Compare $\sqrt{6}-\sqrt{3}$ and $\sqrt{3}-\sqrt{2}$

Question

Compare the numbers $a=\sqrt{5-2\sqrt{6}},b=\sqrt{6}-\sqrt{3}$ and $c=\sqrt{3}-\sqrt{2}$.

We have $$a=\sqrt{5-2\sqrt{2}\sqrt{3}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\sqrt{3}-\sqrt{2},\\b=\sqrt{6}-\sqrt{3}=\sqrt{3}\left(\sqrt{2}-1\right),\\c=\sqrt{3}-\sqrt{2}.$$ We conclude that $a=c$ so we are left with comparing $b=\sqrt{6}-\sqrt{3}$ and $c=\sqrt{3}-\sqrt{2}.$ How can I do that? I have seen that sometimes we can square both sides of an inequality but $1>-3$ is true whereas $1^2=1>(-3)^2=9$ isn't. It's a little confusing for me when we can or cannot. Any other methods?

Answer 1

It's a little confusing for me when we can or cannot.

The rules are:

1. If $0 \le a < b$ then $a^2 < b^2$.

2. If $a < 0 < b$ then .... we don't know. We have to compare the absolute values of $a$ and $b$ which ... we may or may not know anything about.

3. If $a < b < 0$ then $0 < -b < -a$ and $(-b)^2 = b^2 < (-a)^2 = a^2$. The signs flip.

We can put these all into a single rule

4. $|a| < |b| \iff a^2 = b^2$

The only issue is that "negative times negative equals positive" and that can skew our frame of reference.

In this case all terms are positive so we can ass $a^2 < b^2 \implies 0 < a < b$.

SO $b^2 = (\sqrt 6- \sqrt 3) = 6 + 3-\sqrt{18}=9-\sqrt 18$ and $c^2 = (\sqrt 3 -\sqrt 2)^2 = 3 +2 - \sqrt 6= 5-\sqrt 6$

And from here we have $2 < 3$ and $4 < 5$ so $4=2^2 < 6 < 3^2 = 9$ and $4^2=16 < 18 < 5^2 = 25$ so $2 < \sqrt 6 < 3$ and $4 < \sqrt {18} < 5$.

So $9- 5=4 < 9-\sqrt{18} < 9-4= 5$ while $5-3 =2 < 5-\sqrt 6 < 5-2 = 3$.

So $2< c^2 < 3$ and $4< b^2 < 5$. So $b^2 > c^2$ and $b > c$

.....

Alternatively

As $b = \sqrt {6}-\sqrt 3 > 0$ and $c= \sqrt{3} - \sqrt 2> 0$ then

$b > c >0\iff$

$b^2 > c^2; b> 0; c>0 \iff$

$9-\sqrt {18} > 5 - \sqrt 6 \iff$

$4 -\sqrt{18} > -\sqrt 6 \iff$

$4 > \sqrt{18} - \sqrt 6$ and so $\sqrt{18}-\sqrt 6$ is positive

$\iff 4^2=16 > 18 + 6 - 2\sqrt{6\times 18} \iff$

$-8 > - 2\sqrt{6\times 18} \iff$

$8 < 2\sqrt {6\times 18}\iff$

$4 < \sqrt {6 \times 18} \iff$

$4^2 =16 < 6\times 18$

And as the result is certainly true our premise is true.

So $b > a$.

We didn't have to know the result ahead of time. We could reples $>$ with $<$ in all the if and only if statements and conclude

$b < a \iff 16 > 6\times 18$. And the result is certainly false our premise is false. So $b \ge a$.

(And we can rule out $a = b$ easily.

$a = b\iff \sqrt{6}-\sqrt 3 =\sqrt 3 - \sqrt 2 \iff \sqrt 2+\sqrt 6 = 2\sqrt 3 \iff 2+6 + \sqrt {12} = 4\cdot 3 \iff 12=8 + \sqrt {12}$ which is obviously false.)

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In the comments the question "is there a general way to compare $\sqrt a - \sqrt b$ to $\sqrt c - \sqrt d$?"

Consider $\sqrt a - \sqrt b < \sqrt c - \sqrt d \iff$

$\sqrt a +\sqrt d < \sqrt c +\sqrt b$. AS these are both positive...

$\iff (\sqrt a + \sqrt d)^2 < (\sqrt c + \sqrt b)^2 \iff$

$a + d + \sqrt{ad} < c + b + \sqrt {bc} \iff$

$a+d-c-b < \sqrt{bc}-\sqrt{ad}$.

At this point we must note if $a+b -c-d$ is positive or negative or zero and the some comparison for wheter $bc$ is larger or smaller.

If we get get both sides are opposite signs we are done. The positive side is larger than the negative.

If the are the same sign. We take the absolute vales of both sides and flip the inequality and continue. Let's assume $(a+d) > (c+b)$ and $bc > ad$ and we had to continue.

Then

$\iff (a+b-c-d)^2 > bc + ad -2\sqrt{abcd}\iff$

$(a+b-c-d)^2 - (bc+ad) > -2\sqrt{abcd} \iff$

$(bc+ad)-(a+b-c-d)^2 < 2\sqrt{abcd}$.

You'll note I'm not expanding these out. It is assumed that these are actual known integers are will be single values.

If the LHS is $0$ or negative we are done. We know our premise was true (or false) other wise... square

$[(bc+ad)-(a+b-c-d)^2]^2 < abcd$.

So for example to compare $\sqrt{102} - \sqrt {83}$ to $\sqrt{17}-\sqrt{3}$ we note:

$\sqrt{102} - \sqrt {83}>\sqrt{17}-\sqrt{3}\iff$

$\sqrt{102} + \sqrt 3 > \sqrt {17} +\sqrt{83} \iff$

$105 + 2\sqrt{306} > 100 + 2\sqrt{17*83} \iff$

$5 > 2\sqrt{1411} - 2\sqrt{306}\iff$

$25 > 4(1411+306 - 2\sqrt{306*1411} = 6868-8\sqrt{431766}\iff$

$8\sqrt {431766} < 6868-25 = 6843\iff$

$64*431766 < 6843^2 \iff$

$27,633,024 < 46,826,649$.

Which is true. so $\sqrt{102} - \sqrt {83}>\sqrt{17}-\sqrt{3}$ is true.

Answer 2

$\sqrt 2c=\sqrt 6-\sqrt 4=\sqrt 6-2<\sqrt 6-\sqrt 3=b$.

So $c$ is much less than $b$.

Answer 3

If we the first numbers with the (positive) $\sqrt 2+1$, we obtain $$ (\sqrt 6-\sqrt 3)(\sqrt 2+1)=\sqrt 3(\sqrt 2-1)(\sqrt 2+1)=\sqrt 3(2-1)=\sqrt 3$$ and id we multiply the second number with the larger positive factor $\sqrt 3+\sqrt 2$, we obtain $$ (\sqrt 3-\sqrt 2)(\sqrt 3+\sqrt 2)=3-2=1,$$ and yet the product is smaller (but positive). Conclude