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Multiplication can be thought of as repeated addition, where we add something up a certain number of times.

Division can be thought of as repeated subtraction, where we subtract something from the dividend a certain number of times till we get zero or a remainder.

Exponents can be thought of as repeated multiplication.

Roots... Can roots be thought of as repeated division? I don't see how. Can it be thought of as a repetition of any arithmetic function? I don't see it.

E.g., $\sqrt[3]{5} = 5 \div 3 \div 3 \div 3 ...$ ?? No, doesn't work.

Sorry if this is one of those abstract philosophical questions about math, but it's very curious to me.

The question is, can roots be thought of as a repetition of any arithmetic operation?

DrZ214
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  • Yes. Indeed, back in the day before the calculator era, schoolchildren had to be taught root-extraction algorithms, and I seem to recall one that had to do with repeated division, although I can't recall more than that now. You may look up root extraction algorithm by repeated division, or something similar. – Allawonder Apr 24 '19 at 06:43
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    In fact it is logarithms that can be thought of as repeated division: $\log_2 8 = 3$ because $8 \underbrace{{} \div 2 \div 2 \div 2}_{\text{3 times}} = 1$. –  Apr 24 '19 at 06:45
  • @Rahul Yes but something like Log(500) doesn't work because 500 / 10 / 10 with remainder 5 doesn't seem to translate to 2.698970004... – DrZ214 May 05 '19 at 11:40
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    By that logic $8 \div 3$ doesn't work as repeated subtraction either, but in your question that didn't seem to bother you. –  May 05 '19 at 12:18
  • @Rahul Remainders don't bother me, I just don't see how a remainder of 5 somehow gets the decimal part (.69897...) – DrZ214 May 05 '19 at 12:35
  • $2\div3=0.666{\dots}$ so $8\div3=2.666{\dots}$. On the other hand, $\log_{10}5=0.698{\dots}$ so $\log_{10}500=2.698{\dots}$ –  May 05 '19 at 13:06
  • @Rahul Ah I see now. But doing Log(5) = 0.698... Can that process be done with repeated division (or repeating any of the 4 basic ops)? BTW I see your point about 8 / 3 now, but you can still do that process with repeated operations by doing 3 into 8.000... with long division (which uses repeated subtraction, but also has repeated multiplication too). I guess what I'm asking for is a log extraction method, which I suspect is a root extraction method too. I might just have to ask a separate q about it. – DrZ214 May 06 '19 at 11:12

2 Answers2

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Elementary arithmetic consists of the operations $+$, $-$, $\times$, $\div$ and you are correct that you can pair them up so that one operation does the opposite to the other; that is, the pair $+,-$, and the pair $\times, \div$. You can also pair up the analogues, namely $+,\times$, and $-,\div$. This is most apparent when only integers are concerned, since $m\times n=m+m+\cdots+m$ ($n$ times) $=n+n+\cdots+n$ ($m$ times), and $m\div n=m-n-n-\cdots-n$ until we arrive at the minimum non-negative integer.

But $-,\div$ are not commutative, unlike $+,\times$, since $m\div n\ne n\div m$ in general. Repeated multiplication in exponentiation makes sense as $m^a=m\times m\times \cdots\times m$ ($a$ times) and no matter how to bracket each operand, you will get the same result. One difficulty in repeated division is due to its non-commutativity: the expression $a\div b\div c$ is ambiguous without the use of brackets, and this is before we get to radicals.

You can think of roots as the opposite of exponentiation; that is, for what value of $m$ is it true that $m^a=n$? That is, in an integer sense, it is the value of $m$ such that if you multiply itself $a$ times you get $n$. It is impossible to divide anything in $$m\times m\times\cdots\times m=n$$ to find $m$! However, just because exponentiation and radicals have the opposite effect does not mean that they can be defined as the repeated usage of the reverse of the intrinsic operations ($\times$ and $\div$). Otherwise, you can even go further, since for integers, $\times$ is intrinsically linked to addition, surely roots can be done using subtraction?

Now you ask whether radicals can be written in terms of elementary arithmetic operations only. Consider the irrational number $\sqrt2$ which satisfies the equation $m^2=2$. Clearly it cannot consist of a finite combination of such operations as this would imply that $\sqrt2$ is rational. However, it is possible to express it as an infinite series, consisting of the operations used infinitely many times: $$\sqrt2=1+\frac{4}{10}+\frac{1}{10\times 10}+\frac{4}{10\times 10\times 10}+\cdots$$ and can also be expressed through continued fractions, which we will not go into detail here. And it is by this means that we can obtain approximations, as we can never reach every decimal digit of an irrational number.

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yes and no. Yes some roots are as simple as guess,divide by guess, arithmetic mean ( division by 2 for 2 numbers) taken as new guess, repeat. literally repeated division after initial guess. This is called the Babylonian method not all roots are that easy though.