Last week,
a friend (not the hot-air-head friend. He is on vacation.) asked to borrow
my calculator. He was doing some simple arithmetics which involved with
taking the square root of some numbers. I told him that I had left my calculator
in the lab downstairs. After fifteen minutes digging in the lab and two
round trips running between my office (on the third floor) and the lab
(in the basement), he was dead tired, and still, no calculator. I then
told him "Why don't you take a piece of paper and do it by hand?".
"I never learned to do it by hand", came the answer. Finally,
I told him to use the computer. After five minutes writing a two line program,
compiling, and running it, he got the answers. That event reminded me of
a story I read long time ago. It was about a society many years into the
future where people don't use paper anymore. Books are contained in some
computer chips; newspapers are broadcast via the computer network... Every
calculation is done with a computer or a calculator. People don't know
how to calculate in the head or on a piece of paper any more, even the
very simple calculations like 1+1=2. (We, the 20th century people, all
know 1+1=2 in our heads, don't we? Except in love or marriage where 1+1=1
and 1+1+1="one big mess".) A cleaning person becomes a celebrity
when he proves that he can do arithmetics without a computer...

That story
probably has some truth in it. Twenty years ago, most junior-high students
could calculate square roots without a calculator. Some high-school students
today do not know how to divide two numbers without the little hand-held
electronic thing. If this decay rate remains constant, twenty or fifty
years from now, most people may not know how to multiply or divide. It
may even get worse that people forget how to add and subtract. I decided
to try taking the square root of a number on the paper. Guess what? It
took me half hours to fumble through the problem. I guess my mind is very
rusty and I bet that the minds of many of you are rusty, too. When I was
a kid, I could add, subtract, multiply, and divide 2 three digit numbers
in my head easily. Today, I balance my checkbook using neither my head
nor paper but a calculator. Let's refresh our minds, go back to 8th grade,
and re-learn the steps of taking square root and cube root of a number.
It may be useful some day. Like, one sunny day, you decide to go to swim
at the beach. After getting there, you decide to do some statistics (that
means you have to take square root of some numbers) by counting the number
of muscular people, the numbers of people wearing two-piece, one-piece,
zero-piece swimming suits on the beach, etc... And suppose you forget to
bring you lap-top to the beach with you (which is very likely), then what
are you going to do? You probably wish that you have learned how to take
square root, don't you? So here is the direction.

SQUARE ROOT:
Taking the square root of (654.3).

1)Separate
the number into periods of two figures each, beginning at the decimal point.
i.e., 654.3 --> 6 54. 30 00

2)Find the
greatest digit N where M=N^2 <= the 1st left-hand period. Subtract M
from the 1st left-hand period. Bring down the next period to form a new
dividend 2 2 _____________ \/ 6 54. 30 00 - 4 ----- 2 54

3)Trial divisor:
Multiply the root R already found by 10 then by 2, i.e., let I=R*10*2=R*20.
Find the greatest digit N where M=N*(I+N) <= the new dividend. Subtract
M from the new dividend. Bring down the next period to form the next dividend
2 5 2 _____________ \/ 6 54. 30 00 - 4 ----- I=2*10*2=40 ) 2 54 M=5*(40+5)=225
) - 2 25 --------- 29 30

4)Go back
to step 3). Continue the process until the desired decimal place is attained.
Example: 2 5. 5 7 2 _____________ I=0*10*2=0 \/ 6 54. 30 00 M=2*(0+2)=4
) - 4 ----- I=2*10*2=40 ) 2 54 M=5*(40+5)=225 ) - 2 25 --------- I=25*10*2=500
) 29 30 M=5*(500+5)=2525 ) - 25 25 ---------- I=255*10*2=5100 ) 4 05 00
M=7*(5100+7)=35749 ) - 3 57 49 ---------- 47 51

------------------------------

The steps
for calculating the cube roots are similar to those of square roots.

CUBE ROOT:
Taking the cube root of (7654.3).

1)Separate
the number into periods of three figures each, beginning at the decimal
point. i.e., 7654.3 --> 7 654. 300 000

2)Find the
greatest digit N where M=N^3 <= the 1st left-hand period. Subtract M
from the 1st left-hand period. Bring down the next period to form a new
dividend 1 3 _____________ \/ 7 654. 300 000 - 1 ----- 6 654

3)Trial divisor:
Multiply the root R already found by 10, square the product then multiply
by 3, i.e., let I=[(R*10)^2]*3=(R^2)*300. Find the greatest digit N where
M=N*[I+N*(R*10*3)+N^2]=N*(I+N*R*30+N^2) <= the new dividend. Subtract
M from the new dividend. Bring down the next period to form the next dividend.
This step may be done by trial division, i.e., estimate the best value
N and try it. If it doesn't work then try the next number. The estimation
may be done by dividing the new dividend by the value I=(R^2)*300 shown
above. Example, for the new dividend = 6654, I= 1*300=300, then we have
N about 6654/300 = 22 which is impossible. So we try to use the largest
digit which is the number 9 for N. 1 9 3 _____________ \/ 7 654. 300 000
- 1 ----- I=[(1*10)^2]*3=300 ) 6 654 M=9*[300+(9*1*30)+(9^2)]=5859 ) -
5 859 ---------- 795 300

4)Go back
to step 3). Continue the process until the desired decimal place is attained.
Example: 1 9 7 0 3 _____________ I=[(0*10)^2]*3=0 \/ 7 654. 300 000 M=1*[0+(1*0*30)+(1^2)]=1
) - 1 ----- I=[(1*10)^2]*3=300 ) 6 654 M=9*[300+(9*1*30)+(9^2)]=5859 )
- 5 859 ---------- I=[(19*10)^2]*3=108300 ) 795 300 M=7*[108300+(7*19*30)+(7^2)]=786373
) - 786 373 ------------ I=[(197*10)^2]*3=11642700 ) 8 927 000 M=0*[11642700+(0*197*30)+(0^2)]=0
) - 0 000 000 ---------- 8 927 000 ---------------------------------

Here is the
BONUS for the people who finish the course. We all probably know that (3^2
+ 4^2 = 5^2). Now try to find some numbers such that I^N + J^N = K^N, where
I, J, K are integers > 0, and N = integer > 2.

And here is
the EXTRA BONUS: Do you know that (3^3 + 4^3 + 5^3 = 6^3)? Now try to find
some numbers such that I^N + J^N + K^N = L^N, where I, J, K, L are integers
> 0, and N = integer > 3.

EXTRA EXTRA
BONUS: Is there some numbers such that (I^4 + J^4 + K^4 + L^4 = M^4), where
I, J, K, L, M are integers > 0.

Happy number
crunching, everyone!!!

*Duc
Ta Vo, Ph.D.*

ducvo@lanl.gov

`For
discussion on this column, join vacets-tech@vacets.org`

`Copyright
© 1996 by VACETS and Duc Ta Vo`

`:`