Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$?

Question

Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?

To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.

But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?

Thank you.

Am I misunderstanding something here? $1-3\sqrt{2}<0$ and $3\sqrt{3}>0$. — Nick Peterson, Dec 14 '13 at 21:57
Now you just have to compare $8$ to $2\sqrt{18}$, which you can do by squaring. (Or observe that $8=2\sqrt{16}$.) — , Dec 14 '13 at 22:05

score 7 · Accepted Answer · answered Dec 14 '13 at 22:04

7

Hint:

$$\color{blue}8-\color{red}{2\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{4}\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{72}}$$

answered Dec 14 '13 at 22:04

@nolocimatt You're welcome! – Dec 14 '13 at 22:17

Asaf Karagila · Answer 2 · 2013-12-14T22:05:51.510

4

HINT:

Note that $(3\sqrt3)^2-(1+3\sqrt2)^2=27-1-6\sqrt2-18=8-6\sqrt2$. Next note that $\sqrt2>\frac86$.

edited Dec 14 '13 at 22:05

answered Dec 14 '13 at 21:57

Asaf Karagila

393,674

score 3 · Answer 3 · answered Dec 14 '13 at 22:03

3

Hint

$$8^2=64 <72=(2\sqrt{18})^2$$

answered Dec 14 '13 at 22:03

score 2 · Answer 4 · answered Dec 14 '13 at 22:21

You can use an “unknown relation”: denote by ? either $<$ or $>$; so long as you do operations that don't change the direction of inequalities, the symbol will always mean the same; so, when you're done you'll know which one it is. \begin{align} 1+3\sqrt{2}&\mathrel{?}3\sqrt{3}\\ (1+3\sqrt{2})^2&\mathrel{?}(3\sqrt{3})^2\\ 1+6\sqrt{2}+18&\mathrel{?}27\\ 6\sqrt{2}&\mathrel{?}27-1-18\\ 6\sqrt{2}&\mathrel{?}8\\ 3\sqrt{2}&\mathrel{?}4\\ (3\sqrt{2})^2&\mathrel{?}16\\ 18&\mathrel{?}16 \end{align} So ? was $>$.

score 1 · Answer 5 · edited Apr 13 '17 at 12:19

Alternatively $\rm\ \dfrac{2}{a-5} = \color{#c00}4+\color{#0a0}{3\sqrt2}\, <\, \color{#c00}5+\color{#0a0}{3\sqrt3} = \dfrac{2}{b-5}\ $ since $\rm\, \color{#c00}{4<5},\,\ \color{#0a0}{3\sqrt2 < 3\sqrt 3} $

Flipping gives $\rm\ a-5\, >\, b-5 \ \Rightarrow\ a > b\ \ $ QED

Remark $\ $ This is not pulled out of a hat. Rather, it arises from the continued fraction algorithm for comparing real numbers. Namely, first we compare the integer parts of $\rm\,a,b.\,$ Both are easily seen to be $\,5,\,$ so we subtract $5$ from both to get their fractional parts $\rm\,a-5,\ b-5,\,$ then recurse, comparing their reciprocals $\rm\,1/(a-5),\ 1/(b-5),\,$ etc, etc.

Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$?

5 Answers5