# Is it possible to find square root using only rational numbers and elementary arithmetic operators

Suppose I have a number a

How can I find it's square root using only +, -, /, * and rational numbers?

If it is impossible how to prove it?

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Any number obtained from $+,-,\times$ and $/$ and rational numbers is rational, but some squareroots, like $\sqrt 2$, are irrational. – Pedro Tamaroff Jul 17 '14 at 13:02
Is your number $a$ rational? Are you allowing an infinite number of these operations? – user37238 Jul 17 '14 at 13:03

The set of rational numbers is closed under the elementary arithmetic operations (except for division by zero), i.e. if you have two rational numbers and take their sum, difference, product, or quotient, then the result is again a rational number (again, except for division by zero). Since $\sqrt{a}$ can fail to be a rational number (say, $\sqrt{2}$), there is no formula for $\sqrt{a}$ involving only elementary arithmetic operations.

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Thanks! Could you also provide prove of closed under the elementary arithmetic operation. Or share link with prove – c0rp Jul 17 '14 at 13:20
As I said, it just means that the sum, difference, product, or quotient of two rational numbers is again a rational number. I'm sure you can find a proof of this yourself. – Zhen Lin Jul 17 '14 at 13:36
I think the question is interesting if $a$ is assumed to be a perfect (rational) square, i.e. of the form $\frac{p^2}{q^2}$. – R.. Jul 17 '14 at 16:25
– whatever Jul 17 '14 at 21:11
@metacompactness thanks! – c0rp Jul 18 '14 at 4:00

If you allow infinite number of operations, then you can use some algorithm.

One easy example is root searching via Newton's method. Here we do the iteration $$x_{n+1} = \frac{a + x_n^2}{2x_n},$$ which eventually converges to $\sqrt{a}$ if $a$ and $x_0$ are positive.

See https://en.wikipedia.org/wiki/Methods_of_computing_square_roots for other methods.

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In particular, the method that resembles long division requires only the basic operations of +, -, *, and /, and can be used to generate a result with as much precision as the OP desires. Also, it is extensible to 3rd, 4th, etc. roots. en.wikipedia.org/wiki/… – shoover Jul 17 '14 at 16:36

When you use an infinite ammount of operations, you can use:

$$\!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.$$ from here...

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Cool (and all quadratic irrationals have a periodic continued fraction), but the question was a general number $a$, so you need to use some algorithm. – J. J. Jul 17 '14 at 13:10
@J.J. you're right and Google is my friend: Methods of computing square roots... – draks ... Jul 17 '14 at 13:13
Right, but I guess $a$ is an arbitrary real number, not a quadratic irrational. – J. J. Jul 17 '14 at 13:16
It is interesting, in case of infinity, found this. Sum of infinite rational numbers produce irrational – c0rp Jul 17 '14 at 13:42