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Tuesday, September 29, 2009

Funny, Weird, Strange Billboards!

These are pictures of some very funny, weird and/or strange billboards, mostly from the US and Canada. Most of them require no explanation, but it is hard to tell whether some of them are real or just somebody's idea of what a funny billboard should really say! In any case, enjoy!!
































Friday, September 25, 2009

Vedic Mathematics Lesson 22: Vinculums

In this lesson, we are going to step back from the various arithmetic techniques we have discussed so far. Instead of learning a new arithmetic technique, we are going to deal with a technique that can be used to make those arithmetic techniques easier to perform when large numbers are involved.

You can find all the previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2

Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II

We are all familiar with the place value system of modern mathematics where each digit in a multi-digit number takes on a different place value depending on how far to the left or right of the decimal point it is located. The place value is calculated as the product of the face value of the number with an appropriate power of 10. By the way, this place value notation, which makes numbers much easier to handle (rather than roman numerals, for instance) was invented in India in ancient times. Positional notation and the concept of zero were two of the most important contributions of ancient India to modern mathematics.

To put it in more concrete terms, consider a number like 943,875. This number is actually 9 x 100,000 + 4 x 10,000 + 3 x 1,000 + 8 x 100 + 7 x 10 + 5 x 1. In a multi-digit number, the powers of 10 increase steadily from right to left and each power of 10 has a single-digit coefficient associated with it. This single-digit coefficient is almost always a number from 0 to 9.

However, there is nothing inherently wrong with this coefficient being a negative number between 0 and -9. This is usually not very common and may sound very confusing at first, but is actually quite simple to grasp once one becomes familiar with it. We will first explain the concept and then we will deal with some examples of computations that can be simplified significantly by the use of negative coefficients.

The basic concept of using negative coefficients is also referred to as using Vinculums. Vinculum is a latin word that derives from Vincire, to tie. It denotes a bond between two numbers that are linked together by some property.

When we use vinculums in Vedic Mathematics, that bond is usually the 10's complement of individual digits or groups of digits. A number and its 10's complement are tied together because the pair always adds up to a power of 10. This bond is sometimes utilized to rewrite numbers in vinculum form before performing mathematical computations on them.

Let us first take a few examples to see how the concept works. Then we will formalize how to rewrite any number in vinculum form. First, let us take the number 6783. It is easy to see that 6783 = 6 x 1000 + 7 x 100 + 8 x 10 + 3 x 1. It is also easy to see that 6783 = 6 x 1000 + 8 x 100 + (-2 x 10) + 3 x 1. Representing negative numbers by bold, red letters, we can say that 6783 = 6823.

What we have done above is essentially converted 6783 into one of several vinculum forms that are possible for any given numbers. In this case, we converted one of the digits of the original number (8) to its vinculum form (2). Note that 8 and 2 are 10's complements of each other. As mentioned earlier, the bond referred to in vinculum is the 10's complement of a digit or group of digits.

Now, let us take the same example, 6783, and find a different vinculum form of it. It is easy to verify that 6783 = 7383. In this case, we have converted the 7 to its vinculum form. One can also verify that 6783 = 7223. In this case, we have taken the group of digits, 78, and converted the entire group to its vinculum form. Thus, we see that any multi-digit number might have several different vinculum forms, depending on which digit or group of digits is converted to its vinculum form.

But, we are not restricted to converting one digit or a consecutive group of digits to vinculum form to derive one of the many vinculum forms of the whole number. We can convert any number of non-contiguous digits of the original number into their vinculum forms and derive different vinculum forms of the original number also. For instance, 6783 = 7397. In this case, we have taken the vinculum form of the individual digits, 7 and 3 to derive this vinculum form of the original number.

The procedure for deriving a vinculum form of a given number by converting digits and groups of digits into their vinculum form is explained below:

  • Identify the digit to the left of the digit or contiguous set of digits that you want to convert to vinculum form. Call this the left digit. If there are no digits to the left of the digit you want to convert to vinculum form, pad the number with zeroes to the left as needed
  • Increase this left digit by 1. If this number was already a 9, then it becomes 10, which results in that digit becoming 0 and the digit to its left being incremented (normal carryover from right to left applies)
  • Find the 10's complement of the digit or group of digits that you want to convert to vinculum form and substitute them in place of the original digits, but with some indication that they are negative coefficients. Commonly, they are either underlined or a line is drawn on top of them to indicate that they are negative coefficients

That is all there is to it! We will work out a few examples below to illustrate the working of the procedure. In each case, we have colored the digits we are converting to vinculum form in blue and the negative digits in the vinculum form are denoted by red and bold coloring.

83974 = 84026 (026 is the 10's complement of 974, and 3 has been changed to 4 as per the procedure above)

789347 = 811353 (11 is the 10's complement of 89 and 7 has been changed to 8 as per the procedure above. Similarly, 7 has been changed to its 10's complement and the previous digit, 4, has been incremented to a 5)

83974 = 96026 (Note that in this example, the 8 at beginning of the original number was changed to 9 during the conversion of the 3 into its vinculum form. However, instead of 7 (the 10's complement of 3), we have a 6 there because when we convert 74 to its vinculum form, it changes the 9 to a 10, resulting in carryover of a 1 from that digit. This carried over 1 is added to negative 7 to result in the final negative 6)

97886 = 102114 (9 has been incremented to 10 and 7886 has been changed to its 10's complement, 2114)

Obviously, converting a number to one of its vinculum forms really does not accomplish anything significant, so it is not surprising that the method is quite simple. The question still remains: why do we want to convert a number to any of its vinculum forms?

To understand the answer to that question, one first needs to realize that a number and any of its vinculum forms are equivalent when it comes to any mathematical operations. Addition, subtraction, multiplication and division can all be performed exactly as we are used to doing them after converting one of both numbers involved into any of their vinculum forms. The only caveat is that since the vinculum form has some negative coefficients, one has to be careful about the signs of the products of mathematical operations such as multiplications and divisions. For instance, the multiplication of two negative numbers with each other gives a positive answer while the multiplication of 2 numbers of opposite sign gives a negative answer, and so on.

Most of us also intuitively understand that computations with small numbers are much easier to perform than computations that involve large numbers, even when the numbers are just one digit each. Multiplying or dividing by numbers like 1 and 2 is much easier than multiplying or dividing by numbers like 8 or 9. This combination of the equivalency of a number and its vinculum forms, and the ease of computations using small numbers compared with computations using large numbers, gives the vinculum all its power in vedic mathematics.

To convince ourselves of this, we will tackle a few examples dealing with various techniques we have gone through before, and illustrate how one can make a problem simpler simply by working with an equivalent vinculum form of a number.

Let us start with a multiplication problem that we will try to work out using the Urdhva-Tiryak sutra (vertically and cross-wise). We will try to calculate 29 x 48. Using the normal method, we work it out as below:

29
48
---------
072
520
800
---------
1392

Now, let us try it after converting the large numbers, 9 and 8 into their vinculum forms. We get the problem below:

31
52
-------
0002
0110
1500
-------
1412
1392

The answer, as expected is identical, but instead of a vertical product of 9 and 8, we only had to a vertical product of -1 and -2 (which gives us +2). Similarly, instead of 8x2 + 4x9, we had to do -2x3 + -5x1 which gives us -11. The resulting answer is a vinculum number, which can be converted into a normal number simply by following the procedure for producing a vinculum form in reverse.

Let us now try it on 291 x 88. In this case, the normal vertically and cross-wise procedure would be to pad the 88 with a 0 in front and do it as below:

291
088
---------
00008
00800
08800
16000
00000
------------
25608

Now let us try it as below:

311
112
--------
00002
00010
00400
04000
30000
----------
34412
25608

Notice how the large number of carryovers due to the cross-products of large numbers with each other have been eliminated by the conversion of the large numbers into their much smaller vinculum forms! The sums of the cross-products are also easier to calculate because of the smaller numbers involved in the cross-products.

The same kind of time-savings because of having to deal with smaller numbers can be achieved when this method is applied to straight division using the Dhvajanka sutra also. To illustrate, let us consider the example of 81356/29.

•9
2
•|8 1 3 5|6
•| 4 7 1 5
---------------
•|2 8 0 5|11

Notice the large number of times we had to adjust the intermediate quotients and remainders so that the subtraction of the product of the quotient digit with the flag digit does not result in a negative number. Also, it is easy to see that the products themselves are much harder to calculate because of the large numbers involved in the products.

Let us now try the same problem as below:

1
3
•|8 1 3 5|6
•| 2 2 0 0
--------------
•|2 7 0 5|11
•|1
---------------
•|2 8 0 5|11

The answer is once again 2805 with a remainder of 11. A few words of explanation may be necessary in this case to explain the procedure though.

First of all, we converted 29 to its vinculum form 31. So, the flag digit becomes -1. Dividing 8 by 3 gives us a quotient digit of 2 and a remainder of 2. The two digit number we get by putting this remainder next to the second numerator digit gives us 21. Now, we need to subtract the product of the flag digit with the first quotient digit from this 21. The product of the quotient digit (2) with the flag digit (-1) is -2. Subtracting -2 from 21 is equivalent to adding 2 to 21, resulting in an intermediate numerator of 23.

This leads to a quotient digit of 7 and a remainder of 2. Putting 2 in front of the next digit of the numerator gives us 23. Subtracting the product of the flag digit (-1) with the latest quotient digit (7) from this gives us 23 - (-7) = 30. When 30 is divided by 3, we get a quotient of 10 (which is a 2-digit number) and remainder of 0. When we get a quotient that has more than 1 digit, normal carryover rules apply, so, we put the 0 down in its own column and the 1 under the column to the left.

However, remember that the previous quotient was 10, so when it comes time to multiply the quotient digit by the flag digit, use 10 as the quotient digit (even though it is technically not a single "digit"). That is how we get our next intermediate numerator of 05 - (-10) = 15. This gives us the next quotient digit of 5 and remainder of 0. The final remainder is calculated as 06 - (-5) = 11.

Now, add up the numbers under each column to make sure the carryover digit is accounted for correctly, and we get our final quotient of 2805 and final remainder of 11. As you can see, the division by 2 or 3 is of approximately the same level of difficulty. But the products of the quotient digits with the flag digit have been greatly simplified because of the use of the vinculum form of 29. Moreover, it was never necessary to adjust the quotient and remainder of any intermediate division to avoid negative numbers.

Another example of taking advantage of the vinculum form is illustrated below:

•98
3
•|2 3 1 4|23
•| 2 812 0
----------------
•0|0 5 8 4|-1009
•|0 5 8 1|185

02
4
•|2 3 1 4|23
•| 2 3 3 0
----------------
•0|0 5 7 1|185
•|•1
--------------------
•|0 5 8 1|185

On some rare occasions, though, using the vinculum form does not provide as much of an advantage as one might expect. In the example below, the denominator is converted to one of its vinculum forms to replace the 8 with the much smaller 2, but the savings in work is not dramatic.

•81
3
•|5 3 4 9|89
•| 2 3 1 0
-----------------
0|1 4 0 5|-316
0|1 4 0 4|65

21
4
•|5 3 4 9|89
•| 1 3 3 2
-----------------
0|1 3 9 3|446
•|0 0 1
-----------------
•|1 4 0 3|446
•|1 4 0 4|65

Below is another example where converting the denominator to the vinculum form does not provide a huge advantage because the quotient digits become too large, resulting in carryovers (and carryovers imply large quotient "digits" that have to be kept track of carefully during the cross-multiplications). But we do eliminate having to adjust the quotient and remainders to avoid negative results. In general, if the flag digits are all vinculums (negative numbers), one never has to worry about negative results from the subtractions.

•88
1
•|9 8 4 7|94
•| 4 6 5 9
---------------
•0|5 2 3 8|50

12
2
•|9 8 4 7|94
•| 1 0 1 0
-----------------
•0|4 1 1 5|614
•|1 1 2
-----------------
•|5 2 3 5|614
•|5 2 3 8|50

Note that even though we have shown only examples that involve multiplication and division, the use of vinculums is not restricted to these techniques or operations. The vinculum form of a number can be used anywhere where the normal form of the number can be used and vice versa, making the vinculum very powerful and versatile. To demonstrate this, let us work out a simple example as below:

•8
3
•|12 4 3 8|9
•|0 2 1 1
--------------------
•| 4 0 4 5|699
•|1 3 8
--------------------
•| 3 3 4 5|699
•| 3 2 5 5|699
| 3 2 7 3|15

In this case, the division of 12 by 3 first gives us 4 as the first quotient digit and a remainder of 0. The two-digit number that results from placing this remainder and the next digit of the numerator together is 04. The product of the quotient digit and the flag digit is 32. 4 - 32 = -28. Dividing -28 by 3 gives us a quotient of -10 and a remainder of 2 (another way to say this is that -10 x 3 + 2 = -28). Note that -9 can not be considered the quotient in this case because -9 x 3 = -27 > -28.

We put the -10 down as vinculum digits in the quotient row (with the 1 carried over to the next column as appropriate), and the 2 as a remainder next to the next numerator digit, 3. From 23, we need to subtract the product of the latest quotient digit and the flag digit. This product is -10 x 8 = -80, and 23 - -80 = 23 + 80 = 103. So our next intermediate numerator is 103, and the process is continued as in the figure.

At the end, we add up the columns of numbers, making sure to account for vinculum numbers by considering them to be negative numbers. If any digits in the final answer are vinculum numbers, we convert them to normal form by decrementing the previous digit and substituting the vinculum digit(s) with its 10's complement (the reverse of the procedure for converting a digit to its vinculum form).

In this case, the remainder turns out to be larger than the denominator, so we have to go through the extra step of dividing the remainder by the quotient, adding this quotient to the original quotient and keeping the remainder as the final remainder.

As you can see, this procedure can sometimes be more convenient than the normal procedure we have used for straight division because we no longer have to adjust the quotients and remainders of intermediate divisions to make sure that intermediate numerators do not become negative. In this method, if the intermediate numerator does become negative, we just shrug it off, put down a vinculum digit in the quotient and continue on.


Notice that changing the problem so that we make the flag digit a vinculum digit simplifies it much further, as shown below.

2
4|12 4 3 8|9
|0 2 3 2
---------------
| 3 2 6 2|53
|1
---------------
| 3 2 7 2|53
| 3 2 7 3|15

In general, making the flag digits vinculum digits pays good dividends because it enables us to work on the problem without having to adjust the intermediate quotients and remainders very often to avoid negative intermediate numerators.

Hopefully, this digression into the land of vinculums and vinculum digits gives you one more weapon to add to your arsenal. It is important to note, as always, that not all the tools in your toolbox can be used at all times. One has to consider the problem carefully before making a shrewd guess about which tool might be able to solve the problem with the least difficulty. Practice is the key to making this kind of judgment, so hopefully you will find the time to practice working with vinculums. Good luck, and happy computing!

Thursday, September 24, 2009

Michael Moore Kills Capitalism with Kool-Aid

I found this interesting review of Michael Moore's new movie Capitalism: A Love Story by Michael Covel in my mailbox today. More interesting than the review itself is reaction of the audience related by the reviewer. On the one hand, we have people protesting against health-care reform because it is "socialism". And then we have an audience wildly cheering a one-sided, misguided attempt at labeling capitalism as the root of all evil. How did we become so schizophrenic?


Michael Covel is an author, director, and founder of TurtleTrader.com. Covel's books include the best selling Trend Following: Learn to Make Millions in Up or Down Markets, which has sold over 100,000 copies, and The Complete TurtleTrader: The Legend, the Lessons, the Results. Covel also wrote, directed, and produced the documentary film, Broke: The New American Dream. Covel has been interviewed by Bloomberg Radio, Technical Analysis Magazine, Barron's, and others.






Michael Moore Kills Capitalism with Kool-Aid


A friend recently invited me to a private screening of Michael Moore's new film Capitalism: A Love Story. The September 16th invite not surprisingly leaned a certain direction:

"[Michael] Moore takes us into the homes of ordinary people whose lives have been turned upside down; and he goes looking for explanations in Washington, DC and elsewhere. What he finds are the all-too-familiar symptoms of a love affair gone astray: lies, abuse, betrayal and 14,000 jobs being lost every day. Capitalism: A Love Story...is Michael Moore's ultimate quest to answer the question he's posed throughout his illustrious filmmaking career: Who are we and why do we behave the way that we do?"

Considering Moore was going to be there for a Q&A after (moderated by Arianna Huffington), I quickly signed on. Now before painting a picture of Moore's new film let me be honest: my belief set is essentially libertarian ('Government out of my bedroom and my pocketbook'). Not only do government solutions not excite me, they scare the living blank out of me. Remember when George Bush declared, "I've abandoned free- market principles to save the free-market system...to make sure the economy doesn't collapse"? He might as well of said, "Hide your money, kids - 'cause I'm coming to take it!"

Oh sure, in theory I would like to see everyone with their own homestead, money in their pocket for regular shopping frenzies and no health worries despite eating at Burger King 24/7, but arriving at those goals is not exactly doable unless government robs Peter to pay Paul and or starts up the printing press.

And that view of course puts me in opposition to Moore since he has no problem with government as his and our father figure. That is his utopia. He truly believes warehouses of Washington, DC-based federal workers remotely running our lives is the optimal plan. He is an unapologetic socialist who really doesn't care why the poor are poor or the rich are rich, he just wants it fixed. So not surprisingly, and with some generalization as I proffer this, Democrats like Moore and Republicans don't.

However, I was excited to see a 'mainstream' film that was backed by big Hollywood bucks conclude capitalism as 'evil.' Arguably the most successful documentarian ever, a man who has made untold millions of dollars, was going to legitimately make the case that there was an alternative to capitalism. I sat down in a packed Mann's Bruin Theater in Westwood, CA eager to see how his vision could possibly flesh out.

Moore is a rather simple guy. He is likable. He sees the world as good guys (people with no money) and bad guys (people with money). His Flint, Michigan union worker upbringing is his worldview. If you did not have that upbringing or if your life started less severe than his you are an evil capitalist. If on the other hand you were a laid off factory worker with a sixth grade education you are the true hero. I don't care one way or the other that he has that view and I am not knocking union workers, but Moore sees the world through a class warfare lens resulting in a certain agenda: force wealth to be spread amongst everyone regardless of effort. Within minutes it was clear where Capitalism: A Love Story was headed. The 'highlights' included:

  • We listen to heartbreaking stories of foreclosed families across America, but we don't learn why the foreclosures happened. Did these people treat their homes as piggy banks? Were there refis on top of refis just to keep buying mall trinkets and other goodies with no respect to risk or logic? We don't find out.
  • We meet one family who was just foreclosed on so desperate for money that they were willing to accept $1,000 for cleaning out the house that they were just evicted from. Was it sad? Yes. But, should we end capitalism due to this one family in Peoria, IL?
  • We are introduced to a guy whose company is called 'Condo Vultures' buying and selling foreclosed properties. Since he acted like a used car salesman, the implication was that he was an evil capitalist. However, Moore doesn't tell us if his buyers were 'working class' people making smart buying decisions after prices had dropped.
  • We listen to Catholic priests who denounce capitalism as an evil to be eradicated. What they would put in its place and how would the new system work? The priests don't tell us.
  • We learn that Wal-Mart bought life insurance policies on many workers. We are then told to feel outrage when Wal-Mart receives a large payout from an employee death while the families still struggle with bills. I saw where Moore was heading here, but this was a reason to end capitalism?
  • We hear a story from a commercial pilot so low on money that he has to use food stamps. Moore points out that many pilots are making less than Taco Bell managers and then attributes a recent plane crash in Buffalo to underpaid pilots. This one crash is extrapolated out as yet another reason to end capitalism.

I was pleasantly surprised at Moore's attempt at balance. For example, he included:

  • A carpenter, while ply-wooding up a foreclosed home, says, "If people pay their bills, they don't get thrown out."
  • A dressing down of Senator Chris Dodd (D) by name. Moore calling out a top Democrat? He sure did. He nailed him.
  • A lengthy dissertation on the evils of Goldman Sachs. He rips Robert Rubin and Hank Paulson big time and I agree with him. In fact, I said to myself, "Moore you should have done your whole film on Goldman Sachs!"

Throughout the various stories and interviews he also weaves a conspiracy (all Moore films do this). The plot goes something like this: America won World War II and quickly dominated due to no competition (Germany and Japan were destroyed). We had great post-war success where everyone lived in union-like equality. Jobs were plentiful and families were happy. However, things start to go bad in the 1970s, and Moore uses a snippet of President Carter preaching about greed. This clip was predictably building to Moore's big reason for all problems today: the Reagan revolution.

Moore sees Reagan entering the scene as a shill for corporate banking interests. However, everyone is happy as the good times roll all the way through into Clinton times. Moore does take subtle shots at President Clinton, but nails his right hand economic man Larry Summers directly as a primary reason for the banking collapse. So, while Moore sees Japan and Germany today as socialistic winners where corporations benefit workers more than shareholders, he sees America sinking fast.

So is that it? That was the proof that capitalism is an evil to eliminate? Essentially, yes, that's Moore's proof. What is his solution? Tugging on your idealistic heartstrings of course! Moore ends his film with recently uncovered video of FDR talking to America on January 11, 1944. Looking into the camera a weary FDR proposed what he called a second Bill of Rights - an economic Bill of Rights for all regardless of station, race, or creed that included:

  • The right to a useful and remunerative job in the industries or shops or farms or mines of the nation.
  • The right to earn enough to provide adequate food and clothing and recreation.
  • The right of every farmer to raise and sell his products at a return which will give him and his family a decent living.
  • The right of every businessman, large and small, to trade in an atmosphere of freedom from unfair competition and domination by monopolies at home or abroad.
  • The right of every family to a decent home.
  • The right to adequate medical care and the opportunity to achieve and enjoy good health.
  • The right to adequate protection from the economic fears of old age, sickness, accident, and unemployment.
  • The right to a good education.

As FDR concluded and the film ended, I was shocked at the reaction. The theater of 400+ stood and cheered wildly at FDR's 1944 proposal. The questions running through my head were immediate: How does one legislate words like "useful", "enough", "recreation", "adequate", "decent", and "good"? Who decides all of this and to what degree? At past points in history to voice an opposition opinion in the middle of such a single-minded herd would have certainly been my physical demise! Interestingly, during the Q&A Huffington and Moore discussed bank failure fears during the fall of 2008. They asked for a show of hands of how many people moved money around or attempted to protect against a bank failure. I had the only hand that went up.

FDR's plan hauled out by Moore six decades after it was forgotten reminded me of another interchange - this one from the 1970s. Then talk show master, the Oprah of his day, Phil Donahue was interviewing free market economist Milton Friedman and wanted to know if Friedman had ever had a moment of doubt about "capitalism and whether greed's a good idea to run on?"

Friedman was quick in response, "...is there some society you know that doesn't run on greed? You think Russia doesn't run on greed? You think China doesn't run on greed? The world runs on individuals pursuing their separate interests. The great achievements of civilization have not come from government bureaus. Einstein didn't construct his theory under order from a bureaucrat. Henry Ford didn't revolutionize the automobile industry that way. In the only cases in which the masses have escaped from the kind of grinding poverty you're talking about, the only cases in recorded history are where they have had capitalism and largely free trade. If you want to know where the masses are worst off, it's exactly in the kinds of societies that depart from that. So that the record of history is absolutely crystal clear: that there is no alternative way so far discovered of improving the lot of the ordinary people that can hold a candle to the productive activities that are unleashed by a free enterprise system."

Donahue (and the video of this on YouTube is classic) then countered saying that capitalism rewards the ability to manipulate the system and not virtue. Friedman was having none of it, "And what does reward virtue? You think the communist commissar rewards virtue? ...Do you think American presidents reward virtue? Do they choose their appointees on the basis of the virtue of the people appointed or on the basis of their political clout? Is it really true that political self- interest is nobler somehow than economic self-interest? ...Just tell me where in the world you find these angels who are going to organize society for us?"

Friedman's logic was what I was remembering as a theater full of people cheered wildly for a second Bill of Rights. How did this film crowd actually think FDR's 1944 vision could be executed? Frankly, it was clear to me at that moment capitalism was on shaky ground. Starting with Bush 'abandoning' capitalism to bailouts for everyone to Obama gifting away the future - we seriously might be past the point of no return toward a socialization of America.

Figuring someone else must see the problems with this film, I started poking around the net for other views. One critic declared that the value of Capitalism: A Love Story was not in the moviemaking, but in its message that hits you in the gut and makes you angry. This film did not make me angry, but it did punch me in the gut. The people in that theater with me were not bad people, including Moore. They just seem to all have consumed a lethal dose of Kool-Aid! And at the end of his Q&A Moore pushed the audience to understand that while they don't have the money, they do have the vote. He implored them to use their vote to take money from one group to give it another group. Did he really say that openly with no ambiguity? Yes, sadly.

Wednesday, September 23, 2009

Vedic Mathematics Lesson 21: Straight Division II

In the previous lesson, we got our first introduction to straight division using the Dhvajanka sutra. We restricted ourselves to dealing with denominators consisting of 2 digits in that lesson.

You can find all the previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2

Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I

In this lesson, we will extend the straight division method to deal with denominators that may have more than 2 digits. There are two main ways in which straight division can be performed when the denominator contains more than 2 digits.

The first method is very similar to the method we used for 2-digit denominators. We raise only the right-most digit of the denominator "on to the flag", and keep the rest of the denominator as part of the base. Since the number of flag digits is just 1, our method is not much changed from what we saw in the previous lesson. But since the base digits of the denominator now form a much larger number (at least a 2-digit number), we may have to make some adjustments to make sure our intermediate numerators are not negative. As explained in the previous lesson, this involves reducing the quotient and increasing the remainder so that the intermediate numerators are always positive.

To illustrate this method, let us tackle an example. We will start out by trying to find the answer to 8941/121. We divide the denominator into a base part of 12 and a flag part of 1. We then get the figure below:

••1
12 |894|1
•••|
------------

We then proceed with the method as outlined in the previous lesson. Our first division is of 89 by 12. This gives us a quotient of 7 and a remainder of 5. These are written as below:

••1
12 |89 4|1
•••|••5
-------------
•••| 7

Now we find our next intermediate numerator as 54 - 7 = 47 (54 is the number we obtain by putting the remainder of the previous division together with the next digit of the numerator, and 7 is obtained by multiplying our flag digit by the previous quotient digit). Dividing 47 by 12 gives us a quotient of 3 and a remainder of 11. These are written down as shown below:

••1
12 |89 4| 1
•••|••5 11
--------------
•••| 7 3

Since we have now dealt with all the quotient digits of the numerator, we can find the remainder and stop, or proceed further to find a decimal representation of the answer. If we do stop at this point, the answer we obtain is below:

••1
12 |89 4| 1
•••|••5 11
---------------
•••| 7 3|108

We can verify that the answer to the problem is indeed a quotient of 73 and a remainder of 108.

If we proceed further, we obtain the first few digits beyond the decimal point as below:

••1
12 |89 4| 1. 0 0 0 0 000
•••|••5 11 12 4 7 8 3 12 11
---------------------------
•••| 7 3. 8••9 2 5 6 198 . . .

One more example will further solidify this approach in our mind. With that mind, let us take the example of 8878867/1001. We know that it is very easy to deal with this example using the Paravartya method as illustrated below:

1001 0 0 -1
8878|867
0008|
0000|800
0000|070
-------------
8870|-3
8869|998

Now, doing it by the method of straight division, we get the figure below. Once again we have hoisted only 1 digit "on to the flag", and retained the rest of the denominator as base digits.

•••1
100 |887••8••8•••6|••7
••••|•••87•70•100•100
------------------
••••|••8••8••6•••9|998

We get the identical result. With straight division, we don't have to stop with a remainder though. We can carry on the division and find as many digits beyond the decimal point as we want for any division problem.

This method of dealing with multi-digit denominators (retaining all but one digit as base digits and hoisting just one digit on to the flag) works when the base digits are easy to divide by (small numbers like 11, 12, etc., or round numbers like 100, 50, etc.).

Now, let us look at another way of dealing with multi-digit denominators. In this method, we retain the part of the denominator that is easy to divide by as the base digits and hoist all the other digits on to the flag as flag digits. In many cases, this may involve just one base digit and 2 or more flag digits. The procedure for dealing with multiple flag digits is a little different from how we dealt with a single flag digit. We will now go over that procedure in detail as we apply it to a real problem. For the purpose of this illustration, let us take the example of 977995/212.

This time, instead of keeping 2 base digit and 1 flag digit, we will split the denominator into 1 base digit and 2 flag digits. This is illustrated in the figure below. Notice that because there are 2 flag digits, we set aside two digits of the numerator as remainder digits.

•12
2••|9779|95
•••|
-------------

Our first division involves dividing 9 by 2. The quotient is 4 and the remainder is 1. This is written below:

•12
2••|9 779|95
•••| 1
------------
•••|4

The next step in this procedure is not to multiply the first quotient digit by the numbers on the flag. The procedure is very similar to the procedure we use in multiplying vertically and cross-wise using the Urdhva-Tiryak sutra. In this case, the multiplicands are always the flag digits and the last n digits of the quotient, where n is the number of flag digits.

Since the quotient, at this time, contains only one digit, we pad it with as many zeroes to the left as is necessary to make sure that the quotient contains the same number of digits as the flag. In this case, we need to pad the quotient with one zero. In general, this will require padding with (n-1) zeroes where n is the number of flag digits. This is shown in the figure below:

•12
2
•|9 779|95
•| 1
---------------
•0|4

Now, we obtain the sum of the cross-products of 04 and 12 as explained in the lesson on Vertically and Cross-wise. This results in 0 x 2 + 4 x 1 = 4. Subtracting 4 from 17 gives us 13, which becomes our next intermediate numerator. Dividing 13 by 2 gives us a quotient of 6 and a remainder of 1. This is shown in the figure below:

•12
2••|9 7 79|95
•••| 1 1
---------------
••0|4 6

At this point, the 2-digit number we get by putting together the remainder and the next digit of the numerator is 17. From this 17, subtract the sum of the cross-wise products of the 2 flag digits and the last 2 quotient digits. This sum works out to 4 x 2 (first digit of the quotient x second flag digit) + 1 x 6 (second digit of the quotient x first flag digit) = 14 (we don't have to use the zero in front of the first quotient digit anymore since we now have enough real quotient digits to perform our computations with). 17 - 14 = 3. Dividing 3 by 2 gives us a quotient of 1 with a remainder of 1. This is illustrated in the figure below:

•12
2••|9 7 7 9|95
•••| 1 1 1
---------------
••0|4 6 1

Now, the 2-digit number we get by juxtaposing the remainder and the next digit of the numerator is 19. The number to be subtracted from this is the sum of the cross-products of the flag digits and the last 2 quotient digits found so far. The last 2 digits of the quotient are 61 and the flag digits are 12. So, the sum of the cross-products is 6 x 2 + 1 x 1 = 13. 19 - 13 = 6. Dividing 6 by 2 gives us a quotient of 3 and a remainder of 0. This is now added to our figure to give us the following:

•12
2••|9 7 7 9|95
•••| 1 1 1 0
----------------
••0|4 6 1 3

We have now dealt with all quotient digits of the numerator. So, we can now calculate a remainder or we can continue the division by adding a decimal point and proceeding as above.

The calculation of the remainder is also slightly different from what we are used to with a single flag-digit. To calculate the remainder, we now take the last remainder digit and add it to the front of the remainder digits of the numerator. This gives us 095.

Then we calculate the sum of the cross-products of the last 2 digits of the quotient and the 2 flag digits. The sum of the cross products in this case is 1 x 3 + 1 x 2 = 5. We multiply this by the power of 10 that has the same number of digits as the number of flag digits. In this case, that power of 10 is 10 itself. 095 - 5 x 10 = 45.

Now, as in the lesson on vertically and cross-wise, we subtract out from this, the vertical product of the last digit of the quotient and the last flag digit. This vertical product is 3 x 2 = 6. 45 - 6 = 39, and this becomes the remainder for the division problem. This is shown in the figure below:

•12
2••|9 7 7 9|95
•••| 1 1 1 0
------------------
••0|4 6 1 3|39

It is easy to verify that a quotient of 4613 and a remainder of 39 is indeed the correct solution to the divison problem we undertook to solve.

The following steps explain the process to be used when dividing with more than one flag digit:

  • The numerator is divided into quotient and remainder digits such that the number of remainder digits is the same as the number of flag digits
  • The first quotient digit is obtained by dividing the first digit (or digits if necessary) of the numerator by the base digit(s) of the denominator
  • The remainder from the above division is placed in front of the next numerator digit to form a 2-digit number
  • Pad the quotient with zeroes to the left such that the number of quotient digits is the same as the number of flag digits
  • The last n (where n is the number of flag digits) quotient digits are then multiplied and added vertically and cross-wise with the flag digits and this sum of products is subtracted from the 2-digit number we obtained earlier by placing the remainder before the next digit of the numerator. This gives us the next intermediate numerator
  • The intermediate numerator is divided by the base digit(s) of the denominator for the next quotient and remainder. The quotient goes on the quotient line and the remainder goes in front of the next digit of the numerator to form a 2-digit number
  • We now proceed by calculating the sum of cross-products using the last n digits of the quotient (where n is the number of flag digits), and subtracting this from the 2-digit number obtained in the previous step. This gives us the next intermediate numerator
  • This process is continued until we have dealt with all quotient digits of the numerator and our next remainder digit is placed in front of a remainder digit of the numerator
  • If we want a decimal answer, we put a decimal point on the quotient line when our 2-digit number on the numerator line is formed by including one of the remainder digits of the numerator. The method proceeds as outlined in the previous steps, and can be continued to as many digits beyond the decimal place as needed by padding the numerator with zeroes to the right. Do not perform the steps below which are for calculation of a remainder
  • If we want a remainder, then we stop as soon as any step above involves a remainder digit of the numerator. Perform the steps below to calculate the remainder
  • Place the last remainder that was calculated by division of an intermediate numerator by the base digit(s) in front of the remainder digits of the numerator to get an (n+1)-digit number, where n is the number of flag digits (call this the intermediate remainder)
  • Take the last n quotient digits where n is the number of flag digits and calculate cross-products of this number with all the flag digits as in vertically and cross-wise
  • Multiply this cross-product by the power of 10 that has n digits and subtract from the intermediate remainder to get the new intermediate remainder
  • Next, take the last (n-1) quotient digits where n is the number of flag digits and calculate cross-products with the last (n-1) flag digits
  • Multiply this by the power of 10 that has (n-1) digits and subtract from the intermediate remainder to get the new intermediate remainder
  • Proceed this way, winding down the number of digits in the cross-multiplication until you are left with just a vertical multiplication of the last quotient digit with the last flag digit. Subtract this final vertical product from the intermediate remainder calculated through the above steps to get the final remainder

The procedure above looks long and complicated, but is actually quite easy to perform once the basics of vertically and cross-wise are clear. In fact, an astute person would have easily seen that padding the quotient digits with zeroes to the left so that we can perform a full vertically and cross-wise cross-product with the flag digits is actually the same as starting with a vertical product of the first quotient digit with the first flag digit, then in the next step, expanding the cross-multiplication to 2 quotient digits and the first 2 flag digits, and so on until the cross-product expands to all the flag digits. This is the same procedure as is used in vertically and cross-wise to calculate products. And then, the process is wound down by eliminating flag digits from left to right (since we expanded from left to right, we have to wind down from left to right also) until we are left with a vertical product of the right-most quotient digit with the right-most flag digit, just as in vertically and cross-wise. If this is not clear, it is advisable to review vertically and cross-wise once more to be clear on the procedure, especially for multi-digit multiplications, as explained in this past lesson.

We will conclude this lesson with a couple more examples.

64794/423 worked out in decimal form gives us 153.17730496. . . as below (note that in several of the steps below, we have adjusted the division by reducing the quotient and increasing the remainder so that the subtraction step does not produce a negative answer):

•23
4••|6 4 7|9 4 0 0 0 0 0 0 0
•••| 2 2 2 4 5 5 3 3 5 6 6
--------------------------------
••0|1 5 3.1 7 7 3 0 4 9 6 . . .

64794/423 with a remainder gives us a quotient of 153 and remainder of 75 as below:

•23
4••|6 4 7|94
•••| 2 2 2
----------------
••0|1 5 3|75

943809843/2432 with remainder gives us a quotient of 388079 and a remainder of 1715. Note how the quotients and remainders of the individual divisions by the base digit (2) have been adjusted to avoid negative numbers after subtraction. Also note that ultimately, we obtain a negative remainder, which has to be adjusted by subtracting 1 from the quotient and adding the denominator, 2432, to the previous remainder.

•432
2•••|9 4 3 8 0 9|8 4 3
••••| 3 6 6 6 4 1
---------------------------
••00|3 8 8 0 8 0|-717
••00|3 8 8 0 7 9|1715

Hope you will take the time to practice this technique so that it becomes more and more of a habit or second nature rather than a difficult intellectual exercise that requires you to drop everything else to concentrate on this. The whole idea behind mental computation is to be able to perform them "on the side" while trying to think through bigger problems that require answers to these computations only as a part of the whole "bigger picture". After all, I don't think any of us is going to be doing these computations just for the sake of doing them, in real life! So, good luck, and happy computing!!

Monday, September 21, 2009

Vedic Mathematics Lesson 20: Straight Division I

In the previous lessons, we learnt how to divide using two different methods. The Nikhilam method is best suited for division by denominators that are just below a power of 10, while the Paravartya method is best suited for division by denominators that are just above a power of 10.

You can find all the previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2

Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots

However, unfortunately, not all divisions we encounter have denominators that are conveniently close to powers of 10. Fortunately, there are other methods we can use to overcome this limitation of the Nikhilam and Paravartya methods. The method we will learn in this lesson is derived from a sutra which reads Dhvajanka. Literally, this means On Top Of The Flag. This method is also referred to as straight division.

Before we dive into the method let us first review the basics of division by simple single-digit denominators. This review will serve as a foundation for the division method we will move on to.

Let us consider 25/2 for instance. Let us write the problem out as below:

2|25
|
-------

Now, we divide the first digit of the numerator, 2, by the denominator to get 2/2 = 1. Let us put down 1 as the first digit of the quotient. The reminder of the division of 2 by 2 is zero. Let us put this zero just before the 5 as in the figure below:

2|2 5
| 0
--------
|1

Now, divide the two-digit number obtained by placing the remainder and the next digit of the numerator together (in this case, 05), by the denominator, 2. This results in the second digit of the quotient being 2 with a remainder of 1. This is shown below:

2|2 5
| 0 1
---------
|1 2

Since there are no more digits in the numerator, we are left with the final answer. We can write it by moving the last remainder to the answer line as below:

2|2 5
| 0 1
---------
|1 2|1

This gives us a quotient of 12 and a remainder of 1 (which is obviously the correct answer).

We can also choose to proceed further by considering the numerator to be 25.0000... instead of just 25 with no digits after the decimal point. If we do, then, we have to follow the same procedure as before with the new remainder we have found. The two-digit number we get by placing the remainder next to the next digit of the numerator is 10 (place the remainder 1 before 0 in the numerator after the decimal point). Since the remainder has been attached to the first digit of the numerator beyond the decimal point, we also put a decimal point in the answer line. Then, we do the division by 2 to 10/2 = 5 as the next digit of the quotient. The remainder is zero. This is illustrated below:

2|2 5.0000
| 0 1
-----------
|1 2.5

Since the remaining numbers are all uniformly zero, we know that there will be no further addition to the answer line except zeroes (and since these zeroes will be after the decimal point, they do not change the final answer). That completes this division.

To tackle a slightly more interesting case, let us consider 85/7. The figure below illustrates what happens when we stop with a quotient and remainder:

7|8 5
| 1 1
----------
|1 2|1

The quotient of 12 and remainder of 1 can be verified to be correct. When we proceed with the division instead of stopping with a remainder we get the figure below:

7|8 5.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
| 1 1 3 2 6 4 5 1 3
-----------------------------------------
|1 2.1 4 2 8 5 7 1 4 . . .

We see that since the remainders repeat themselves, we know that the answer digits will also repeat themselves. Instead of continuing the calculations, we stop at that point.

These examples form the basic foundation on which straight division using the Dhvajanka sutra rests. To illustrate straight division, let us now take the example of 562/41. First we separate the denominator into two parts, the base and the flag digits. Usually, the base consists of just the left-most digit of the denominator. Sometimes, it helps to make more digits part of the base if division by the base will remain easy. The rest of the digits become flag digits (hence the name of the sutra: On Top Of The Flag).

The numerator is also divided into two parts. A quotient part and a remainder part. The remainder part consists of as many digits as there are on the flag. In this particular problem, we will consider 4 to be the base digit of the denominator and hoist 1 on top of the flag. Similarly, we will consider 56 to be the quotient part of the numerator and 2 to be the remainder digit. We now get the figure below:

1
4 |56|2
••|
------------

We begin by trying to divide the first digit of the numerator by the base digit of the denominator. We get 5/4, which has a quotient of 1 and a remainder of 1. We write them as below:

1
4 |5 6|2
••| 1
------------
••|1

The difference between the single-digit division we reviewed earlier and straight division comes at this step: Instead of trying to divide the 2-digit number obtained by putting the remainder and the next digit of the numerator together (16 in this case), we do something slightly different. We take the 16 we obtain by putting the remainder digit together with the next digit of the numerator and subtract from it the product of the previous quotient digit and the flag digit. That product is 1 x 1 = 1 in this case. The subtraction gives us 15. This now becomes the next dividend for division by the base digit of the denominator. 15/4 is 3 with a remainder of 3. This is written in the figure below:

1
4 |5 6|2
••| 1 3
------------
••|1 3

We now repeat the same procedure with this remainder digit as we did in the previous step. We take the 2-digit number we get by putting it together with the next digit of the numerator (32 in this case), and subtract from it the product of the previous quotient digit by the flag digit (3 x 1 = 3, in this case). We get 29. Since this 29 is obtained after taking into account the remainder digit of the numerator, we can stop here with a remainder of 29 if we want to, giving us the answer as below:

1
4 |5 6|2
••| 1 3
-------------
••|1 3|29

Thus, we find the answer to our problem to be a quotient of 13 with a remainder of 29. This can be verified to be true. The advantage with straight division using the Dhvajanka sutra is that we don't have to stop with a remainder. We can proceed as far as we want by finding digits beyond the decimal point using the same procedure as we followed above. This is illustrated below:

1
4 |5 6|2.0 0 0 0 0 0 0 0 0
••| 1 3 1 3 2 1 3 1 3 2 1
--------------------------
••|1 3.7 0 7 3 1 7 0 7 3 . . .

We see that the digits start repeating, so there is no point in continuing the division further. The general method can be summarized in the following steps:

  • Split the denominator into one or more base digits and one or more flag digits (in most cases, it makes the most sense to have one base digit and all the rest of the digits raised on to the flag)
  • Split the numerator into quotient digits and remainder digits. The number of remainder digits has to be the same as the number of flag digits
  • Now perform division of the first digit of the numerator by the base digit of the denominator (if this is not possible, take the next digit of the numerator also to perform the division)
  • Put the quotient on the answer line and put the remainder just before the next digit of the numerator in the remainder line
  • Take the 2-digit line formed by the combination of this remainder digit with the next digit of the numerator and subtract from it the product of the flag digit(s) and the last quotient digit (on the answer line). Call this sum the intermediate numerator
  • Perform division of this intermediate numerator by the base digit of the denominator and put the quotient on the answer line as the next digit of the quotient. Put the remainder just before the next digit of the numerator in the remainder line
  • Now, perform the process in the above 2 steps over and over again until you have taken into account the first of the remainder digits of the numerator
  • At this point, one can stop, and consider the number on the answer line to be the quotient, and the intermediate numerator to be the final remainder of the division problem
  • If one wants to find a decimal answer to the problem, add a decimal point to the answer line as soon as one of the remainder digits of the numerator is involved in the calculation of the intermediate numerator
  • Then, add a decimal point to the numerator, pad it with zeroes to the right and continue this process until either the intermediate numerators become zero, or the quotient digits and intermediate numerators start repeating

That is all there is to this method! There is no multiplication by multi-digit denominators, there is no complicated multi-digit subtractions, and most importantly, no trial-and-error in figuring out the next quotient digit. If you can do simple division by single-digit denominators, you can master this method and perform division by much higher denominators.

Let us now deal with a special case that one might encounter when using this method. At the step where one subtracts the product of the quotient digit and the flag digit from the two-digit number formed by the remainder and the next digit of the numerator, you may get a negative number as the intermediate numerator. We haven't made any provisions for such a contingency in the procedure above, so we need a method to deal with it if the situation arises.

The solution to this problem is actually quite simple: we simply have to remember that even though we are used to thinking that only one quotient and remainder exist for a given division problem, actually an infinite number of them exist. Take the case of 9/2 for instance. The traditional answer to this problem is a quotient of 4 and remainder of 1. However, technically, once can consider a quotient of 3 and a remainder of 3 to be a perfectly valid answer, as are a quotient of 2 and remainder of 5, and so on.

So, the solution to our special case is simply to reduce the quotient by 1 to get a higher remainder which will then make the negative answer go away. Let us illustrate this by finding the answer to 819/49. After following the procedure through the first few steps, we get the figure below:

9
4 |8 1|9
••| 0
-----------
••|2

Now we find the 2-digit number formed by the remainder and the next digit of the numerator to be 01. When we try to subtract the product of the quotient digit and the flag digit (2 x 9 = 18) from 01, we get -17, which threatens to throw a wrench into the works here! As explained above though, the way out of this predicament is to consider the answer to the first division to be a quotient of 1 and remainder of 4 rather than a quotient of 2 and remainder of 0. When we do that, the answer flows naturally based on the steps outlined in the procedure:

9
4 |8 1|9
••| 4 0
---------
••|1 8

Now, we find the 2-digit number to be 41, and it is easy to subtract 1 x 9 = 9 from 41. We get 32 as our intermediate numerator. At this step, if we consider 8 to be our quotient and 0 to be our remainder from the division of 32 by the flag digit, 4, we encounter the same problem as before. Our 2-digit number becomes 09, and the product of 8 and 9 is much higher. So, let us try the same method as before and reduce the quotient to 7 with a remainder of 4. We get the figure below:

9
4 |8 1|9
••| 4 4
-----------
••|1 7

But, we find that this small concession is not actually sufficient to get over the problem yet. We find our 2-digit number to be 49 and our product of the quotient digit with the flag digit to be 63. To get over the problem, we need to make a bigger concession. We finally solve the problem by considering the result of the second division to be a quotient of 6 with a remainder of 8. We illustrate the final solution below:

9
4 |8 1|9
••| 4 8
-----------
••|1 6|35

We can verify that a quotient of 16 and a remainder of 35 is indeed the correct (and most widely accepted!) solution to our problem.

In future lessons, we will learn how to extend this method to perform divisions by denominators that have more than just 2 digits. In the meantime, hope you will find time to practice this method so that you can do it rapidly without errors. Practice makes perfect, so good luck, and happy computing!

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