# Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator

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).

• sorry, multiply both sides by $6+2\sqrt{3}$ and use $(a-b)(a+b)=a^2-b^2$ – luka5z Jun 10 '17 at 19:00
• \begin{eqnarray*} \sqrt{a}+ \sqrt{b}= \sqrt{a+b+2 \sqrt{ab}} \end{eqnarray*} might be helpful. – Donald Splutterwit Jun 10 '17 at 19:01
• @Learner132 A lot of the solutions here add or subtract from both sides then square then compare to infer about the direction of the original identity. This is only guaranteed to work if both values are positive before squaring. E.g. consider the counterexample: $2.5>1 \rightarrow 2.5-2>1-2 \rightarrow 0.5 > -1 \rightarrow 0.25 > 1$. – Ian Miller Jun 11 '17 at 11:49
• I would just guesstimate. $\sqrt 3 \approx 1.75$, $\sqrt 2 \approx 1.4$, so $6 - 2 \sqrt 3 \approx 2.5$ and $-2 + 3 \sqrt 2 \approx 2.2$. – Robert Soupe Jun 15 '17 at 0:39

$6-2\sqrt 3 \gtrless 3\sqrt 2-2$

Rearrange: $8 \gtrless 3\sqrt 2 + 2\sqrt 3$

Square: $64 \gtrless 30+12\sqrt 6$

Rearrange: $34 \gtrless 12\sqrt 6$

Square: $1156 \gtrless 864$

• You might simplify by $2$ before squaring… – Bernard Jun 10 '17 at 19:11
• Thank you, good answer, but I doubt if math teacher is okay with this (> or <) symbol – Learner132 Jun 10 '17 at 19:29
• What's important about any symbol is its mathematical meaning. If you can the mathematical meaning of this symbol to your math teacher, I'll bet he/she is open to its usage. – Lee Mosher Jun 10 '17 at 19:57
• You should probably mention that the squaring steps are valid (both forward and backward) since the values on both sides are clearly positive. – Rory Daulton Jun 11 '17 at 0:33
• Yes, ordinarily you would start at the last step with a definite operator there (> in this case) and then work towards the inequality given to you. However, to find the necessary steps, it is often easier to work from what you’re given to something known, i.e., backwards. – BallpointBen Jun 11 '17 at 2:47

We have $\sqrt{3}\leq 1.8$ so $6-2\sqrt{3}\geq 2.4$, whereas $\sqrt{2}\leq 1.42$ so $3\sqrt{2}-2\leq 2.26$.

• I love your answer, I'll try to see if math teacher is okay with estimation – Learner132 Jun 10 '17 at 19:31
• @Learner132: Note that the estimates are easy to prove by squaring, if necessary. – user14972 Jun 10 '17 at 22:06

$$6-2√3 \sim 3√2-2\\ 8 \sim 3√2 +2√3 \\ 64 \sim 30+12√6\\ 34 \sim 12√6\\ 17 \sim 6√6\\ 289 \sim 36 \cdot 6\\ 289 > 216$$

• Thanks, quite understandable, but how I can explain ~ symbol to math teacher in this? – Learner132 Jun 10 '17 at 19:27
• This $\sim$ simbol is a placeholder for an undefined relationship, you can use whatever other symbol you prefer, while you dont multiply by -1 – Brethlosze Jun 10 '17 at 19:29
• I remember using that during school first courses without trouble – Brethlosze Jun 10 '17 at 19:29
• I understand you. Well, I hope math teacher is okay with it although it is not covered in the lecture. – Learner132 Jun 10 '17 at 19:40
• Teachers know how this work and what you thought... rewritting everything is not needed. – Brethlosze Jun 11 '17 at 6:17

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 $$2 < \sqrt {18} - 2 < \frac{9}{4}$$

$$\color{red}{ 2 < \sqrt {18} - 2 < \frac{9}{4} < \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 }$$

• Thank you, but I'll study it one day – Learner132 Jun 10 '17 at 19:51

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 $6=2×3 < (5/2)^2$

$a-b > 8-8=0, a>b$

• Thank you, it took me a while to understand 2×3<(5/2)^2 but it is easier to explain to math teacher. – Learner132 Jun 10 '17 at 19:45

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}

I'll use >=< to represent the unknown comparison.

$6-2\sqrt{3} >=< 3\sqrt{2}-2$

Lets start by adding two to both sides to reduce the number of numbers. This doesn't change the comparison result.

$8-2\sqrt{3} >=< 3\sqrt{2}$

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.

$15 >=< 8\sqrt{3}$

Square again.

$225 >=< 64 \times 3$

$225 > 192$

Therefore

$6-2\sqrt{3} > 3\sqrt{2}-2$

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 $6-2√3 > 3\sqrt 2 - 2$.

• see this answer same concept but better (less digits, can be easily verified by hand, using lower and upper bounds instead of $\approx$)) and 12 hours earlier than your answer – miracle173 Jun 11 '17 at 14:16
• @miracle173 What you saw was only a part of what I intended to do. I needed some more time to think about the problem so I thought I had deleted everything. Obviously, I hadn't. What you see above is what I intended to do. – steven gregory Jun 11 '17 at 19:57