Why would you need this??? We have calculators!

Anyway, thought some of you guys would appreciate this

edit: holy, massive picture

Ah yes, this is a very nice method. I've been doing this since 7th grade I happened to have a trig book from the 50's, no electronics, and a whole summer vacation of having fun with math.

In binary:

Let:
A=Result
B=Remainder

Initial:
d=0
b=01 01.00 00
a=0

Cycle 1
;shift upper two bits in b to D
  b=01 00 00 00
  d=00 00 00 01
2*A
 ;A=00 00 00 00
A→C
 ;C=00 00 00 00
2*C+1
 :C=00 00 00 01
cp C,D
 if C is less than or equal to D
  D-C→D
  ; D= 00 00 00 00
  A+1→A
  ;A=00 00 00 01
Repeat
;shift upper two bits in b to D
  b=00 00 00 00
  d=00 00 00 01
2*A
 ;A=00 00 00 10
A→C
 ;C=00 00 00 10
2*C+1
 :C=00 00 00 11
cp C,D                              ;C is greater than D
x if C is less than or equal to D   ;False
x  D-C→D                            ;Not Executed
x  ; D= xx                           ;""
x  A+1→A                            ;""
x  ;A=xx                            ;""
Repeat

The first two bits calculated in this case is the integer part and after that all the bits are the fractional part (in binary).

EDIT: Those are supposed to be ; D. grrr...

The square root one is a specific example of the Shifting nth root algorithm
You can also find any higher roots, and in any base!