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VACETS Regular Technical Column

"Science for Everyone"

"Science for Everyone" was a technical column posted regularly on the VACETS forum. The author of the following articles is Dr. Vo Ta Duc. For more publications produced by other VACETS  members, please visit the VACETS Member Publications page or Technical Columns page.

The VACETS Technical Column is contributed by various members , especially those of the VACETS Technical Affairs Committe. Articles are posted regulary on vacets@peak.org forum. Please send questions, comments and suggestions to vacets-ta@vacets.org

October 17, 1994

Roots, Roots, and Roots!!!

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

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