Is there a formula for the square root of a number, that only uses addition, subtraction, multiplication, or division?
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11$\begingroup$ If you allow infinite steps, then yes. Otherwise no, there is no exact equation for a square root. $\endgroup$– KaynexMar 9, 2016 at 1:19
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$\begingroup$ Kaynex may be referring to the Taylor series of $\sqrt{1 + x}$, which is addressed in this question. $\endgroup$– Austin MohrMar 9, 2016 at 1:23
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$\begingroup$ You can approximate certain square roots by evaluating their continued fractions and get precision to any number of finite decimal digits. $\endgroup$– AnotherPersonMar 9, 2016 at 1:23
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9$\begingroup$ If it were possible then $\sqrt2$ would be a rational number $\endgroup$– IvoMar 9, 2016 at 8:38
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6 Answers
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There is not. The numbers that can be formed from rational numbers using a finite sequence of $+,\times,-$, and $\div$ are rational numbers, but many square roots of rational numbers ($\sqrt{2}$ for example) are irrational.
However, with an infinite number of steps it is possible. See Methods of computing square roots. A popular one is the Babylonian method: $$ x_{n+1} = \frac{1}{2}\left( x_n + \frac{x_0}{x_n} \right) $$ Let $x_0$ be the number you want to find the root of. Iterating this formula results in a sequence $x_0, x_1, \dots$ which approaches $\sqrt{x_0}$.
answered Mar 9, 2016 at 1:22
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$\begingroup$ $@Daniel : $ x_1 = 1$ will work. $\endgroup$ Mar 9, 2016 at 20:18
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Back in the Olden Days, when I was in school, we learned an algorithm for square-root extraction similar to the long division algorithm. But that topic is long gone: there is no need to learn to do this by hand, since computers and calculators do it quite well.
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Here is the method.
Examples from Wikipedia:
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1$\begingroup$ I like the algorithm, especially in binary. en.wikipedia.org/wiki/… $\endgroup$ Mar 9, 2016 at 1:30
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4$\begingroup$ Could you pull in a summary of the relevant info from the link? $\endgroup$– jpmc26Mar 9, 2016 at 4:55
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$\begingroup$ I like this. I think I’m older than you, but this was never taught in my school. $\endgroup$– LubinMar 9, 2016 at 14:29
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$\begingroup$ This algorithm provides an approximation, you should address this in your answer $\endgroup$– seldonMar 9, 2016 at 18:26
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No, because square root is not a rational function.
answered Mar 9, 2016 at 1:21
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With those operations, no. But if you add the square root (which is ironic).. well I like this formula, which is a cute tool for mental square roots (a bit approximate but it may works for daily use).
Call $N$ the number you have to root and think about the nearest perfect square $s^2$. Then you can always write
$$N = s^2 \pm q$$
There is the approximation:
$$\sqrt{N} = \sqrt{s^2 \pm q} \approx s \pm \frac{q}{2s}$$
Example
Suppose we want to find an approximation of the square root of $87$ (which is $9.3274..$). Then the nearest perfect square could be $81$:
$$\sqrt{87} = \sqrt{81 + 6} \approx 9 + \frac{6}{18} = 9 + \frac{1}{3} = 9.3333333...$$
On the same way on could take $100$ as the nearest perfect square:
$$\sqrt{87} = \sqrt{100 - 13} \approx 10 - \frac{13}{20} = 10 - 0.65 = 9.35$$
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1$\begingroup$ Have you noticed in your example that both estimates are above the real root. This is always true. The estimate based on the highest square below $N$ is lower than the estimate based on the lowest square above, unless $N$ is $1$ less than a square. Therefore, if you are not $1$ below a square, you are always better off using the lower square estimate to the upper square estimate. $\endgroup$ Mar 9, 2016 at 5:29
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$\begingroup$ Also, having found a first estimate $n_1$ to the root, you can use it as the "nearest square" and apply the same algorithm again (high side this time, since $n_1 > \sqrt N$) to produce a better approximation $n_2$. And so on. Of course, this is just Newton's method. Not easy to do many iterations if you don't have a calculator around, and if you do have one, it'll probably have a square root key. So, maybe if you are desparate and the only working calculator in the vicinity is some cheapo with only the most basic operations.... $\endgroup$ Mar 9, 2016 at 5:34
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There is something called the fixed point theorem, which helps us find a way to calculate square roots by hand. Although it will take some time, we can do it.
The fixed point theorem says that if everything goes good, and a sequence that is recursively defined below converges:
$$f{(x_n)}=x_{n+1}$$
Then it's limit $L$ will be a solution to:
$$f(L)=L$$
So let us consider the equation:
$$L^2=y$$
In other words $L$ is square root of $y$. Then,
$$2L^2-L^2=y$$
$$2L^2=y+L^2$$
$$L=\frac{y+L^2}{2L}$$
So now it's in the form $L=f(L)$, and now we can use a sequence to find $L$.
$$x_{n+1}=\frac{y+{x_n}^2}{2x_n}$$
And,
$$L=\lim_{n \to \infty}x_n$$
Given a $x_0$ that allows the sequence to converge.
After simplification:
$$x_{n+1}=\frac{1}{2}\left(\frac{y+{x_n}^2}{x_n}\right)$$ $$x_{n+1}=\frac{1}{2}\left(\frac{y}{x_n}+x_n \right)$$
One may get this, which is the Babylonian method, as mentioned earlier. You may come up with many other methods using the fixed point theorem, just play around with the algebra and overcomplicate the equation as I did when I replaced $L^2$ by $2L^2-L^2$.
answered Mar 14, 2016 at 4:10
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There is an algorithm- for example , calculate the square root of 1212. Write 1212 as $1212.00$. Group digits on the left of the decimal in two's starting from right; and on the left in two's starting from left. Thus the number we have is $12- 12.00-00$ start with the first 2 digits-$12$. Closest square is 9. Write divisor as 3 and quotient as 3, and carry out long division. The new remainder is $12-9=3$. Bring down the next cluster of digits-$12$, so that the number in remainder is now $312$. In the divisor, add the units digit of the divisor to the divisor. Here. units place of divisor (i.e. $3$) is 3 itself, so the new divisor is $3+3=6$. Now think of a number which starts with 6 and when multiplied by its units place digit gives a number close to the remainder i.e. $312$. such a number is $64$, and $64*4=256$. Then the new remainder is $312-256=56$. Put the decimal in numerator and bring down the next cluster-00. SO the new remainder is 5600. Also, adding the units digit of divisor to itself $64+4=68$ is the new divisor. Repeat the process till any degree of precision. The quotient is the answer.
