How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably. Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$. Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).
$6-2\sqrt 3 \gtrless 3\sqrt 2-2$
Rearrange: $8 \gtrless 3\sqrt 2 + 2\sqrt 3$
Square: $64 \gtrless 30+12\sqrt 6$
Rearrange: $34 \gtrless 12\sqrt 6$
Square: $1156 \gtrless 864$
We have $\sqrt{3}\leq 1.8$ so $6-2\sqrt{3}\geq 2.4$, whereas $\sqrt{2}\leq 1.42$ so $3\sqrt{2}-2\leq 2.26$.
$$ 6-2√3 \sim 3√2-2\\ 8 \sim 3√2 +2√3 \\ 64 \sim 30+12√6\\ 34 \sim 12√6\\ 17 \sim 6√6\\ 289 \sim 36 \cdot 6\\ 289 > 216 $$
Define
$a=6-2\sqrt 3>0$
$b=3\sqrt 2-2>0$
$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$
$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6) < 60 < 64$ because $6=2×3 < (5/2)^2$
$a-b > 8-8=0, a>b$
Using simple continued fractions for $\sqrt {12}$ and $\sqrt {18}.$ Worth learning the general technique...Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 12} = 3 + \frac{ \sqrt {12} - 3 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{3 } = 2 + \frac{ \sqrt {12} - 3 }{3 } $$ $$ \frac{ 3 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{1 } = 6 + \frac{ \sqrt {12} - 3 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 3 & & 2 & & 6 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 7 }{ 2 } \\
\\
& 1 & & -3 & & 1
\end{array}
$$
$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 12 \cdot 0^2 = 1 & \mbox{digit} & 3 \\ \frac{ 3 }{ 1 } & 3^2 - 12 \cdot 1^2 = -3 & \mbox{digit} & 2 \\ \frac{ 7 }{ 2 } & 7^2 - 12 \cdot 2^2 = 1 & \mbox{digit} & 6 \\ \end{array} $$
Continued fraction convergents alternate above and below the irrational number, we get $$ \frac{ 3 }{ 1 } < \sqrt {12} < \frac{ 7 }{ 2 } $$ Your first number was $6 - \sqrt {12},$ $$ 3 > 6 - \sqrt {12} > \frac{ 5 }{ 2 } $$ $$ \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 $$
Next 18......................========================================
$$ \sqrt { 18} = 4 + \frac{ \sqrt {18} - 4 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{2 } = 4 + \frac{ \sqrt {18} - 4 }{2 } $$ $$ \frac{ 2 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{1 } = 8 + \frac{ \sqrt {18} - 4 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 4 & & 4 & & 8 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 17 }{ 4 } \\
\\
& 1 & & -2 & & 1
\end{array}
$$
$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 18 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 18 \cdot 1^2 = -2 & \mbox{digit} & 4 \\ \frac{ 17 }{ 4 } & 17^2 - 18 \cdot 4^2 = 1 & \mbox{digit} & 8 \\ \end{array} $$
This time the number is $\sqrt {18} - 2.$
It is enough to use $$ 2 < \sqrt {18} - 2 < \frac{9}{4} $$
$$ \color{red}{ 2 < \sqrt {18} - 2 < \frac{9}{4} < \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 } $$
In general we can solve problems like this with the following approach.
I'll use >=< to represent the unknown comparison.
$ 6-2\sqrt{3} >=< 3\sqrt{2}-2$
We start by adding 2 to both sides. This isolates one of the radicals.
$ 8-2\sqrt{3} >=< 3\sqrt{2}$
Both sides are clearly positive ( $ 2\sqrt{3} < 6 $ ) so we can square both sides without changing the comparison result.
In a more maginal case where we were unsure if the left hand side was positive we could have compared the two terms in the left hand side by squaring both of them and hence determined whether the left hand side was positive or negative.
$ 64 -32\sqrt{3} + 12 >=< 18$
Now lets collect terms.
$ 60 >=< 32\sqrt{3}$
Divide by four (not strictly needed but keeps the numbers smaller).
$ 15 >=< 8\sqrt{3}$
Square again.
$ 225 >=< 64 \times 3 $
$ 225 > 192 $
Therefore
$ 6-2\sqrt{3} > 3\sqrt{2}-2$
Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
$$\begin{align} \left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\ \frac{3}{2} & \gt \sqrt 2\\ \frac{9}{2} & \gt 3\sqrt 2 \end{align}$$
And $$\begin{align} \left(\frac{7}{4}\right)^2 = \frac{49}{16} & \gt 3\\ \frac{7}{4} & \gt \sqrt 3\\ \frac{7}{2} & \gt 2\sqrt 3 \end{align}$$
Adding those two approximations, we get $$\begin{align} \frac{9}{2} + \frac{7}{2} = 8 & \gt 3\sqrt 2 + 2\sqrt 3\\ 6 + 2 & \gt 3\sqrt 2 + 2\sqrt 3\\ 6 - 2\sqrt 3 & \gt 3\sqrt 2 - 2 \end{align}$$
There is still some hope in taking the first minus the second in this case: $$6-2\sqrt{3} - (3\sqrt{2}-2) = 8 - (2\sqrt{3} + 3\sqrt{2})$$
So now the question boils down to if the expression with the square root exceeds $8$. We know that $8^{2} = 64$ and: $$(2\sqrt{3}+3\sqrt{2})^{2}=4*3+2(2\sqrt{3})(3\sqrt{2})+9*2 =30+12\sqrt{3}\sqrt{2}$$
Conversely: $$8^{2} = 64 = 28+36=28+6*6 = 28+6*\sqrt{6}\sqrt{6}$$
Subtracting them yields: $$ 28+6*\sqrt{6}\sqrt{6}-30+12\sqrt{3}\sqrt{2} = -2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2}$$
Thus this is only positive if the square root terms are positive and exceed 2. Comparing the square root terms: $$6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = 6*\sqrt{2}\sqrt{3}(\frac{\sqrt{2}}{\sqrt{2}})\sqrt{6}-12\sqrt{6}=12\frac{\sqrt{3}}{\sqrt{2}}\sqrt{6}-12\sqrt{6}=12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1)$$
We conclude that the square root term is positive since $\sqrt{3}>\sqrt{2}$ . But is it greater than $-2$? Or rather, how much do we need to multiply to the expression $\frac{\sqrt{3}}{\sqrt{2}}-1$ for it to be greater than $2$? $$n*(\frac{\sqrt{3}}{\sqrt{2}}-1)>2$$ $$n>\frac{2}{\frac{\sqrt{3}}{\sqrt{2}}-1}*\frac{\frac{\sqrt{3}}{\sqrt{2}}+1}{\frac{\sqrt{3}}{\sqrt{2}}+1}\approx\frac{2*(1.5+1)}{0.5} = 10$$ So the factor in front of the term $\frac{\sqrt{3}}{\sqrt{2}}-1$ should be at least 10. But since $12>10$ and $\sqrt{6}>1$, we conclude that $12\sqrt{6}>10$. Thus, substituting back to the original equation: $$-2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = -2+12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1) > 0$$ And so we conclude that: $$64>(2\sqrt{3}+3\sqrt{2})^{2}$$ and so, for positive root, $$8>(2\sqrt{3}+3\sqrt{2})$$ and $$ 8 - (2\sqrt{3} + 3\sqrt{2}) > 0 $$
You may use other approaches. Notice that I split $8^{2} = 28+36$. Equally viable is to split $8^{2} = 30+34$, and you might have to use a similar (Squaring than rooting) trick to show that $34 > 12\sqrt{6}$, and then finally conclude that $8>(2\sqrt{3} + 3\sqrt{2})$. This alternative approach should be able to give you a smaller threshold compared to the approximation I made midway.
$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align} \dfrac{49}{4} &> 12 \\ \dfrac 72 &> 2\sqrt 3 \\ \dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\ 6 - 2\sqrt 3 &> \dfrac 52 \end{align}
and
\begin{align} \dfrac{81}{4} &> 18 \\ \dfrac 92 &> 3\sqrt 2 \\ \dfrac 52 &> 3\sqrt 2 - 2 \end{align}
It follows that $6-2√3 > 3\sqrt 2 - 2$.
<<<<< added 11/23/2043 >>>>>
What I am looking for is a simple, improvable, method for finding nice bounds for $6-2√3$ and $3\sqrt 2 - 2$. First pass:
$$ \left( 2\sqrt 3 \right)^2 = 12 $$ $$ 9 < 12 < 16 $$ $$ 3^2 < \left( 2\sqrt 3 \right)^2 < 4^2 $$ $$ 3 < 2\sqrt 3 < 4 $$ $$ -4 < -2\sqrt 3 < -3 $$ $$ 2 < 6 - 2\sqrt 3 < 3$$
and
$$ \left( 3 \sqrt 2 \right)^2 = 18 $$ $$ 16 < 18 < 25 $$ $$ 4^2 < \left( 3 \sqrt 2 \right)^2 < 5^2 $$ $$ 4 < 3 \sqrt 2 < 5 $$ $$ 2 < 3 \sqrt 2 - 2 < 3 $$
And all we have accomplished is to show that both quantities are between 2 and 3. So we double the two quantities; that is, we compare $12-4√3 $ and $6√2-4$
$$ \left( 4\sqrt 3 \right)^2 = 48 $$ $$ 36 < 48 < 49 $$ $$ 6^2 < \left( 4 \sqrt 3 \right)^2 < 7^2 $$ $$ 6 < 4\sqrt 3 < 7 $$ $$ -7 < -4\sqrt 3 < -6 $$ $$ 5 < 12 - 4\sqrt 3 < 6 $$ $$ \dfrac 52 < 6 - 2\sqrt 3 < 3 $$
and
$$ \left( 6 \sqrt 2 \right)^2 = 72 $$ $$ 64 < 72 < 81 $$ $$ 8^2 < \left( 6 \sqrt 2 \right)^2 < 9^2 $$ $$ 8 < 6 \sqrt 2 < 9 $$ $$ 4 < 6 \sqrt 2 - 4 < 5 $$ $$ 2 < 3 \sqrt 2 - 2 < \dfrac 52 $$
And we conclude that
$$ 2 < 3 \sqrt 2 - 2 < \dfrac 52 < 6 - 2\sqrt 3 < 3 $$
