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Some ]psau~ to make some multiplications easier still (did you believe that was even possible?).
yaavadUnaM tavadUnaIÌ%ya vaga- ca yaÜjayaot i.e. whatever the deficiency, lessen it further to the same extent, and setup its square.
This is useful for squaring certain numbers.
Let us see how to get the sq. of 8.
Working base: 10
8 is 2 less than 10. Reduce it further by 2 i.e. 8-2=6(LHS)
set up sq. of this difference: 2*2=4 (RHS)
Answer: 64
Sq. of 13
LHS=13+3=16
RHS=3*3=9
Answer:169
Sq of 109
Base 100
LHS=109+9=118
RHS=9*9=81
answer: 11881
sq. od 111
Base 100
LHS=111+11=122
RHS=11*11=(1)21 (1 is carried over as base hundred and hence RHS has 2 digits)
Answer: 12321
sq. of 88
LHS=88-12=76
RHS=12*12=(1)44
Answer:7744
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The algebraic trickbehind the previous posting is simple: (x+a)*(x-a)=x*x-a*a
Thus, to get x*x, get (x+a)*(x-a) and add a*a to it. 'a' can be any number: to make things convenient, choose it such that (x+a) or (x-a) is a power of 10.
e.g. x=97 choose a=3 so that (x+a)=100.
(x-a) is then 94. (x+a)*(x-a)=9400; a*a=3*3=9
answer: 9409
ekaiQakona pUvao-Na i.e. 'one more than the previous one' takes it to the limit.
It is used for numbers that end in 5. Then use a=5 too.
So 65*65=(65-5)*(65+5)+5*5=60*70+25=6*7/25=4225
The procedure is simple, whatever is to the left of 5, multiply it by one more, and put 25 in front of it. (No worries about carries or padding by 0)
115*115=11*12/25=13225
75*75=7*8/25=5625
195*195=19*20/25=38025
Try out the squares of the following numbers:
85, 95, 98, 104, 105, 997, 1005, 295
Also try out the following multiplications:
93x97, 142x148, 154x156
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This second trick was known to us in the scholl itself. The first one is brilliant. I mean, though the algebraic solution was known, the ease in doing this is new.
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brilliant brilliant
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Vedic Maths baddal 'Q & A' saathi separate BB aahe kaa?
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Yogi, Q and A sathi separate suru karu shakatos (literature madhye - admin can move it if needed).
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Ashish, shaLet asha reeteene shikawake asate tar mala jarase gaNit jamale asate ki re! kewaL uchcha ahe he sarw!
Mee nidan Aksharnandansarkhya shaLet taree waidik gaNit shikawa ashee soochana karaNar ahe.
BTW, please, too ullekh kelyapramaNe apalya widnyan sanskruteebaddal tula asalelee mahitee paN ek nawa BB suroo karoon ka det nahees? It would be very useful.
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In school I once wrote answers directly and was given zero marks. Our teacher wanted ALL steps to be writeen down :-(
Anyway, here's today's dose:
Yesterday's formula can also be used when the LHS is same and the sum of last digits is 10 e.g.
93x97 (LHS=9 3+7=10)
As before, 9*10=90 (LHS)
3*7=21 (RHS)
answer: 9021
142*148=14*15/2*8=21016
154*156=15*16/4*6=24024
31*39=12/09
Extending further:
146*154=1*2/46*54=2/(50-4)*(50+4)=2/50*50-4*4
=22486
2997*2003=2*3/002991=6002991
When the multiplier contains all 9s, you can use another ]psau~ viz.eknyaUnaona pUvao-Na i.e. one less than the multiplicand (=pUva-).
7*9=(7-1)/RHS=6/(9-LHS)=63
RHS is obtained using inaiKlama ... or by subtracting LHS from all 9s
76*99=(76-1)/(99-LHS)=75/24
all from 9, last from 10 for 76 is 24
so also is 99-75
812*999=811/188
13423*99999=13422/86577
22*999=022*999=021/978=21978
Should there be more digits in the multiplicand (than there are 9s in the multiplier), the LHS is obtained by doing ekanyunen i.e. subtracting 1, and then further subtracting the number made by the excess part of the multiplicand. RHS is obtained as before.
723*99=7/23*99=(722-7)/(99-22)=71577
8912*99=89/12*99=(8911-89)/(99-11)=882288
8888*999=8887-8/112=8879112
Next time we will get on to division but that could be after 4-5 days as I proceed for an observing run. I may keep on giving some exercises though :-)
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wonderful Ashish !!!!!
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Finally on to division. I will start with the same sutra used for multipli
cation viz. urdhvatiryak.. giving us straight division. This division invo
lves only small multiplications and subtractions (subtractions are left as
an exercise).
However, division is not as straightforward as multiplication (or rather,
a single sutra is not best for all situations). So try examples out and yo
u will hit roadblocks. That will help us discuss other sutras for specefic
situations.
I will demonstrate examples such that there is a quotient and a reminder.
If you wish to obtain answers in decimal fractions, simply append so many
zeroes to the number being divided (dividend) before starting off.
Consider 7415/37. Rather than dividing by 37, we chose to divide by 3 and
MULTIPLY by 7. Thus the divisor is split into an LHS (viz. 3) and an RHS (
viz. 7). The dividend is also split into 2 parts, the RHS contains as many
digits as in the RHS of the divisor (in this case 1). Thus we state our p
roblem as:
_7|741:5
3
In the first step, we take first one or two digits of the dividend and div
ide it by 3. Thus 7/3 gives 2 with a reminder of 1.
_7|7 4 1 : 5
3___1
---------------
___2
We wrote the 2 down as the first digit of our answer and the 1 before the
next digit of the dividend viz. 4 to give us our next temporary dividend v
iz. 14. But before we proceed, we have to take into consideration the 7 th
at we left out. Multiply it by 2 (the first digit of the answer) and subtr
act that from 14 (14-14=0). So our next dividend is really 0. 0/3 gives 0,
the next digit of the answer with 0 as the reminder. prefix it to the nex
t digit viz. 1
_7|7 4 1 : 5
3___1 0
--------------
___2 0
Our new dividend is now (01-0*7=1)
1/3=0 reminder 1. Prefix 1 to the last digit viz 5
_7|7 4 1 : 5
3___1 0 1
---------------
___2 0 0
Our quotient is 200 and remainder is (15-0*7=15).
--------------------------------------------
If the medicene seems as bad as the disease, its just due to lack of pract
ise. Get some and you will be an expert.
----------------------------------------
Let us see another example viz. 97498/72
_2|9 7 4 9 : 8
7___2
-----------------
___1 (next div. is 27 - 1*2 = 25)
_2|9 7 4 9 : 8
7___2 4
-----------------
___1 3 (next div. is 44-3*2=38)
_2|9 7 4 9 : 8
7___2 4 3
------------------
___1 3 5 (next div is 39-10=29)
_2|9 7 4 9 : 8
7___2 4 3 1
------------------
___1 3 5 4 Q = 1354 R=18-8=10
---------------------------------------------
A bigger example: 674355/4321
__21|67__4__3| 5 5
43_____24
------------------------
______1 next div= 244-1*2=242
__21|67__4__3| 5 5
43_____24 27
------------------------
______1__5 next div is 273-5*2-1*1=262
__21|67__4__3| 5 5
43_____24 27__4
------------------------
______1__5__6 next div is 45-6*2-5*1=28
But since we have reached the end, we finish with R = 285-6*1 = 279 Hence
A = 156 R =279
Notice that we did not do any big divisions or multiplications.
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ashish, Good Work. thanks.
do you know this division is based on any mathematical expression?
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I think I mentioned somewhere in the beginning of the posting that this division method is based on urdvatiryagbhyam on which the general multiplication method was based, but perhaps it is not equally evident:
(ax+b)*(cx+d)=ac(x*x)+(ad+bc)x+bd
Then, when we are given l(x*x)+mx+n and asked to divide it by (px+q) (where x is 10 to suit our decimal system). Our task is to solve the following equations:
l=ac; m=ad+bc and n=bd with the additional knowledge that a=p and b=q
We may have to bring in an additional constant, t, to account for the remainder that we get.
Hence we start out dividing l(=ac) by a(=p) to get c (obviously). Then, the remainder is prefixed to the next term viz. m(=ad+bc) and now subtract bc (the product of the first digit that we got and the units place of the divisor) to get ad. This we now divide by a(=p) to get d i.e. the units place of our quotient. If it is a complete division, n(=bd) is what remians (which we need to now subtract out due to urdhvatiryakbhyam).
It extends straightforwardly to divisions of multiple numbers:
(ax^4+bx^3+cx*x+dx+e)/fx+g
Divide a by f to get l as the first digit of quotient (it will be 1000's place since x^4/x=x^3).
Add the remainder to b, and subtract from it l*g (because now we are dealing with x^3 and that is what l is associated with). Haveing done that, divide the difference by a to get the x^2 digit m and so on and on till you reach the units place and the remainder.
I hope I have made it clear now.
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How do I retain proper spaces between digits in my posting? Is there something I can use to quote the maths? Like the <pre></pre> tages in HTML? (crossposting this question to I need help).
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Have been behind schedule a little bit. Will resume after I figure out the formatting stuff and once I am back from a tour. Expect a posting after 14th Jan. Happy new year and happy mathing to everybody.
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Ashish,
I know its very annoying to have the spaces. but I experimented a bit and found a method which worked :o)
actually the posting process is combining all white spaces into a single space for conserving, so I tried this method.
ASCII 32 is the normal space bar character
ASCII 255 is the 'blank' character
you can alternate between these two to get required amount of spacing. start with blank character (ASCII 255). I don't know on Unix how to get that character, for pc users, press and hold ALT key while pressing 255 on numeric key pad gets you the character.
finally use courier new font to get the fixed width font using \font{courier new}
lets see
Put Alt+255 in place of the '.'
\font{courier new,
0
.1
. 2
. .3
. . 4
}
like this
0
1
2
3
4
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Bhm (Bhm)
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| | Tuesday, February 19, 2002 - 1:56 am: | 
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ekaiQakona pUvao-Na naamak saU~aivaYayaI vaacat hÜtÜ. (a saU~anao 1/19, 1/29, 1/39 AXaa saM#yaaMcao dXamaana pwtIt ApUNaak ilaihlao Asata yaoNaaro AakDo AÜLKta yaotat
jasao
1/19 = 0.052631578947368421052631578947368421...
AXaa AMkat zraivak AakD\yaaMcaI AÜL punha punha Aavata-cyaa sva$pat yaot rhato jasao
1/19 maQyao 052631578947368421 (a AakD\yaaMcaI saaKLI punha punha yaoto. yaalaa Aavat- mhNaU yaa. [qao (a Aavata-caI laaMbaI 18 Aaho. 1/29 cyaa baabatIt laaMbaI 28 Aaho. pNa 1/39 caI maa~ 6 Aaho. tr (a Aavata-caI laaMbaI kXaI XaÜQaayacaI (acao ]<ar (a saU~anao kLto kaÆ kuNaalaa maaiht Asalyaasa saaMgaa.
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It is not a single or simple sutra. 19 and 29 are prime and so the repeated fractions go all the way to n-1 i.e. 18 and 28 (because 1/19 through 18/19 have disticnt leading digits)
39 is not a prime and has two factors viz. 3 and 13 and hence 3/39=1/13, 6/39=2/13 ... should lead to 12 rather than 38 distinct leading digits.
But again, since 13/39 and 26/39 should provide the other series of leading digits, they end up factoring 12 by 2 leading to 6 digits.
I will try to conjure a more complete answer, but the short explaination is that the prime factors of your divisor decides the number of digits in the decimal fraction.
I should really really restart on this BB.
Oh, but for time. cash, strength and patience ...
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Ashish, GREAT WORK BY YOU.
I posed this question long long ago to many ppl, but still havent been replied satisfactorily.
I believe, 1757051 is the largest prime number. I tried to figure out a larger number using my pc, but failed. I therefore would like to know from you, whether vedic maths does say anything abt. such largest prime numbers?
Any clues?
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Ajay, all I am doing is typing here what I have learnt from the books. So the great work was really done by others. Writing here helps me revise stuff. Thus it is totally selfish :-)
BTW, 1757051=1291*1361. Thus it is not a prime.
In fact there is NO largest prime. And it is trivial to prove.
Take all known primes. Multiply them with each other. Add 1 to that. Now this number is not divisible by any of your earlier primes. So either it is a prime, or it is divisible by a number which was not in your primes list. Either way, you have a new prime. And this process continues ad infinitum.
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Aschiq,
I am reading your postings and learning a lot from it. Your efforts of writing all you know is really appericiable.
But as you said regarding prime numbers:
If I know Prime numbers say
3, 5 => 3 * 5 = 15 + 1 = 16 (not a prime)
5, 7 => 5 * 7 = 35 + 1 = 36 (not a prime)
So how can I claim a larger prime number than the largest known?
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Boli.. I think you should take all the prime numbers.. 2 is a prime number so if you take
2 * 3 * 5 = 30 + 1 = 31( prime number)
2 * 3 * 5 * 7 = 210 + 1 = 211 (Prime Number)
1 * 2 = 2 + 1 = 3 (prime number)
Hope this is right..
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Anilbhai,
Thanks! I got your point... I was considering random Prime Numbers.
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You can take random primes (all primes known to you) e.g. 3 and 5 then as you said 16 is not prime but as Anilbhai pointed out it is divisible by 2 which is not in your prime's list, so now you know one more, Now take the product of all primes you know ie 2,3 and 5 to get 31.
Thus this method helps you find all primes below a specific number ( but not ALL primes since there are an infinity of them).
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Friends,
Recently there was a big debate over the NET, on whether moments like 20/02/2002 at 20:02 hrs (perfect symmetry)are unique. I dont know whether this has any context to vedic maths, but I remember my grandfather saying that such symmetrical numbers DO HAVE significant importance in Vedic maths (am i correct Ashish?)
Any one has any idea abt that?
BTW, i figured out that 21/12/2112 at 21:12 hrs too would be a perfect moment to cherish........
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What date it is depends on two things:
(1) your offset i.e. when you start counting e.g. it is 2002 AD, and
(2) the base you use e.g. binary, decimal etc.
For people of different continents it also matters whether you write the date first or the month.
Thus, in an absolute sense dates have no meaning. They are simply references. As far as references go, the timings you mention are indeed remarkable.
At the start of this year, a friend of mine emailed: welcome to the last palindromic year of our life (a palindrome is the same both ways - the next will be 2112). immediately a friend replied that in hexadecimal this year is 7D2. So 7D7 is coming up in five years!
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Aschig
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| | Saturday, April 06, 2002 - 6:42 pm: |
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The book vedic maths mentions of a vedic numerical code that was used to help students learn long numbers in the form of verse. The code goes thus:
kaid nava , Taid nava , paid pMcak , yaaVXTkÊ xaÁ Xaunyama i.e.
k = 1, K = 2, ... Ja = 9
T = 1, z = 2, ... Qa = 9
p = 1, f = 2, ... ma = 5
ya = 1, r = 2, ... h = 8
xa = 0
Then you could compose your verses with the possibility of using any of
k , T , p , ya for 1 and so on - the vowels did not matter.
When computers that can "memorize" millions of bits were not available, this would indeed be a very good mnemonic.
The author gives four examples in the book including a verse in Anaustuba CMd that gives the value of pi (the ratio of circumference to diameter and which is popularly attributed to Aryabhatta of 5th century AD) to 31 decimal places:
gaÜpIBaagyamaQauva`at - EaUD.igaXaÜdiQasainQaga ||
KlajaIivatKatava galahalaarsaMQar ||
giving pi/10=0.314159265358979323846264338327892
The author goes on to say that this verse, looked at from its sanskrit meaning, can be treated as in praise of lord krishna, or, separately, as in praise of lord shiva. He goes on to say that it has a hidden key that reveals the value of pi to as many digits as one wishes.
This last statement makes me very skeptical since pi is a transcendental number i.e. a number that can not be described as a ratio of two integers (other examples are sq. root of 2, that of 3 etc.). Thus the digits in its decimal "equivalent" do not come in a deterministic fashion (unlike a lot of other ratios - those that recur after a set number). There is absolutely no pattern in the digits in the decimal fraction.
Anyway, it is a good mnemonic to remember the value of pi to 31 places (more than we typically care).
later I will give the other three examples mentioned by te author.
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Aschig
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| | Sunday, April 07, 2002 - 10:57 pm: |
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Continuing on my post from yesterday, I give here some tricks about recurring decimals and the remaining three examples of the vedic numerical code.
Fractions can be denoted in the vulgar notation e.g. 2/9 or in the decimal notation e.g. 0.222...
In the decimal notation, one sees that the decimal often repaets itself as in 2/9 above, the digit 2 repeats itself ad infinitum. We see below how to convert vulgar fractions to decimal and also determine when the decimal starts repeating itself.
1/2=0.5 (no repeats and 1 digit after the decimal point)
1/4=1/(2x2)=0.25 (no repeat, 2 digits)
1/8=1/(2x2x2)=0.125 (no repeat, 3 digits)
.
.
.
1/128=1/2^7 has 7 non-repeating digits
(left as an exercise)
1/5=0.2 (1 digit, no repeats)
1/25=1/5^2=0.04 (2 digits, no repeat)
1/625=1/5^4 has 4 non-repeating digits (exercise)
1/10=0.1
1/10^4=0.0001 (4 digits, non-repeating)
Thus, 2, 5 and 10 behave in similar fashions.
1/20=1/(2x10)=0.05 (2 non-repeating digits:
1 because of the factor 2 and one due to 10)
1/(2x10x2x2)=1/80 should have 4 non-repeating digits (exercise)
Now we come to the interesting recurring part:
1/3=0.33333333 (1 digit keeps repeating)
1/9=0.11111111 (as above)
1/11=0.090909090909 (2 digits keep repeating)
1/13=0.076923 (6 digits keep repeating)
1/17=0.0588235294117647 (16 digits repeat)
1/19=0.052631578947368421 (18 digits repeat)
When the factors are mixed, the results are mixed:
1/6=1/(2x3)=0.16666666... (1 fixed digit viz. 1 due to the factor 2 which gives a fixed digit, and 6 recurs due to the factor 3)
1/15=0.06666666...
1/18=0.05555555...
Thus factors 3,9 contribute one recurring digit, 11 contributes 2, 7 6, 19 18, 29 28 and so on, which we will discuss later.
TO BE CONTINUED ...
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Hello Aschiq,
Is there any method in Vedic Mathematics to solve eigenvalue problems
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Pooh
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| | Wednesday, June 25, 2003 - 5:53 pm: |
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This is not really related to Vaidik Ganit, however, this is an excellent site for history of mathematics. Piece on Indian Mathematics is pretty good.
http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Indian_mathematics.html
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Mona
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| | Wednesday, June 25, 2003 - 6:25 pm: |
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Aschig - wonderful BB. I haven't read in detail yet but will take a printout and read in leisure. Pray tell me, what is the right age to digest Vedic Mathematics ? My son is 3 years old (I know its too early), but I would like to start him with a solid Maths foundation (if he has the inclination, that is), when should children be introduced to Vedic Maths ?
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Anilbhai,
1 is not a prime no. Prime nos start from 2,3....
pan mala mait nahi ka te..mhnaje mi visaralo... kadachi Aschig tyavar prakash taku shakel.
(Ref: Anilbhai (Anilbhai)
Posted on Friday, February 22, 2002 - 9:27 am: )
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Anilbhai
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| | Thursday, June 26, 2003 - 11:58 am: |
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Thanks Mandar
I found this link which explains this..
http://www.utm.edu/research/primes/notes/faq/one.html
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Aschig
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| | Friday, August 01, 2003 - 11:21 pm: |
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Trying to continue where I left off:
Let us see how the recurring digits in 1/19 can be obtained.
We use ekadhikenpurveN again when the last digit is 9.
one more than the previous one is 1+1=2
so 1/19 = (1/2)
1/2 = Q0 R1 (meaning quotient 0 remainder 1)
So put the 0 down as the first digit and the 1 BEFORE it as a carry
Answer so far: 0.0 next part is (10/2) where 1 is the carry, 1 is the last digit
we obtained and 2 is our standard divisor
10/2 = Q5 R0 so put 5 down and carry 0 over
Answer so far: 0.05 next part is (05/2) 0 is carry, 5 is last digit of answer
and 2 is the selfsame divisor
5/2 = Q2 R1 giving us 0.052 and 12 as the next number to be divided
0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1
_1 0 1 0 0 1 1 1 1 0 1 0 1 1 0 0 0 0
and we are back to the first point (1/2) and the decimal repeats here after.
Another example: 1/29 = (1/3)
Q0 C1 (10/3)
Q3 C1 (13/3)
Q4 C1 (14/3)
Q4 C2 (24/3)
Q8 C0 (8/3)
Q2 C2 (22/3)
........
1/29=0.0344827586206896551724137931
You can just write it down digit after digit as you think.
This of course also works for things like 2/19, 3/29 etc
3/29:
(3/3)
1 0 (1/3)
0 1 (10/3) and then it will be the same digits as in 1/29 i.e.
3/29 = 0.1034482.....
Therefore you can use it for problems like 1/17
1/17 == 7/119 (multiplying 7 by 7 to get 9 as the last digit)
7/119 = (7/12)
Q0 C7 (70/12)
Q5 C10 (105/12)
Q8 C9 (98/12)
Q8 C2 (28/12) .....
1/17=0.0588235294117647
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Aschig
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| | Friday, August 01, 2003 - 11:24 pm: |
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Mona, once your son starts understanding numbers and how to read them (rather than just counting) you could try the most basic stuff. Initially of course it is better to start with geometric concepts of squares and cubes using blocks.
ashishdighe, I am not aware of specific methods for eigenvalues, but there are some methods related tomatrices. Perhaps those could be used.
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