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I did not even remember how to do square roots from high school, but I vaguely recall it was similar to long division. (Thankfully I remember how to do long division.) I just went to youtube and refreshed my memory on how to do square roots.

Is there a "long root" process to do nth roots? N can be any integer of 2 or more. (N could be 1 too but that is trivial.) The radicand can be any positive number. The radicand does not hafta be a perfect square, perfect cube, or perfect n-power of anything.

Btw, I am aware of factorization. So $\sqrt{153} = \sqrt{9 * 17} = 3\sqrt{17}$. In this case, what I want to do is something like 2 into 17, but instead of long division, use a "long root" process for putting 2 into 17. The process would go on forever, much like 25/7 goes on forever because the remainder never "settles". I would just stop when I get, say, 3 decimal places or however many I think is accurate enough.

Examples:

  • $\sqrt{68}$

  • $\sqrt[3]{401}$

  • $\sqrt[7]{50}$

  • $\sqrt[21]{675}$

  • $\sqrt[n]{x}$

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    Perhaps you mean something like [here](http://www.homeschoolmath.net/teaching/square-root-algorithm.php)? – 2017-02-25
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    @S.C.B. Yes but that appears to be a method only for square roots. I want n roots. So, say, $\sqrt[5]{167}$. – 2017-02-25

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In a nutshell, the algorithm for square root works because of the identity: $$ y =(a+x)^2 = a^2 + 2ax + x^2 = a^2 + (2a+x) x $$ In the algorithm you iteratively determine increasing (adding one decimal at the time) values for $a$ by looking for the largest value of $x'$ at the appropriate decimal place so that $$y - a^2 \geq (2a+x') x'$$ You then essentially replace $a$ by $a+x'$ and continue for the next decimal place.

I suppose you could do this for e.g. cube roots by looking at $$ y = (a+x)^3 = a^3 + (3 a^2+3x+x^2) x $$ but it looks rather nasty to carry out in practice.

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    Does x' mean the derivative of x with respect to x? Or does x' mean x to some power? – 2017-02-25
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    No I just wrote $x'$ to distinguish it from the exact value of $x$ for which you have equality. The description I give is like e.g. http://www.homeschoolmath.net/teaching/square-root-algorithm.php The method is very similar to a Newton iteration (in which you solve $2ax = y-a^2$, so without the term $x^2$). The Newton method is probably preferable for many reasons and works easily for any $n$'th root. – 2017-02-25
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I'm having trouble understanding exactly what you are asking about but I'll focus on the following "The process would go on forever, much like 25/7 goes on forever because the remainder never "settles". I would just stop when I get, say, 3 decimal places or however many I think is accurate enough."

You might want to grab a book on introductory real analysis. R. P. Burn's Numbers and Functions has very good chapters on sequences and completeness that will satisfy your needs.

For example, for any $a$, $0\leq a-\frac{\lfloor a10^n \rfloor}{10^n}<\frac{1}{10^n}$ ($\lfloor x \rfloor$ here is called floor $x$, which is the greatest integer less than $x$). From this we can deduce that for any number $a$ there is a sequence of rational numbers which tends to it. This is because the sequence $(1/10^n)$ tends to zero and so by the squeeze rule for null sequences, the sequence $(\frac{\lfloor a10^n \rfloor}{10^n}-a)$ tends to zero, so by the definition of limit we see how the result is proved. Remember that $\frac{\lfloor a10^n \rfloor}{10^n}$ is always rational.

The chapter on completeness will go over how every nth root of a number exists and is unique.

Hope that points you in the right direction.

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    The process of long division: pick a number, write it on the top, multiply that number by the divisor, write the product under the dividend, subtract, bring down the next number. I left out some special things like make sure the product is as close to the goal without going above it, sometimes you hafta write a 0, and other stuff. The process works with any 2 numbers, including divisors with 2 or more digits. **But now I want a process for roots.** I want to know the process for the square root, cube root, 4th root, 5th root, and in general, n root, even multi-digit roots like 21, 150, etc. – 2017-02-25
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    The thing you wrote seems to prove that a real root exists for any n and radicand that are real numbers. Just knowing it exists does not help me find a way to extract the root, at least, not that I can see. – 2017-02-25
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    @DrZ214, what I wrote does not prove real root exists for any number. It is a sketch for a proof that shows every real number is the limit of a sequence of rational numbers with terminating decimals. I apologise if I have misunderstood your question. Maybe the definition of $nth$ root will help you? for example $3.3=3^2$ so $\sqrt {3^2}=3$, $7.7.7=7^3$ so $\sqrt[3]{7^3}=7$. – 2017-02-25
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    I want any roots of any number, not just perfect powers. $\sqrt{7} = 2.645...$ according to calculator, but what process do you apply to get that? $\sqrt[3]{11} = 2.223...$, but what process do you apply? – 2017-02-25
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    You might find this helpful http://math.stackexchange.com/questions/1102865/calculating-the-nth-root-in-simple-calculator – 2017-02-25