8
$\begingroup$

This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and compute the nominator with the value of square root 2 and 3 memorized. And as to $e\cdot 2^{1/3}$, I just don't see how to compare it to other two items without take cubic and compute. This whole process is very time consuming.

I have seen some tricks to compare say $2^\pi,\pi^2$. But the technique does not seem to apply here.

$\endgroup$
  • 1
    $\begingroup$ I remember $\pi^2\approx10$ and $e^3\approx20$; thus $\pi^6\approx1000$, whereas $(e\cdot2^{1/3})^6\approx1600$, which implies $\pi<e\cdot2^{1/3}$ $\endgroup$ – J. W. Tanner Aug 26 '19 at 2:40
  • $\begingroup$ In fact, we can establish the inequality without relying on the approximations being good enough if we know $\pi^2 < 10$ and $e^3 = 20$. (This inequality turns out to be the unhelpful one, though, since the third quantity is the one in between these two.) $\endgroup$ – Travis Willse Aug 26 '19 at 2:42
  • 1
    $\begingroup$ Notice that $e=2.7128...>2.71$ and ${1.25}^3=\frac{125}{64}<2 \rightarrow 2^{1/3}>1.25$, so $e \times 2^{1/3}> 2.71 \times 1.25 >3.2 >\pi$ $\endgroup$ – Isaac YIU Math Studio Aug 26 '19 at 2:50
  • $\begingroup$ @Travis: Good point. Of course you meant $e^3>20$ $\endgroup$ – J. W. Tanner Aug 26 '19 at 2:54
  • $\begingroup$ @J.W.Tanner Oops, yes, that's what I meant. $\endgroup$ – Travis Willse Aug 26 '19 at 2:55
8
$\begingroup$

Here's a dirty decimal arithmetic method that presumes knowledge only of the bounds $1.41 < \sqrt{2} < 1.42$, $1.73 < \sqrt{3} < 1.74$---which you probably know if you're taking the GRE subject test---and the not-too-obscure fact $e^3 > 20$: Since $\sqrt 6 = \sqrt 2 \sqrt 3$ multiplying gives $2.43 < \sqrt{6} < 2.47$ Then, using the rationalization $$\frac{1 + \sqrt 2}{\sqrt 3 - 1} = \frac{1}{2} (1 + \sqrt 2 + \sqrt 3 + \sqrt 6) ,$$ and substituting the decimal values gives $$\pi < 3.29 < \frac{1 + \sqrt 2}{\sqrt 3 - 1} < 3.32 .$$ Now, $3.32 < \frac{10}{3}$, so $$\left(\frac{1 + \sqrt 2}{\sqrt 3 - 1}\right)^3 < \left(\frac{10}{3}\right)^3 < 40 = 2 \cdot 20 < (\sqrt[3]{2} e)^3,$$ establishing

$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$

Alternatively, here's a version that uses only estimates using fractions with small denominators (which themselves follow from the decimal bounds above): Since $$\frac{7}{5} < \sqrt{2} < \frac{10}{7} \qquad \textrm{and} \qquad \frac{12}{7} < \sqrt{3} < \frac{7}{4} ,$$ we have $$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} > \frac{1 + \frac{7}{5}}{\frac{7}{4} - 1} = \frac{16}{5}.$$ (Of course we can verify the bounds on $\sqrt{2}, \sqrt{3}$ without knowing anything about the numbers' decimal representations---just square all of the numbers, which reduces the problem to comparing rational numbers.) This is $3.2 > \pi$, but we can avoid decimal representations using $\frac{16}{5} > \frac{22}{7} > \pi$.

On the other hand, $$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \frac{1 + \frac{10}{7}}{\frac{12}{7} - 1} = \frac{17}{5} .$$ Since $e^3 > 20$, we have $(\sqrt[3]{2} e)^3 > 40$, but $$\left(\frac{17}{5}\right)^3 < 40,$$ giving the order $$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$

See this this follow-up question that discusses methods for deriving the inequality $e^3 > 20$ by hand.

$\endgroup$
  • $\begingroup$ Just do 2.718 cubed. It’s not that hard by hand. $\endgroup$ – Gabe Aug 26 '19 at 14:56
  • $\begingroup$ The bound is very tight---the relative difference is $< 0.5\%$---so I'm not sure how efficient one can be. Naively, $e = \sum_{k=0}^\infty \frac{1}{k!} > \sum_{k=0}^5 = \frac{163}{60}$, and so we need only show $20 < \left(\frac{163}{60}\right)$, which is equivalent to $4\,320\,000 < 4\,330\,747$. $\endgroup$ – Travis Willse Aug 26 '19 at 15:15
  • $\begingroup$ @TheSimpliFire I was surprised that hadn't been asked on this site before, so I wrote it up as a question, giving two elementary (but not terribly enjoyable) methods, including the one in my previous comment: math.stackexchange.com/questions/3335059/show-that-e3-20 . $\endgroup$ – Travis Willse Aug 26 '19 at 17:40
  • $\begingroup$ (The quantity in parentheses in my previous answer should be raised to the third power.) $\endgroup$ – Travis Willse Sep 21 '19 at 5:09
0
$\begingroup$

If you raise them to the third power you get

$\pi^3 \approx (3+\frac 17)^3 \approx 3^3 + 3*3^2*\frac 17 \approx 27*(1\frac 17)$

$e^3*2 \approx 2(3-0.29)^3 \approx 2(3^3 - 3*3^2*0.29)\approx 27*2*(0.71)\approx 27*1.42$ so $e*2^{\frac 13} > \pi$. (And as a weird unexpected bonus I get that $e^3*2 \approx 27*\sqrt 2$.)

$(\frac {1 +\sqrt 2}{\sqrt 3-1})^2 =\frac {3+2\sqrt 2}{4- 2\sqrt 3}$ and

$(\frac {1 +\sqrt 2}{\sqrt 3-1})^3=\frac {3+2\sqrt 2}{4- 2\sqrt 3}\frac {1 +\sqrt 2}{\sqrt 3-1}=\frac {3+4+5\sqrt 2}{-6-4+6\sqrt 3}$

$=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}\frac {6\sqrt 3 + 10}{6\sqrt 3 + 10}$

$=\frac{30\sqrt 6 +42*\sqrt 3 + 50\sqrt 2 +70}{36*3-100}$

$\approx \frac 18(30*1.7*1.4 + 42*1.7 + 50*1.4 + 70)\approx$

$\frac 18(51*1.4 + 7*10.2 + 70 + 70)\approx$

$\frac 18(70*4+1.4+1.4) \approx \frac {70.7}2 \approx 35.35$.

$\approx 27 + 8 \approx 27(1\frac 14)$

And $1\frac 17 < 1\frac 14 < 1.42$ so

So $\pi < \frac {1 +\sqrt 2}{\sqrt 3-1} < e*2^{\frac 13}$

.... which.... was really way too much work for too little result but.... well, it's one way to do it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.