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This is a variation from my personal favorite 'A Concise Introduction to Pure Mathematics' by M. Liebeck (Chapter 3, Exercise 4).

Without using a calculator, find the cube root of 2, correct to two decimal places. This is how I work. (Only odd numbered exercises have either hints or solutions.)

Let $\sqrt[3]{2}=a_0{.}a_1a_2a_3\ldots$. First, since $1^3=1$ and $2^3=8$, $\sqrt[3]{2}$ lies between 1 and 2, and hence $a_0=1$. Next, $1{.}2^3=1{.}728$ and $1{.3}^3=2{.}197$, so $a_1=2$. Likewise, $1{.}25^3\approx 1{.}953$ and $1{.3}^3>2$, so $a_2=5$. Hence, a decimal representation of $\sqrt[3]{2}$ is $1.25$.

I am sure that there are better ways than above method always in the confines of number theory (no root finding algorithms). Thanks a lot for any help!

Bill Dubuque
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Dimitris
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  • Sure, there are multiple methods you can use, google “root finding algorithm” for a more general way. There are so many ways that this question is too broad to be answered. – Vivaan Daga Feb 24 '23 at 11:45
  • What you have found is the first three digits of the non-terminating decimal representation, which is different from what is normally meant by "correct to two decimal places." Have you been given a definition of what it means for an approximation to be correct to $n$ decimal places? – David K Feb 26 '23 at 21:53
  • how about newtons method or variants ? – mick Feb 26 '23 at 22:17
  • Or you could try the mediant methods – mick Feb 26 '23 at 22:18
  • You're looking for an algorithmic way to find a root of $x^3-2$, without using a root-finding algorithm. Hmmm. – Gerry Myerson Feb 26 '23 at 23:07

1 Answers1

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We have (5/4)^3 = 125 / 64 slightly smaller than 2. So the solution is a bit larger than 5/4.

(5/4 + d)^3 = 125/64 + 3 * 25/16 * d + 3 * 5/4 * d^2 + d^3. Ignoring the last two terms, 3 * 25/16 * d = 3 / 64, so d = 1/100. The cube root of 2 is 1.26. Actual value is about 1.2599.

gnasher729
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