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Recently, the following question was asked: Without calculator, find out what is larger: $60^\frac{1}{3}$ or $2+7^\frac{1}{3}$. (Apologies; I don't know how to link to that question, but it is not essential for the question I am asking.)

Most people would not be able to extract cube roots without a calculator, unless the numbers were particularly easy, such as $64^\frac{1}{3}$ or $2+8^\frac{1}{3}$. But not using a calculator does not rule out doing some calculation.

As it turns out, the numbers in this case lend themselves to reasonably calculable approximations, which many people could perform in their heads, but might prove daunting to less experienced individuals.

So my question is, are the calculations I made reasonably within the intention of the restriction "without calculator?"

Please consider the difficulty of the calculations, not the total amount of calculation performed. Here is what I did, emphasizing the arithmetic calculation aspects: I cubed both quantities, leaving me to compare $60$ with $8+12(7^\frac{1}{3})+6(7^\frac{2}{3})+7$.

Collecting terms and rearranging, the original question becomes one about a quadratic equation:

Is $$x^2+2x-7.5$$ greater or less than $0$ when $x=7^\frac{1}{3}$?

This in turn becomes: Is $7^\frac{1}{3}$ greater or less than $r$, the positive root of $$x^2+2x-7.5=0$$.

By the quadratic formula $$r=(-2+\sqrt{4+30})/2$$.

Although the square root of $34$ may look like one of those calculations that would require a calculator, it turns out that determining a precise value would just make subsequent calculations dependent on a calculator as well.

By good fortune (or the cleverness of the original poser of the question), $34$ is close to $36$, so we may approximate $\sqrt{34}$ as $(6-a)$.

Thus we look for $$34=36-12a+a^2$$.

But since $a$ will be small compared to $6$, we can approximate by ignoring the $a^2$ term and calculate $a=\frac{1}{6}$. it is easy to see that $(6-\frac{1}{6})^2$ exceeds $34$ by $\frac{1}{36}$. Again, by seeming good fortune, the next reasonable fraction greater than $\frac{1}{6}$ is $\frac{17}{100}$.

$$(6-\frac{17}{100})^2$$ is also calculable as $$36-\frac{204}{100}+\frac{289}{10000}$$. Since the second term decrements $36$ by $2.04$ and the third term only restores $0.0289$, we see that $(6-\frac{17}{100})^2$$ is less than $34$. So $$(6-\frac{1}{6})>\sqrt{34}>(6-\frac{17}{100})$$, hence $$(2-\frac{1}{12})>r>(2-\frac{17}{200})$$.

What remains is to cube the numerical values bracketing $r$ and compare the results to $7$.

$$(2-\frac{1}{12})^3=8-1+\frac{1}{24}-\frac{1}{1728}$$ which is greater than $7$ by observation.

$$(2-\frac{17}{200})^3=8-\frac{204}{200}+6(\frac{289}{40000})-(\frac{17}{200})^3=8-\frac{204}{200}+(\frac{1734}{40000})-(\frac{17}{200})^3$$.

The arithmetic is a little harder here, but the first and second terms are less than $7$ by $0.02$ and the third term is reasonably seen to be greater than $0.04$, making the sum of the first three terms greater than $7$ by at least $0.02$. The last term is certainly smaller than $(\frac{20}{200})^3$ which is $0.001$, so the sum of the terms is greater than 7.

This means that $r^3>7$ or $$r>7^\frac{1}{3}$$. From this, the original question can be answered. In performing calculations, no roots were extracted, but binomial expressions up to cubes involving fractions were calculated. I personally found the numbers in the numerators and denominators tractable, but would this be considered by the community as being in the spirit of "without calculator?"

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    $\begingroup$ Consider using linebreaks to improve readability. $\endgroup$ – JMoravitz Jul 10 '18 at 19:27
  • $\begingroup$ I've voted to close this question, as I think that potential answers are a matter of opinion. I'm also not convinced that this is entirely on-topic (couched in the right way, it might be more reasonable for the MathEd SE, but the focus of the question would need to change a bit, I think). Plus, what @JMoravitz said. $\endgroup$ – Xander Henderson Jul 10 '18 at 19:30
  • $\begingroup$ Since this is entirely subjective and the context was not explicitly stated I'm going to give my answer before this questions is closed.. no it is not reasonable in general $\endgroup$ – Attack68 Jul 10 '18 at 19:34
  • $\begingroup$ No calculator or no paper? The question is okay for no calculator but not for no paper. $\endgroup$ – fleablood Jul 10 '18 at 19:42
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    $\begingroup$ What you state can not be done in ones head. But $f(x)=x^{\frac 13}$ then $f'(x) = \frac 1{3x^{\frac 23}}$ so $f(x) = f(x_0)+*f'(x_0)(x-x_0)$ so $60^{\frac 13}\approx 4-4*\frac 1{3*16}$ and $2 + f(x)\approx 4-\frac 13$. So $60^{\frac 13}> 2 + 7^{\frac 13}$ isn't. Although I don't trust the accuracy that the margin of error isn't greater than the result. $\endgroup$ – fleablood Jul 10 '18 at 19:56

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