A TRICK FOR SHORTCUTTING SQUARE-ROOTING BY SUBTRACTION

If you've performed SQUARE-ROOTING a number, such as 144, by a standard method, you've learned to parse the digits of the number in pairs, starting from the right: 144 -> 1'44.

Why? Because each single-dig number has a SQUARE OF ONE OR TWO DIGITS. Numbers 1, 2, 3, have 1-digit squares -- respectively, 1, 4, 9 -- but the remaining numbers up to 9 have 2-digit squares (16, 25, 36, 49, 64, 81). So, use work from PAIRS OF DIGITS IN THE SQUARE TO FIND A SINGLE DIGIT IN THE SQUARE-ROOT. Dig?

You will see the same SUBALGORITHM enter below.

The square root of 225 is 15. Conversely, we can find the square-root of 225 by subtracting the first 15 odd numbers, namely, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29. Thus, 225 - 1 = 224 - 3 = 221 - 5 = 216 - 7 = 209 - 9 = 200 - 11 = 189 - 13 = 176 - 15 = 161 - 17 = 144 - 19 = 125 - 21 = 104 - 23 = 81 - 25 = 56 - 27 = 29 - 29 = 0. That is, SUBTRACTING THE FIRST 15 ODD NUMBERS FROM 225 ARRIVES AT 0; HENCE, THE SQUARE ROOT OF 225 IS 15.

But this is TEEEEEEEJUS! And finding the SQUARE ROOT OF 1225 = 35 X 35, BY SUBTRACTING THE FIRST 35 ODD NUMBERS, WOULD BE EVEN MORE TEEEEEEEEEEEEJUS!

Not to worry. The TRICK I'll show you will find 35 as SQUAREROOT of 1225 in just 8 SUBTRACTIONS! BECAUSE 3 + 5 = 8.
  1. Parse 1225 in PAIRS OF DIGITS FROM RIGHT: 12'25.
  2. Start SUBTRACTING FROM THE LEFT PAIR: 12 - 1 = 11 - 3 = 8 - 5 = 3. STOP, because we can't subtract the NEXT ODD NUMBER: 7. But we SUCCEEDED in 3 SUBTRACTIONS WITH THE LEFT PAIR. So, we keep 3 as PART OF THE SQUAREROOT ANSWER.
  3. The REMAINDER was also 3, so we ADJOING THAT WITH THE RIGHT PAIR, to obtain 325, which we'll now use as the MINUED OF OUR PICKUP SUBTRACTION. But WHAT'S THE SUBTRAHEND for this? Ah, that's the REAL TRICK!
  4. Taking that 3 remainder, we MULTIPLY IT BY 2 AND BY 10 AND ADD 1: (3 X 2 X 10) + 1 = 61. (Why? Patience! Explanation, later.)
  5. With 325 as our MINUEND, and 61 as our FIRST SUBTRAHEND, be CONTINUE THE SUBTRACTION OF SUCCESSIVE ODD NUMBERS: 325 - 61 = 264 - 63 = 201 - 65 = 136 - 67 = 69 - 69 = 0. FINISHED -- in 5 SUBTRACTIONS. We RIGHT-CONCATENATE that 5 with the 3 on hold, to obtain: 35, as the SQUAREROOT of 35. QED.

    But wherefore that TRICK of

    1. taking the REMAINDER of a PARTIAL SUBTRACTION PROCESS,
    2. multiplying it by 2
    3. and multiplying by 10
    4. and adding 1 -- to OBTAIN THE CONTINUING SUBTRAHEND?
    Taking the REMAINDER should be obvious. To explain the 2-Multiple, show you an "algebraic binomial": (a + b)2 = a2 + 2ab + b2. I've UNDERLINED that 2-multiplier to show whence the 2-multiplier in the above process.

    But the 10-multiplier? Ah! That REMAINDER was from the LEFT PAIR, which CORRESPONDS TO A 10S-DIGIT IN THE SQUARE-ROOT. Hence, we MULTIPLY THE REMAINDER by 10, TO GIVE IT THE CORRECT "PLACE-VALUE". That gives us, in the present case, 2 x 3 x 10 = 60. So, we ADD 1, because that obtains for us the ODD NUMBER BEGINNING THIS DECADE, as it will do in THE GENERAL CASE. Dig?

    And notice that this TRICK obtained the SQUAREROOT 35 in 3 + 5 = 8 SUBTRACTIONS.

    I'll do one more, to enlighten you. 45 x 45 = 2025. Conversely, we'll find the SQUAREROOT of 2025 to be 45 in 4 + 5 = 9 ODD-NUMBER SUBTRACTIONS.
    1. Parsing 2025: 20'25. So we start with the left pair: 20.
    2. 20 - 1 = 19 - 3 = 16 - 5 = 11 - 7 = 4. Can't subtract the next odd number, 7. But we SUCCEEDED in 4 PARTIAL SUBTRACTIONS, so we put 4 on hold.
    3. We ADJOIN the 4 REMAINDER to the RIGHT PAIR, obtaining 425, as NEW PARTIAL MINUEND.
    4. The REMAINDER WAS 4, so we form: (4 x 2 x 10) + 1 = 81, and we CONTINUE ODD-NUMBER SUBTRACTING with FIRST SUBTRAHEND of 81 from MINUEND 425.
    5. 425 - 81 = 344 - 83 = 261 - 85 = 176 - 87 = 89 - 89 = 0. FINISHED, in 5 ODD-NUMBER SUBTRACTIONS.
    6. So, we RIGHT-CONCATENATE this 5 with the 4 on hold, to obtain: 45 as SQUAREROOT of 2025.
    7. And we found this in 4 + 5 = 9 ODD-NUMBER SUBTRACTIONS. Dig?

      All this follows from PYTAGORUS DISCOVERING, around 2500 years ago, THAT ODD NUMBERS ARE THE GNOMONS (BULDING-BLOCKS) OF SQUARES. ASSIGNMENT: BUILD A SQUARE FROM UNIT BLOCKS. If you build a 3 x 3 SQUARE, you have 3 ROWS, and COLUMNS of blocks, OR 3 X 3 = 9 BLOCKS. To INCREASE this to a 4 x 4, you must ADJOIN a 3-BLOCK-ROW and a 3-BLOCK-COLUMN, for 2 x 3 = 6 BLOCKS. But to FILL IN THE CORNER FOR A SQUARE STRUCTURE, YOU MUST ADJOIN ANOTHER BLOCK FOR 6 + 1 = 7 ADJOING BLOCKS. 9 + 7 = 16 = 4 x 4. In GENERAL, (n + 1)2 = n2 + 2n + 1 = (n2) + (2n + 1) -- the left term is a square, (n2), which is transformed into the next square, (n + 1)2, by adding the ODD NUMBER, (2n + 1). Dig?