math-history
We were taught the long division method of finding square root in junior classes. The logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.
You can find information about the history of the usage in Jeff Miller's page here: Earliest Uses of Symbols of Operation.
Quoted:
Square root. The first use of a capital R with a diagonal line was in 1220 by Leonardo of Pisa in Practica geometriae, where the symbol meant "square root" (Cajori vol. 1, page 90).
The radical symbol first appeared in 1525 in Die Coss by Christoff Rudolff (1499-1545). He used the symbol
(without the vinculum) for square roots. He did not use indices to indicate higher roots, but instead modified the appearance of the radical symbol for higher roots.
It is often suggested that the origin of the modern radical symbol is that it is an altered letter r, the first letter in the word radix. This is the opinion of Leonhard Euler in his Institutiones calculi differentialis (1775). However, Florian Cajori, author of A History of Mathematical Notations, argues against this theory.
In 1637 Rene Descartes used
, adding the vinculum to the radical symbol La Geometrie (Cajori vol. 1, page 375).
To solve $450\div 3$ we are asking how many groups of 3 there are in 450. The obvious way to solve this is to subtract groups of 3 from 450, one at a time, and count each group as it is subtracted. Thus:
450
- 3
----
447
- 3
----
444
- 3
----
441
- 3
.
. (do this many many times)
.
----
6
- 3
----
3
- 3
----
0
We have to repeat this 150 times, after which the 450 has been reduced to 0. So the answer is that 450 contains 150 groups of 3.
It should be clear that there is an easy way to speed up this process: instead of removing a single group of 3 each time, we could remove 10 or even 100 groups of 3 at once. So for example:
450
- 300 (100 groups)
-----
150
- 30 (10 groups)
-----
120
- 30 (10 groups)
-----
90
- 30 (10 groups)
-----
60
- 30 (10 groups)
-----
30
- 30 (10 groups)
-----
0
Here we have removed 150 groups of 3 as before, but instead of doing them one at a time, we removed 100 right away, and then five sets of 10 groups each, for a total of 150 groups.
We can abbreviate this further:
-------
3 ) 450
We begin by removing 100 groups of 3, as before:
1
-------
3 ) 450
300 (100 groups)
---
150
Now we want to remove sets of 10 groups from 150. Removing a set of 10 groups of 3 will remove 30. Instead of removing the sets one at a time, we apply our knowledge of multiplication to see that 150 is big enough to remove 5 sets of 10 groups of 3, for a total of 150:
15
-------
3 ) 450
300 (100 groups)
---
150
150 (5 sets of 10 groups = 50 groups)
---
0
Since the total has reached 0, we do not need to remove any single groups, so we fill in the final 0 in the answer:
150
-------
3 ) 450
300 (100 groups)
---
150
150 (5 sets of 10 groups = 50 groups)
---
0
Added 2014-04-23: I learned today that my daughter is being explicitly taught this method in the fourth grade. The example she showed me was:
--------
7 ) 182
Here, she said, you might happen to know that $7\times 20=140$, so you remove the 140 from the dividend:
--------
7 ) 182
- 140 20
-----
42
Then perhaps you remember that $7\times 5 = 35$, so you remove the 35:
--------
7 ) 182
- 140 20
-----
42
- 35 5
-----
7
Then you remove the remaining 7:
--------
7 ) 182
- 140 20
-----
42
- 35 5
-----
7
- 7 1
-----
0
The remainder is now 0, so you add the right-hand column, $20+5+1$, to obtain the quotient 26.
Best Answer
For $3$ or $4$ digit numbers you were using $$(10a+b)^2=100a^2+(20a+b)b$$
For $5$ or $6$ digit numbers you would have used the messier $$(100a+10b+c)^2=10000a^2+(2000a+ 100b)b+(200a+ 20 b+c)c$$
So you would separate the number to be square-rooted into $100$s, in your example $27,04$
and it looked like
The largest square less than or equal to $27$ is $25=5^2$ so you write
You then double the $5$ you have written at the top to give $10$ and then add an extra digit $X$ so $10X \times X$ is a large as possible but does not exceed $204$: $102 \times 2= 204$ works exactly
So $\sqrt{2704} =52$
If the result had not been exact, you could have continued the same way, bringing down two more digits (possibly $00$).
My school thought this a waste of time and instead taught us to use logarithm tables.