# 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 $(a-b)(a+b)=a^2-b^2$ Jun 10, 2017 at 19:00
• \begin{eqnarray*} \sqrt{a}+ \sqrt{b}= \sqrt{a+b+2 \sqrt{ab}} \end{eqnarray*} might be helpful. Jun 10, 2017 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: $2.5>1 \rightarrow 2.5-2>1-2 \rightarrow 0.5 > -1 \rightarrow 0.25 > 1$. Jun 11, 2017 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 $-2 + 3 \sqrt 2 \approx 2.2$. Jun 15, 2017 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… Jun 10, 2017 at 19:11
• Thank you, good answer, but I doubt if math teacher is okay with this (> or <) symbol Jun 10, 2017 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. Jun 10, 2017 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. Jun 11, 2017 at 0:33
• I think its worth including $\iff$ symbols for each line with the accompanying justification as Rory Daulton says. Jun 11, 2017 at 3:22

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 Jun 10, 2017 at 19:31
• @Learner132: Note that the estimates are easy to prove by squaring, if necessary.
– user14972
Jun 10, 2017 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? Jun 10, 2017 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 Jun 10, 2017 at 19:29
• I remember using that during school first courses without trouble Jun 10, 2017 at 19:29
• I understand you. Well, I hope math teacher is okay with it although it is not covered in the lecture. Jun 10, 2017 at 19:40
• Teachers know how this work and what you thought... rewritting everything is not needed. Jun 11, 2017 at 6:17

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. Jun 10, 2017 at 19:45

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 Jun 10, 2017 at 19:51

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}

In general we can solve problems like this with the following approach.

1. Isolate a radical on one side of the ineqality.
2. Check both sides of the inequality are positive. If one side is positive and the other is negative then you have your answer. If both sides are negative then multiply both sides by -1 and make a note of this since it means that the manipulated inequality will be the opposite of the un-manipulated one.
3. Square both sides.
4. Repeat until no radicals are left.

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

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

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 $$6-2√3 > 3\sqrt 2 - 2$$.

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

$$2 < 3 \sqrt 2 - 2 < \dfrac 52 < 6 - 2\sqrt 3 < 3$$

• 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 Jun 11, 2017 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. Jun 11, 2017 at 19:57