn, one of the pioneer mathematician of the history

Nutrition and Food Engineering / exercise your brain
« on: August 04, 2013, 03:11:35 PM »
A committee of three is chosen from five councilors - Adams, Burke, Cobb, Dilby and Evans.

What is the probability Burke is on the committee?

Faculty Forum / math problem of the day
« on: July 18, 2013, 02:31:08 PM »
L= {-6, -5, -4, -3, -2}
P = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set L above and an integer is to be randomly selected from set P above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5

Science Discussion Forum / math problem of the day
« on: July 10, 2013, 12:54:39 PM »
Pat, kate and mark charged a total of 162 hours to certain project.If pat charged twice as much time to the project  as kate and 1/3 as much time as mark , how many more hours did mark charge to the project than kate?

Ans: 90 hours

Faculty Forum / Farmat'e Little Theorem
« on: May 23, 2013, 02:45:23 PM »
Fermat's Little Theorem

It comes from observation of multiplication tables modulo prime number p that all rows are nothing but a permutation of the first row {1, 2,... , p-1}. The same is true for the columns. Here I wish to verify that this is indeed so. The proof depends on the Euclid's Proposition VII.30
    If two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers.

Let a be one of the positive remainders of division by p: 0 < a < p. I wish to prove that [a]p, [2a]p, [3a]p, ..., [(p-1)a]p are all different. Which, in terms of remainders, claims that in the sequence {a, 2a, 3a, ..., (p-1)a} no two numbers are congruent modulo p. Assume the opposite: let there be two numbers 1 ≤ m < n < p such that na = ma (mod p). This would imply that p|a(n-m). By the Proposition VII.30 either p|a or p|(n-m). But both a and (n-m) are positive integers less than p. So it can't divide either of them. Contradiction.

It indeed follows that the set {[a]p, [2a]p, [3a]p, ..., [(p-1)a]p} is just a permutation of the set
    {[1]p, [2]p, [3]p, ..., [p-1]p},

or that rows in the multiplication tables are just permutations of the first row.

If two sets are permutations of each other, then products of their elements are clearly equal:
[(p-1)!]p   = [1]p·[2]p·[3]p·...·[p-1]p
    = [a]p·[2a]p·[3a]p·...·[a(p-1)]p
    = [ap-1(p-1)!]p
    = [ap-1]p·[(p-1)!]p

Now, dividing by [(p - 1)!]p (which is not 0 by Euclid's Proposition VII.30) gives 1 = [ap-1]p. Or, in terms of remainders,
    ap-1 = 1 (mod p)

Going over the proof we may notice that it's an overkill to require a to be less than p. The proof remains valid for any a not divisible by p.

The statement first appeared without proof in a letter dated October 18, 1640 that Fermat wrote to Frenicle de Bessy . The first proof was given by Leibniz (1646-1716) and the one above was found by Ivory in 1806. Euler proved the theorem in 1736 and its generalization in 1760. The theorem is now known as the Fermat's Little Theorem to distinguish it from the Fermat's Last or Great Theorem. The latter has been finally established by the Princeton mathematician Andrew Wiles (with assistance from Richard Taylor) in 1994.

The set {{[0]N, [1]N, [2]N, ..., [p-1]N} is an additive group. The set {[1]p, [2]p, [3]p, ..., [p-1]p} is a multiplicative group. For the latter we saw that to every element [a]p in the group, there corresponds a permutation
    {[a]p, [2a]p, [3a]p, ..., [(p-1)a]p}

of its elements. This relation is a group isomorphism: it preserves the group operation and is 1-1. A general statement, known as the Cayley's Theorem, asserts that this is a rule:
    Every group is isomorphic to a group of permutations.

Departments / Chinese Remainder theorem
« on: May 23, 2013, 02:40:26 PM »
Chinese Remainder Theorem, CRT, is one of the jewels of mathematics. It is a perfect combination of beauty and utility or, in the words of Horace, omne tulit punctum qui miscuit utile dulci. Known already for ages, CRT continues to present itself in new contexts and open vistas for new types of applications. So far, its usefulness has been obvious within the realm of “three C's”. Computing was its original field of application, and continues to be important as regards various aspects of algorithmics and modular computations. Theory of codes and cryptography are two more recent fields of application.

According to D. Wells, the following problem was posed by Sun Tsu Suan-Ching (4th century AD):

There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder is 2. What will be the number?

Oystein Ore mentions another puzzle with a dramatic element from Brahma-Sphuta-Siddhanta (Brahma's Correct System) by Brahmagupta (born 598 AD):

An old woman goes to market and a horse steps on her basket and crashes the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time they came out even. What is the smallest number of eggs she could have had?

Problems of this kind are all examples of what universally became known as the Chinese Remainder Theorem. In mathematical parlance the problems can be stated as finding n, given its remainders of division by several numbers m1, m2, ..., mk:
(1)   n = n1 (mod m1)
n = n2 (mod m2)
n = nk (mod mk)

The modern day theorem is best stated with a couple of useful notations. For non-negative integers m1, m2, ..., mk, their greatest common divisor is defined as

gcd(m1, m2, ..., mk) = max{s: s|mi, for i = 1, ..., k},

where, as always, "s|m" means that s divides m exactly. The least common multiple of k numbers is defined as

lcm(m1, m2, ..., mk) = min{s: s > 0 and mi|s, for i = 1, ..., k},

Both gcd() and lcm() are symmetric functions of their arguments. They are complementary in the sense that, for k = 2,

gcd(m1, m2)·lcm(m1, m2) = m1· m2.

(A proof and an interactive illustration for this identity appears elsewhere.)

However, for k > 2 a similar identity does not in general hold. For an example, consider two triplets: {2, 4, 16} and {2, 8, 16}. Both have exactly the same gcd and lcm but obviously different products. On the other hand, both gcd and lcm are associative:

gcd(m1, (gcd(m2, m3)) = gcd(gcd(m1, m2), m3)

and, both equal gcd(m1, m2, m3). Similarly,

lcm(m1, (lcm(m2, m3)) = lcm(lcm(m1, m2), m3)

If, for a prime p, pαi|mi, with αi being the largest exponent with that property, then pα|lcm(m1, m2, ..., mk), where α = max{αi} and α is the largest exponent with that property. Similarly, the greatest common divisor of several numbers is the product of the largest powers of the primes that divide all the given numbers.

Associativity allows one to proceed a step at a time with an inductive argument without putting all eggs into a basket at once. Jumping at the opportunity I'll prove the most basic case of k = 2.
Theorem 1

Two simultaneous congruences

n = n1 (mod m1) and
n = n2 (mod m2)

are only solvable when n1 = n2 (mod gcd(m1, m2)). The solution is unique modulo lcm(m1, m2).

(When m1 and m2 are coprime their gcd is 1. By convention, a = b (mod 1) holds for any a and b.)

By a generalization of the Euclid's algorithm, there are integers s and t such that

tm1 + sm2 = gcd(m1, m2).

Since n2 - n1 = 0 (mod gcd(m1, m2)), for some, possibly different t and s,

tm1 + sm2 = n2 - n1.

Then n = tm1 + n1 = -sm2 + n2 satisfies both congruences in the theorem. This proves the existence of a solution.

To prove the uniqueness part, assume n and N satisfy the two congruences. Taking the differences we see that

N - n = 0 (mod m1) and N - n = 0 (mod m2)

which implies N - n = 0 (mod lcm(m1, m2)).

As was previously stated, a more general theorem can now be proved by induction.
Theorem 2

The simultaneous congruences (1)

    n = n1 (mod m1)
n = n2 (mod m2)
n = nk (mod mk)
are only solvable when ni = nj (mod gcd(mi, mj)), for all i and j, i ≠ j. The solution is unique modulo lcm(m1, m2, ..., mk).

Theorem 1 serves the initial step verification. Assume the theorem holds for k congruences and consider k + 1 of them.

n = n1 (mod m1)
n = n2 (mod m2)
n = nk (mod mk)
n = nk+1 (mod mk+1)

Let s be a solution to the first k equations. Then the congrurence n = s (mod lcm(m1, m2, ..., mk)) has a solution. (Observe that every solution of the latter also satisfies the first k congruences.) To be able to apply the already proven Theorem 1, we need to show that

s = nk+1 (mod gcd(lcm(m1, ..., mk), mk+1)).

Let's write g u = gcd(m u, mk+1), u = 1, ..., k. Then we know that n u = nk+1 (mod g u) for these values of u. But n u = s + t um u, for some t u, implying nk+1 = s + t um u (mod g u) so that

s = nk+1 (mod g u), u = 1, 2, ..., k.

If so,
    s   = nk+1 (mod lcm(g1, ..., gk))
        = nk+1 (mod lcm(gcd(m1, mk+1), ..., gcd(mk, mk+1))).
        = nk+1 (mod gcd(lcm(m1, ..., mk), mk+1))

because lcm(gcd(m1, mk+1), ..., gcd(mk, mk+1)) = gcd(lcm(m1, ..., mk), mk+1).

Thus the system

n = s (mod lcm(m1, m2, ..., mk))
n = nk+1 (mod mk+1)

has a solution which is unique modulo lcm(lcm(m1, m2, ..., mk), mk+1) = lcm(m1, m2, ..., mk, mk+1). It also satisfies the whole set of k + 1 congruences.

The simultaneous congruences (1)

n = n1 (mod m1)
n = n2 (mod m2)
n = nk (mod mk)

where all mi's are pairwise coprime has a unique solution modulo m1·m2· ... ·mk.

If some mi's are not mutually prime, a solution may not exist unless the corresponding congruence agree. For example, the system n = 1 (mod 2) and n = 2 (mod 4) has not solution, while the system n = 1 (mod 2) and n = ±1 (mod 4) does.

Let's now solve the two problems we started the page with.
Problem #1

p1:   x = 2 (mod 3)
p2:   x = 3 (mod 5)
p3:   x = 2 (mod 7)

From p1, x = 3t + 2, for some integer t. Substituting this into p2 gives 3t = 1 (mod 5). Looking up 1/3 in the division table modulo 5, this reduces to a simpler equation

t = 2 (mod 5)

which, in turn, is equivalent to t = 5s + 2 for an integer s. Substitution into x = 3t + 2 yields x = 15s + 8. This now goes into p3: 15s + 8 = 2 (mod 7). Casting out 7 gives s = 1 (mod 7). From here, s = 7u + 1 and, finally, x = 105 u + 23.

Note that 105 = lcm(3, 5, 7). Thus we have solutions 23, 128, 233, ...
Problem #2

q1:   x = 1 (mod 2)
q2:   x = 1 (mod 3)
q3:   x = 1 (mod 4)
q4:   x = 1 (mod 5)
q5:   x = 1 (mod 6)
q6:   x = 0 (mod 7)

With the experience we acquired so far, the combination of q1-q5 is equivalent to


x = 1 (mod 60),

x = 60t + 1. Plugging this into q6 gives 60t = -1 (mod 7). Casting out 7 simplifies this to 4t = 6 (mod 7) and then 2t = 3 (mod 7). From the division tables modulo 7, 3/2 = 5 (mod 7). Therefore, t = 7u + 5. Finally, x = 420u + 301. Allowing for an average size farmer, the most likely number of eggs she might expect to be compensated for is 301.

Note: Chinese Remainder Theorem finds an important application in the clendrical calculations.

Hadith / Recently solved problems
« on: April 27, 2013, 12:46:15 PM »

    Gromov's problem on distortion of knots (John Pardon, 2011)
    Circular law (Terence Tao and Van H. Vu, 2010)
    Hirsch conjecture (Francisco Santos Leal, 2010[12])
    Serre's modularity conjecture (Chandrashekhar Khare and Jean-Pierre Wintenberger, 2008[13])
    Heterogeneous tiling conjecture (squaring the plane) (Frederick V. Henle and James M. Henle, 2007)
    Road coloring conjecture (Avraham Trahtman, 2007)
    The Angel problem (Various independent proofs, 2006)
    The Langlands–Shelstad fundamental lemma (Ngô Bảo Châu and Gérard Laumon, 2004)
    Stanley–Wilf conjecture (Gábor Tardos and Adam Marcus, 2004)
    Green–Tao theorem (Ben J. Green and Terence Tao, 2004)
    Cameron–Erdős conjecture (Ben J. Green, 2003, Alexander Sapozhenko, 2003, conjectured by Paul)[14]
    Strong perfect graph conjecture (Maria Chudnovsky, Neil Robertson, Paul Seymour and Robin Thomas, 2002)
    Poincaré conjecture (Grigori Perelman, 2002)
    Catalan's conjecture (Preda Mihăilescu, 2002)
    Kato's conjecture (Auscher, Hofmann, Lacey, McIntosh, and Tchamitchian, 2001)
    The Langlands correspondence for function fields (Laurent Lafforgue, 1999)
    Taniyama–Shimura conjecture (Wiles, Breuil, Conrad, Diamond, and Taylor, 1999)
    Kepler conjecture (Thomas Hales, 1998)
    Milnor conjecture (Vladimir Voevodsky, 1996)
    Fermat's Last Theorem (Andrew Wiles and Richard Taylor, 1995)
    Bieberbach conjecture (Louis de Branges, 1985)
    Princess and monster game (Shmuel Gal, 1979)
    Four color theorem (Appel and Haken, 1977)

Hadith / Riemann Hypothesis
« on: April 27, 2013, 12:45:10 PM »

Riemann Hypothesis

Formulated in his 1859 paper, the Riemann hypothesis in effect says that the primes are distributed as regularly as possible given their seemingly random occurrence on the number line. Riemann's work gave an 'explicit' formula for the number of primes less than x in terms of the zeros of the zeta function. The first term is x/log(x). The Riemann hypothesis is equivalent to the assertion that other terms are bounded by a constant times log(x) times the square root of x. The Riemann hypothesis asserts that all the 'non-obvious' zeros of the zeta function are complex numbers with real part 1/2.

Hadith / How to make saline solution
« on: April 27, 2013, 12:36:13 PM »
Saline solution is used to cleanse the mouth when you have a canker sore, a soft tissue laceration, are healing after oral surgery, and so on. Rinsing with warm salt water two to three times a day may help to relieve a toothache, because salt water works as an antiseptic to remove bacteria from the infected area.

Saline solution is easily prepared in your own home, and only requires two to three ingredients that almost everyone has on hand.
Prep Time: 12 minutes
Total Time: 12 minutes

    1 Teaspoon Salt

    8oz. Warm Water

    2 Teaspoons Baking Soda (Optional)


Start by bringing the water to a rolling boil, for 10 minutes. After boiling, leave the water to stand until it is cool enough to rinse with, but warm enough to completely dissolve the next two ingredients.

When cooled accordingly, place the salt in the water while gently siring until he salt has been completely dissolved.

As an option, dissolve the baking soda in the water along with the salt.

Use the saline solution as directed, and discard any leftover solution.

Hadith / math problem of the day
« on: April 27, 2013, 12:33:14 PM »
What is the value of the expression (1 + 1/2)(1 + 1/3)(1 + 1/4).....(1 + 1/100)?
solution: 50.5

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