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How to find the square root

  1. Dec 10, 2008 #1
    Hi,
    I was taking a square root of a fraction and I was wondering how to get a fraction result after I've taken the square root of it. And it also made it think; How do I take a root of a number in pen and paper, without the use of calulators?

    Can anyone help me?

    Thanks
     
  2. jcsd
  3. Dec 10, 2008 #2
    If you mean numerically finding the value of a root you can use Newton's Method; say you want to find the square root of two numerically, then whatever that number is can be treated as a variable x as such...

    [tex]x=\sqrt{2} \Rightarrow x^2=2[/tex]

    and by Newtons method we use the recursive formula,

    [tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]

    You must choose a starting point [itex]x_0[/itex] that is close to what you suspect the root to be and in this case we can see that [itex]f(x)=x^2-2[/itex] so [itex]f'(x)=2x[/itex] so your first iteration will be (say we choose 1.5 to be the starting point)

    [tex]x_{n+1}=\frac{3}{2}-\frac{\left(\frac{9}{4}\right)-2}{3}[/tex]

    and you can see how it just continues from here. It actually converges rather quickly to a good approximation of the number.
     
  4. Dec 11, 2008 #3
    If you want to try to find the square root in a manner similar to long division, Wikipedia describes a couple.
     
  5. Dec 11, 2008 #4

    uart

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    Hi Razza, you've asked two questions here but let me just address the first one.
    Do you know about irrational numbers? They cannot be expressed as a fraction (that's pretty much the definition of an irrational number btw). In general the square root of a rational is not another rational, it can of course be rational but more often it's not.

    In a way it's a bit like asking the question "I've just taken the square root of a whole number and I want to express the answer as another whole number". I'm pretty sure you'd realize that this is simply not possible in general. It's similar with fractions, the sqrt of a fraction can only be expressed as another fraction if the first fraction, in it's simplist form, has both the numerator and the denominator as an exact square.

    For example sqrt(9/4) = 3/2, sqrt(25/9) = 5/3 etc, but in every case other than this type then you'll get an irrational answer, so no fraction possible.

    I hope that makes sense to you. :)
     
    Last edited: Dec 11, 2008
  6. Dec 11, 2008 #5
    However, if finding a fraction, regardless of rationality is what you want, you can do this.

    Remember that a square root can also be described as an exponent of 1/2. And like other exponents, it can be 'distributed' to each number that is being multiplied or divided. Of course, if you can, you should simplify so that we don't have a square root in the bottom, so we multiply by root b over root b. So this is the process to use, in order to simplify square roots of fractions.

    [tex]\sqrt{\frac{a}{b}}=\left(\frac{a}{b}\right)^{1/2}=\frac{a^{1/2}}{b^{1/2}}=\frac{\sqrt{a*b}}{b}[/tex]
     
    Last edited: Dec 11, 2008
  7. Dec 11, 2008 #6
    With great pain.

    As explained above, you can use a numerical approximation. Alternatively, you can use the relation:

    [tex]\sqrt{x} = x^{\frac{1}{2}} = e^{\ln x^{\frac{1}{2}}} = e^{\frac{1}{2} \ln x}[/tex]

    Then, if you have a table of logs and exponents, you can calculate it like that. (Or any other root for that matter).

    If you wanted to do it completely by hand, without any tables at all, you could crank it out by expanding [tex]e^\dots[/tex] and [tex]\ln \dots[/tex] with a series:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/cseri.html#c2
    http://hyperphysics.phy-astr.gsu.edu/Hbase/math/lnseries.html

    Using standard rational-number arithmetic, you can approximate exp and ln to any accuracy you want, and thus, sqrt as well. It just might take you a couple hours. That is why, before the invention of calculators, people used slide rules and tables.
     
  8. Dec 19, 2008 #7
    Here's an even easier way to get a fractional approximation to sqrt(a/b).

    1. Select some error tolerance... call it epsilon.
    2. Take 0 and a/b as endpoints... call them x and y.
    3. Compute z = (x+y)/2.
    4. If |a/b - z^2| < epsilon^2, then z is your answer.
    5. Otherwise, if a/b > z^2, let x = z and return to step 3.
    6. If a/b < z^2, let y = z and return to step 3.

    This will halt when z is within epsilon of sqrt(a/b), and it must a fortiori yield a fractional answer. Also, it will finish after approximately t = log_2[ (a/b) / epsilon ] steps in the worst case... so for 23/7, with epsilon = 1/100, we have to iterate at most 9 times. Not to shabby...

    Another method is using differentials. So say you want the square root of 23/7. Well, we notice that 23 / 7 equals 161 / 49, and 161 ~ 169. If we want the square root of x, y = x^(1/2), we find dy = [(1/2) / sqrt(x)] dx and can approximate y = 13/7 - [(1/2)/(13/7)](8/49) = 13/7 - (104)/(686) which comes out to something a bit higher than 12/7. And do you need much more accuracy than that?
     
  9. Dec 20, 2008 #8

    Gib Z

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    9 iterations seems quite long for 2 decimal places.. and the Differentials method yields 1.70554, whose square is about 2.9, whilst 23/7 ~ 3.3 . I would say many people do need more accuracy than that.

    Newton's Method in jeffreydk's post works extremely well however - a single iteration with a_0 = 1.8 yields ~1.8127, which is already 3 d.p. Another iteration gives 9 d.p.
     
  10. Dec 21, 2008 #9

    uart

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    That method is know as "halving the interval". It's much slower to converge than Newtons method.

    BTW. For solving x^2 = y Newtons method can be compacted to :

    [tex]x_{n+1} = \frac{1}{2}(x_n + y/x_n)[/tex]
     
  11. Dec 27, 2008 #10
    Slower, but easier. And both are good enough, depending upon how much accuracy you need.
     
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