# Square root of 0.9

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Does anybody have any ideas on how to work out [imath]\sqrt{0.9}[/imath] without a calculator (to approx 10 d.p.)?

I know there are a few methods for working out for working out square roots by hand, but i was wondering if there was a cleverer/easier way of doing it.

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it's irrational, no?

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Use a binomial expansion for [imath](1+x)^{\frac{1}{2}}[/imath]. Since this converges for |x| < 1, you can happily slap in x = -0.1, work out the first 4 terms of the series and get a pretty accurate answer.

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the thing is, this was set as a sort of 'brainteaser' question, so i was wondering if there was a simple, almost 'sneaky' way of doing it...

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If you only need a close approximation, you can approximate it linearly pretty easily.

Let f(x)=x^(1/2), then f'(x)=1/2 x^(-1/2).

Since .9 is pretty close to 1, and we can find f'(1)=1/2, we know that the slope of the line passing through (1,1) and tangent to the curve of f(x) is 1/2. Then, use y=mx+b to find the equation of the tangent line. we know m=1/2, and (1,1) is a point on the line. Plug this in, and find b=1/2. So, y=1/2x+1/2. Let x=.9, and we get y=.9/2+1/2, which is .95. That's the linear approximation.

Does anyone know how to input formulas in these forums? I'm new.

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[MATH]\sqrt{0.9}=\sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}[/MATH], which is irrational, so you have to stick with the approximations.

An elementary method is to start with some intervall [MATH]a < \sqrt{0.9} < b[/MATH], and then to decrease the width of the intervall by picking [MATH]c, a < c < b[/MATH] and checking whether [MATH]c^2 < 0.9[/MATH] or [MATH]c^2 > 0.9[/MATH]. In either case, we have reduced the intervall to either [MATH]a < \sqrt{0.9} < c[/MATH] or [MATH] c < \sqrt{0.9} < b[/MATH]. Then you pick another number [MATH]d[/MATH] and do it all over again to make your new intervall even smaller. The method is very easy to understand, although there are stronger methods out there.

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• 2 weeks later...

Hey this one I found little easier to find in case u have access to logarithmic tables.

x^2 = 0.9

2log(x)= log(0.9)

log x = 1/2 log(9) - 1

log x = 1/2 (0.95424 -1)

x = 0.948683

In case you have a better method and quicker + accurate then please reply soon.

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• 4 months later...

The best way is to use binomial expansion .

EXpand (x+1)^0.5 untill u get the term aX^0.5

to get the most accurate 10 dp answer ..

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cant you just do the SQR of 9 and adjust the decimal place after?

btw, I try the 0.9 V- = on my one calc, I get 0.9486832 !?????

surely THAT cant be right can it?

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It is correct. Why do you think it cannot be? Because the sqr root is bigger than the number?

That is correct though because these are fractions. Think about it, 0.5² is 0.25, so the sqr root of 0.25 is 0.5.

Think of it this way:

$\sqrt{0.1} = \sqrt{\frac{1}{10}} = \frac{\sqrt{1}}{\sqrt{10}} = \frac{1}{\sqrt{10}}$

And we know the $\sqrt{10}$ roughly equals 3 so $\frac{1}{\sqrt{10}} = \frac{1}{3}$ so $\sqrt{0.1} > 0.1$

The same works for all not-top-heavy fractions (or decimals) because the denominator (bottom part) is bigger than the numerator (top part) so reduces by a larger percentage than the numerator so the overall fraction becomes smaller, (as the denominator becomes bigger).

=====

Back to the original question I think that dave's binomial expansion method is the best.

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cant you just do the SQR of 9 and adjust the decimal place after?

btw' date=' I try the 0.9 V- = on my one calc, I get 0.9486832 !?????

surely THAT cant be right can it?[/quote']

It's right for the reason to which Algebracus alluded - there's a $\sqrt{10}$ involved. To "adjust the decimal," you need to go in jumps of 100, for which the square root is 10, i.e. it would work for the square root of .09

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• 2 years later...

I am 80 years old and for the last year have not enjoyed very good health. I am still interested in all things scientific but I am not aiming to take over where Einstien left off.

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• 2 weeks later...

I can show you how to calculate square roots by hand if you are interested.

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wouldnt it be 0.3? {as the square root of nine}

It's right for the reason to which Algebracus alluded - there's a involved. To "adjust the decimal," you need to go in jumps of 100, for which the square root is 10, i.e. it would work for the square root of .09

wouldnt that mean the square root of 10 is 1?

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wouldnt it be 0.3? {as the square root of nine}

wouldnt that mean the square root of 10 is 1?

This thread has been dead for three years, so the original participants aren’t going to answer you.

But no, the square root of 0.9 is not 0.3. It is $\frac{3}{sqrt(10)}$.

And no, the square root of 10 is not 1. 1^2 = 1. The square root of ten is roughly 3.16.

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wouldnt it be 0.3? {as the square root of nine}

You should be able to answer this yourself. if x is the square-root of y, then by definition, x*x = y.

So, what does 0.3*0.3 equal?

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.09

as for the 3 year old thing, i just saw it at the top so i gave it a try.

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I was trying to figure out how to show you the square root method, but it is easy to diagram and a little tricky for me to explain so I think I'll write up a web page and send in the link.

But it is possible to directly calculate square roots one digit at a time without using a method that uses refinement.

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Why calculate the square one digit at a time when Newton's method more than doubles the number of digits each step? Newton's method to compute the square root of some number a is a iterative algorithm,

$x_{n+1} = \frac 1 2 \left(\frac a {x_n} + x_n\right)$

You need an initial guess, and it does not need to be particularly good. (Newton's method will converge for any initial guess.)

For example, $\surd 0.9=0.948683298505138154$.

1. x1=0.95. The square root of 0.9 is obviously somewhere between 0.9 and 1, so a good place to start is the average of these two. This happens to be a very good initial guess. Let $e_n=-\log_10\left(\left|1-x_n/\surd 0.9\right|\right)$, which is essentially the number of significant digits in the nth[/su] estimate. So, right off the bat, the initial guess is good to 2.86 decimal places.
2. x2=(0.9/0.95+0.95)/2=0.948684210526315769. This next step more than doubles the precision, as e2=6.02.
3. x3=(0.9/0.948684210526315769+0.948684210526315769)/2=0.9486832980509526865. The precision metric is now 12.33.
4. x4=(0.9/0.9486832980509526865 + 0.9486832980509526865)/2=0.948683298505138154. This nails the result. I would have to use an extended precision calculator beyond this point.

Starting with a semi-reasonable initial guess of 0.9 or 1.0 merely adds one step (both give 0.95 as the next guess). Suppose we start with a lousy initial estimate.

1. x1=0.3. Oops. The error metric is 0.17, not even one significant digit.
2. x2=(0.9/0.3+0.3)/2=1.65, with an error metric of 0.13. In terms of the error metric, this is a step away from the solution.
3. x3=(0.9/1.65+1.65)/2=1.09772727. The error metric is now 0.80; we almost have one significant digit.
4. x4=(0.9/1.09772727+1.09772727)/2=0.958801524. The error metric is now 1.97, so we are once again in the regime of more than doubling the precision with each step.
5. x5=(0.9/0.958801524 + 0.958801524)=0.948736687, for which e5=4.25
6. And so on.

It took quite a few steps to get to the stage where quadratic convergence kicked in, but once there the technique zeros in on the answer.

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The question is cost of the computation. The original question is how to do this without a calculator. The method I propose generates the correct next digit at each step. If you need 10 digits of precision, then 10 steps are required. Each step is simple and direct.

You used a calculator. The question says "without a calculator". For hand calculations my method is likely to be simpler - less prone to errors.

Merged post follows:

Consecutive posts merged

Isn't it nice when someone else has already posted the information. Here is the method I was going to post and it is already done:

No calculator needed. No divisions.

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cant you just do the SQR of 9 and adjust the decimal place after?

btw, I try the 0.9 V- = on my one calc, I get 0.9486832 !?????

surely THAT cant be right can it?

LOL!!!! No YD you always amaze me with your funny 'answers' NO, you can not JUST take root 9 and adjust decimals (see earlier replies .. sorry didn't read all the way down).

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