# Determine a fractional square root without calculator

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• vcsharp2003
In summary, the conversation discusses the best methods for computing square roots without a calculator, with the option of using the binomial theorem as a quick approximation. The suggested method involves using the binomial theorem to expand the expression and taking the first two terms for a more accurate approximation. However, it is important to remember that the answer must always be greater than 1. The
vcsharp2003
TL;DR Summary
Solving a numerical problem without a calculator
I have to solve a certain numerical problem without using calculator and furthermore, there is a time limit for solving this problem.

The answer I have got so far is ## \sqrt{\frac{100}{99}}##

I know I can reduce the numerator to 10 but then I am stuck with square root of denominator which is not a perfect square.

Question: How would I determine in shortest possible time the above square root without any calculator or tables?

jedishrfu said:
There are several choices outlined here:

https://en.wikipedia.org/wiki/Methods_of_computing_square_roots
Is it possible to use Binomial theorem to do a quick approximation of the fractional square root? This might be better than finding square root of denominator since then I will need to do two approximations, first for denominator square root and then find decimal equivalent for fraction. Binomial theorem might involve a single operation.

vcsharp2003 said:
Is it possible to use Binomial theorem to do a quick approximation of the fractional square root?
How could that not be possible?

PS note that ##(\frac{100}{99})^{1/2} = (\frac{99}{100})^{-1/2}##

docnet and vcsharp2003
PeroK said:
How could that not be possible?
That might be easier. I will try expanding the binomial expression on RHS in equation below and maybe take only the first two terms.

$$\sqrt{\frac{100}{99}} = {(1 +\frac{1}{99})}^{\frac{1}{2}}$$

vcsharp2003 said:
That might be easier. I will try expanding the binomial expression below and maybe take only the first two terms.

$${(1 +\frac{1}{99})}^{\frac{1}{2}}$$
Okay, but see above for a useful idea.

docnet
PeroK said:
Okay, but see above for a useful idea.
I saw your last post. Once I get ##(\frac{99}{100})^{-1/2}##, then how would I proceed?

vcsharp2003 said:
I saw your last post. Once I get ##(\frac{99}{100})^{-1/2}##, then how would I proceed?
Use the binomial theorem. This time you have ##(1 - 0.01)^{-1/2}##

docnet and vcsharp2003
Of course, you could always approximate ##\frac 1 {99} \approx 0.01##, but inverting the fraction feels neater to me.

docnet and vcsharp2003
PeroK said:
Of course, you could always approximate ##\frac 1 {99} \approx 0.01##, but inverting the fraction feels neater to me.
Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of ##\frac{1}{99}##. After taking only first two terms of binomial expansion I get an approximation of 1.005.

$$\sqrt{\frac{100}{99}} = (1 - 0.01)^{-1/2} \approx 1.005$$

First two terms of binomial expression are obtained as below.
$$(1 - 0.01)^{-1/2} = 1^{-\frac{1}{2}} + (-\frac{1}{2}) 1^ {(-\frac{1}{2} - 1)} (-.01)^1 + \frac{(-\frac{1}{2}) (-\frac{1}{2} -1)}{(1)(2)} 1^ {(-\frac{1}{2} - 2)} (-.01)^2 + ...$$

Taking the first two terms in binomial expression since the terms starting from third term are going to very small ( even smaller than .0001) we get ##1+.005 = 1.005## which seems correct. My initial approximation of .995 was not correct.

Last edited:
vcsharp2003 said:
Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of 199. After taking only first two terms of binomial expansion I get an approximation of 0.995.

10099=(1−0.01)−1/2≈.995
That can't be right. The answer must be greater than ##1##.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$

docnet
PeroK said:
That can't be right. The answer must be greater than ##1##.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$
You are correct. I have edited my last post to reflect this. Thanks.

PeroK said:
That can't be right. The answer must be greater than ##1##.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$
Your formulas are awesome. I can memorize these formulas for quicker calculations and not even bother to use Binomial expansion. Thanks.

vcsharp2003 said:
Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of ##\frac{1}{99}##. After taking only first two terms of binomial expansion I get an approximation of 0.995.

$$\sqrt{\frac{100}{99}} = (1 - 0.01)^{-1/2} \approx .995$$

First two terms of binomial expression are obtained as below.
$$(1 - 0.01)^{-1/2} = 1^{-\frac{1}{2}} + (-\frac{1}{2}) 1^ {(-\frac{1}{2} - 1)} (-.01)^1 + \frac{(-\frac{1}{2}) (-\frac{1}{2} -1)}{(1)(2)} 1^ {(-\frac{1}{2} - 2)} (-.01)^2 + ...$$

Taking the first two terms in binomial expression since the terms starting from third term are going to very small ( even smaller than .0001) we get ##1+.005 = 1.005## which seems correct.
The usual binomial expansion has a ##1## in the first position. I would always take the factor out first. E.g.
$$(a + \epsilon)^{1/2} = a^{1/2}(1 + \frac{\epsilon}{a})^{1/2}$$ In any case, if you already have ##1## here then: $$(1 + \epsilon)^{1/2} = 1 + \frac 1 2 \epsilon + \frac{1}{2!}(\frac 1 2)(-\frac 1 2)\epsilon^2 + \frac{1}{3!}(\frac 1 2)(-\frac 1 2)(-\frac 3 2)\epsilon^3 \dots$$ Also, to avoid missing minus signs, I would tend to do: $$(1 - \epsilon)^{1/2} = (1 + (-\epsilon))^{1/2} = 1 + \frac 1 2 (-\epsilon) + \frac{1}{2!}(\frac 1 2)(-\frac 1 2)(-\epsilon)^2 + \frac{1}{3!}(\frac 1 2)(-\frac 1 2)(-\frac 3 2)(-\epsilon)^3 \dots$$

docnet and vcsharp2003
PeroK said:
That can't be right. The answer must be greater than ##1##.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$
I think these approximation formulas are only valid if ##\epsilon## is very small i.e. as close to 0 as possible on the real number line. Is that correct?

vcsharp2003 said:
I think these approximation formulas are only valid if ##\epsilon## is very small i.e. as close to 0 as possible on the real number line. Is that correct?
Yes, that's why I used ##\epsilon##. The series converges for any ##-1 < \epsilon < 1##, but it's only useful as an approximation for small ##\epsilon##.

docnet and vcsharp2003

## 1. How do I determine a fractional square root without a calculator?

To determine a fractional square root without a calculator, you can use the long division method or the Babylonian method. Both methods involve breaking down the number into smaller factors and using basic arithmetic operations to find the square root.

## 2. What is the long division method for determining a fractional square root?

The long division method involves dividing the number into smaller factors and finding the square root of each factor. Then, the square roots are multiplied together to get the final answer. This method can be time-consuming but is relatively easy to understand and perform.

## 3. How does the Babylonian method work for determining a fractional square root?

The Babylonian method is an iterative algorithm that involves repeatedly guessing and refining the estimate for the square root of a number. It is based on the fact that the square root of a number is the average of the number and its reciprocal. This method can be more efficient than the long division method but requires a bit more mathematical knowledge.

## 4. Can I use any number for the fractional square root?

Yes, you can use any number for the fractional square root. However, some numbers may have irrational square roots, which means the decimal representation of the square root will never end or repeat. In such cases, it is best to round the answer to a certain number of decimal places.

## 5. Is it necessary to determine a fractional square root without a calculator?

No, it is not necessary to determine a fractional square root without a calculator. With modern technology, we have access to calculators and other tools that can quickly and accurately find the square root of any number. However, understanding the concept and being able to perform the calculation manually can be a useful skill to have.

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