Squaring:

any number can be approximated to include a five in the digits; first step in squareing 36 is to approximate it as 35

to find 35^2

simply ignore the 5; you are left with 3

add one to 3 : 1+3 =4 and multiply back by itself : 3*4 = 12

add a 25 to the end of the nmber : 12,25

so

35^2 = 12,25 (the answer is within 10% of this answer)

36^2 > 35^2 > 12,225

since 36 is one more than 35 so we have 36^2 = (35 +1)^ 2 = 35^2 + 1 + 70 = 12,25 + 1 + 70 = 12,96 ANSWER

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Another example :

634^2

634 is closest to 635 so find 635^2

1) ignore the 5, you have 63
2) add one to 63 + 1 = 64
3) 63 * 64 = 4032
5) answer is 403225 (this answer is within 10% of this value)
6) since 634^2 = (635 - 1)^2 = 635^2 + 1 - 1268 = 403225 + 1 + 1268 = 401958 ANSWER


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where calculators fail for quantum physicists and large numbers:

(456430590686058698686965066868969560463463636363636363636)^2 = Z

first approximate to include a five in the units:

1)456430590686058698686965066868969560463463636363636363635
2) ignore the five and we have:
45643059068605869868696506686896956046346363636363636363 and add one to this number and you have
45643059068605869868696506686896956046346363636363636364

3) multiply the two together:

45643059068605869868696506686896956046346363636363636364 * 45643059068605869868696506686896956046346363636363636363

as you can see the two numbers are very close to each other and we can simply replace 3, and 4 in the ending digit units and replace them with 5, so we have:

45643059068605869868696506686896956046346363636363636365^2 = Y

again we repeat the step above but now for our new number:

since we already have converted the 3 or 4 into 5 , we simply ignore the 5 so we have

1)4564305906860586986869650668689695604634636363636363636 , and again
2) we add 1 to the number above, and we have:

4564305906860586986869650668689695604634636363636363637
3) we multiply this back by the value in 1) so we have:

4564305906860586986869650668689695604634636363636363637 * 4564305906860586986869650668689695604634636363636363636

again like before we ignore the 7, or 6 digit and assume they are five so we have to find :

4564305906860586986869650668689695604634636363636363635 ^ 2 = X


again this number ends in 5 we ignore the 5 and we add 1 to the reminder and we multiply them so we would end with (I dont include the steps since you know them by now):

456430590686058698686965066868969560463463636363636365 ^2 = W

again we repeat this and we have:

45643059068605869868696506686896956046346363636363635 ^ 2 = V

and again:

4564305906860586986869650668689695604634636363636365 ^ 2 = U

and again:

456430590686058698686965066868969560463463636363635 ^ 2 = T

and again:

45643059068605869868696506686896956046346363636365 ^ 2 = S

and again:

4564305906860586986869650668689695604634636363635 ^ 2 = R

and again:

456430590686058698686965066868969560463463636365 ^ 2 = Q

and again:

45643059068605869868696506686896956046346363635 ^ 2 = P

and again:

4564305906860586986869650668689695604634636365 ^ 2 = O

and again:

456430590686058698686965066868969560463463635 ^ 2 = N

and again:

45643059068605869868696506686896956046346365 ^ 2 = M

and again:

4564305906860586986869650668689695604634635 ^ 2 = L

and again:

456430590686058698686965066868969560463465 ^ 2 = K

and again:

45643059068605869868696506686896956046345 ^ 2 = J

and again:

4564305906860586986869650668689695604635 ^ 2 = I

and again:

456430590686058698686965066868969560465 ^ 2 = H

and again:

45643059068605869868696506686896956045 ^ 2 = G

and again:

4564305906860586986869650668689695605 ^ 2 = F

and again:

456430590686058698686965066868969565 ^ 2 = E

and again:

45643059068605869868696506686896955 ^ 2 = D

and again:

4564305906860586986869650668689695 ^ 2 = C

and again:

456430590686058698686965066868965 ^ 2 = B

and again:

45643059068605869868696506686895 ^ 2 = A

and again:

4564305906860586986869650668685 ^ 2 = A'

and again:

456430590686058698686965066865 ^ 2 = B'

and again:

45643059068605869868696506685 ^ 2 = C'

and again:

4564305906860586986869650665 ^ 2 = D'

and again:

456430590686058698686965065 ^ 2 = E'

and again:

45643059068605869868696505 ^ 2 = F'

and again:

4564305906860586986869655 ^2 = G'

and again:

456430590686058698686965 ^ 2 = H'

and again:

45643059068605869868695 ^ 2 = I'

and again:

4564305906860586986865 ^ 2 = J'

and again:

456430590686058698685 ^ 2 = K'

and again:

45643059068605869865 ^ 2 = L'

and again:

4564305906860586985 ^ 2 = M'

and again:

456430590686058695 ^ 2 = N'

and again:

45643059068605865 ^ 2 = O'

and again:

4564305906860585 ^ 2 = P'

and again:

456430590686058 ^ 2 = Q'

and again:

45643059068605 ^ 2 = R'

and again:

4564305906865 ^ 2 = S'

and again:

456430590685 ^ 2 = T'

and again:

45643059065 ^ 2 = U'

and again:

4564305905 ^ 2 = V'

and again:

456430595 ^ 2 = W'

and again:

45643055 ^ 2 = X'

and again:

4564305 ^ 2 = Y'

and again:

456435 ^ 2 = Z'

and again:

45645 ^ 2 = A''

and again:

4565 ^ 2 = B''

and again:

455 ^ 2 = C''

and again:

45 ^ 2 = D''

and again:

5 ^ 2 = E''

Ok, now that we have finished all the 55 digits of the number lets see what we have:

The final answer we wish to find is that of

(456430590686058698686965066868969560463463636363636363636)^2 = Z

In the above I equalled the answer to Z which we wish to find. We also know that the number has 56 digits. As we shown in the beginning of this thread the approximated 5 digited number is within 10% of the actual answer so we have:

Z = Y25 + .1(Y25)

and

Y = (X25) + .1 (X25)

so

Z = X25 + .1 X25 + .1 ( X25 + .1 X25) = 1.1X25 + .2 X25 (1)

and since

X = W25 + .1W25

and

W = V25 + .1V25

and so

X = V25 + .1V25 + .1( V25 + .1V25) and along with (1) we have :

Z = 1.1V25 + .11V25 + .11V25 + .011V25 + .2 V25 + .2V25 + .02V25 + .022V25 = 1.8 V25

**Note: V25 means if V = 111111 for instance the actual value is 11111125

and do so and simplify until you reach a number that is doable using a calculator and then replace it in the formula to find Z. In this case I did not write the whole thing so Hypothetically my calculator could calculate V. In real life however this would take you a few hours until you get to some value like U' that fits into your calculator screen. Anyhow this is how you find the square of very large numbers.



Testing for validity:

we know that 5 ^ 2 = E'' and also that 45 ^ 2 = D''

so based on my assumption : D'' = E''25 + .1E''25

since E'' = 25 we then have

D'' = 2525 + .1(2525) = 2272
and D'' = 45 ^ 2 = 2025 !!

as you can see the method works but with a slight error.

QUESTION: Wont the slight error increase as we progress up to the larger number?
Yes, but the point is that not all number are ten percent larger and some are smaller, usually for very big numbers the error cancells itself out and the result is a beautiful, 5% error only and for a number with 55 digit that is very accurate!
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some more samples:

65^ 2 =?

ignore 5

6 +1 = 7

6 * 7 = 42

add 25 : 42,65

65^ 2 = 42,65

there you have it. You can have a lot of fun with this:

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in general for a number like "abcdefg" ^ 2 simply ignore g and you have

"abcdef"

add 1 to abcdef and you have abcde(f+1)

multiply the two numbers:

abcdef * abcde( f+1) = X

now add 25 to the front of X

so your answer is


abcdef ^ 2 = [abcdef * abcde( f+1)]25.

Is there any authority backing this up? Is it accepted by mathematicians as the guess and babylonian methods for approximating square root?

No. Drude was lying.

Specifically two claims were made

10n + d can be well approximated by 10m + 5 which is not true for all purposes

and

(10m + 5)^2 = 100(m+1)m + 25

which is true and can be shown as
(10m + 5)^2
= 100m^2 + 100m + 25 # By the binomial theorem or multiplication of polynomials
= 100(m^2 + m) + 25
= 100(m + 1)m + 25

This generalizes as
(2 k m + k)^2 = 4 k^2 (m+1)m + k^2

630 = 2 * 30 * 10 + 30
630^2 = 4 * 900 * 110 + 900 = 396900 # Which makes more sense in base-60 math

But as an approximation scheme, it only makes sense if you are stuck with fixed-point math, when floating point systems already allow you to get approximate answers and computer systems already allow you to square ridiculously large integers exactly.