Notesale: Turn your study into money

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Square Roots by Hand
Instructions:
1 Count the number of digits your number has that are on the left
side of the decimal point
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Obviously that doesn't
change your number's value, but it's important because now you
need to break the number into groups of two digits, and it's
important that you break your number up on the proper
boundries
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Find the
largest number whose square is equal to or less than that
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3 Square the first digit of the square root and subtract the result
from the first group of two digits
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5 Now multiply what you have of the square root so far by 2, and
discard any decimal points
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6 Divide the partial divisor into the partial dividend and take the
first digit of the result
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Since we worked with the first digit after the decimal in our
original number (the 6 in 28
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So far, our square root
is:
7 Take the new digit from step 6 and append it to the partial
divisor from step five
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9 Subtract that from the old partial dividend
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At this point the algorithm starts repeating steps:
11 Repeat step 5: Multiply what you have of the square root by
two and discard decimal points, to get a new partial divisor
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Tack the new digit onto the end of the square root, to get:
13 For more digits of accuracy, just repeat steps 7 through 10, and
then loop back to step 5 as many times as you want
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Example:
28
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However, 9
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The first group is 28,
which gives us 5
since 52 = 25, which
is less than 28
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3
"10" + "3" yields
103
103 x 3 = 309
360 - 309 = 51
"51" + "00" yields
5100
53 x 2 = 106
5100/106 = 48
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,
which yields 4
5
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In fact, if you
tried to follow that set of instructions with some other number than 28
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Well, since the results of step 9 are the basis for the
next partial dividend, getting a negative number means having a negative partial
dividend
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If you think back to when
you were in the first or second grade when your math teacher explained the
decimal notation system to you, you'll remember that decimal notation is a (very
convenient) shorthand for a sum
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6 is 1 in the 100's place, plus
2 in the 10's place, plus 7 in the one's place
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1
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So with
that in mind, we see that having a negative digit is no big deal (aside from being
awkward to write)
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For example, what if instead of
127
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1, which is
100 - 20 + 7 +
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6
So if the result of step 9 is a negative number, don't worry about it
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Let's look at an example that has this phenomenon (I'll just show the
steps here, I won't repeat the descriptions of them):
The square root of 77
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15) =
?
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15 yields "77" "15" "00"
2 82 = 64, and 64 < 77
8
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8?? (= 8 + 8x10-1)
7 "16" + "8" = "168"
8 168x8 = 1344
9 1315 - 1344 = -29
10 "-29" + "00" = "-2900"
11 88x2 = 176
12 -2900/176 yields -1
8
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Remember that you have to

append the new digit to the partial divisor, then multiply the result by the new digit
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The minus signs nicely cancel out,
which is good
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Then multiply by the new
digit as normal
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14 -2900 - 1761 = -4661
15 "-4661" + "00" = "-466100"
16 879x2 = 1758
17 -466100/1758 yields -2
8
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7882 = 77
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078944
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You may wonder if this business with negative digits ever stops
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The reason is because of the nature of the algorithm
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So if your initial approximation is too small, the algorithm will
have to add a small positive amount at each step in order to make progress
towards the correct answer
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Otherwise, the approximation would just get worse
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They are engineered to always give you a positive first digit
that is an under-approximation of the true answer
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Therefore, that number's square root is less than the square root
of the first group of digits
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Since it is easy to figure out that the first digit of the answer is an underapproximation of the true answer, it is also pretty easy to see that the second
digit of the answer will be positive
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However, it is just part of the
nature of the algorithm that the second digit can "overshoot" the true answer
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I leave it
to cleverer minds than my own to figure out why it is possible for this second digit
to overshoot the true answer (and hopefully to fix the algorithm)
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In the event that you choose to find the square root of a
number that has a rational square root (such as 17
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2), you will find that at some instance of step 9 the subtraction comes
out to exactly zero and there are no more groups of digits left to deal with that
aren't "00" themselves
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This is easy to prove; if you assume that the result of step 9 is exactly 0, and you
follow that zero through the calculation, you'll see that all other digits produced by
the algorithm will also be 0

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