Why must rational $x$ and $y$ satisfy $x+y=2$ and $2\sqrt{3xy}=3$ in order for $2\sqrt{3}-3=(x+y)\sqrt{3}-2\sqrt{3xy}$ to be true?

Question

I am going through a maths book and unable to understand the following:

Given that $x$ and $y$ are rational numbers, then if the equation $$2\sqrt{3}-3=(x+y)\sqrt{3}-2\sqrt{3xy}$$ is true, then we must have $x+y=2$ and $2\sqrt{3xy}=3$.

I don't understand this. We cannot reach such a conclusion with equations involving only rational numbers. For example, let $a$, $b$, $c$, and $d$ be rational numbers, and $a - b = c - d$; then it does not mean that $a = c$ and $b = d$ always for the equation to be true.

How can we be sure that we must have $x + y = 2$ and $2 \sqrt{3xy} = 3$ for the equation to be true?

Answer 1

For better clarity, we can transform the given equation as:

$$2-\sqrt{3} = (x+y) - 2\sqrt{xy}.$$ Set $z=x+y$ for time being. Then

$$2-z= \sqrt{3}-2\sqrt{xy}.$$ Squaring both sides, we get $$4+z^2-4z -2-4xy= -4\sqrt{3xy}.$$

Note that the left hand side is rational number. Therefore right hand side should also be rational, implying that $\sqrt{3xy}$ is rational. Say $\sqrt{3xy}=r$, a rational.

Now, get back to your original equation:

$$2\sqrt{3}-3 = (x+y)\sqrt{3} -2r.$$ This must imply that $x+y=2$ and $3=-2r$.If not, then $$\sqrt{3} = \frac{2r-3}{x+y-2},$$ would be rational!

Answer 2

Consider $2 \sqrt{3} - 3 = (x + y) \sqrt{3} - 2 \sqrt{3xy}$. If $xy=a^2$ for some non-negative rational $a$ then we get

$$2 \sqrt{3} - 3 = (x + y) \sqrt{3} - 2a \sqrt{3}$$

$$\sqrt{3}= \frac3{2-x-y+2a}$$

which can’t be true. If $xy=3a^2$ for some non-negative rational $a$ then we get

$$2 \sqrt{3} - 3 = (x + y) \sqrt{3} - 6a$$

$$\sqrt3\cdot (2-x-y)=3-6a.$$

Obviously, $x+y=2$ and $3=6a\iff 2\sqrt{3xy}=3$. If $xy$ is of any other form, then $3xy$ is equal to $pa^2$, where $a$ is non-negative rational and $p$ is such a non-negative rational that every prime is included once in its either numerator or denominator, or not included at all. And there is at least one such prime other than $3$. The equality becomes:

$$2 \sqrt{3} - 3 = (x + y) \sqrt{3} - 2a \sqrt{3p}.$$

Divide by $\sqrt3$:

$$2 - \sqrt{3} = (x + y) - 2a \sqrt{p}$$

$$2 - (x + y) = \sqrt3 - 2a \sqrt{p}$$

$$\implies (2 - x - y)^2=3+4a^2p-4a\cdot \sqrt{3p}.$$

But it can’t be true since $\sqrt{3p}$ is not rational due to structure of $p$.

Answer 3

Hmm... it's not as obvious as I thought

See, my first thought was this as a case of if $W$ is irrational and $a,b,c,d$ are rational and $a+bW = c+dW$ then $b=d$ and $a=c$.

Why? Because $a-c = (d-b)W$. The left hand side is rational so the right hand side must be rational. But $W$ is irrational and the only way an irrational times a rational can be rational is if we are multiplying by zero. So $d-b =0$ and $a-c=0$ and so $a=c$ and $b=d$.

But for us to use this on $-3 +2\sqrt 3 = -2\sqrt{3xy} + (x+y)\sqrt 3$ and reach the conclusion $-3 =-2\sqrt{3xy}$ and $2=x+y$, we have to have to be assuming $-2\sqrt{3xy}$ is rational. And we can't assume that, can we?

Well, lets muck about:

$2\sqrt 3 -3 = 2 \sqrt{3} - 3 = (x + y) \sqrt{3} - 2 \sqrt{3xy}$ we can divide both sides by $\sqrt 3$ to get

$2 -\sqrt{3}= (x+y) - 2\sqrt{xy}$. Put the nasty irrationals to one side

$2-(x+y) = \sqrt 3 - 2\sqrt{xy}$ and square both sides

$4 -4(x+y) + (x+y)^2 = 3 + 4xy -4\sqrt{3xy}$

Now the LHS is rational so the RHS must be rational and therefore $\sqrt{3xy}$ is rational.

And that breaks it wide open!

Go back to the very beginning.

$\underbrace{-3}_{rational} + \underbrace{2}_{rational}\sqrt 3=\underbrace{- 2 \sqrt{3xy}}_{rational}+ \underbrace{(x + y)}_{rational} \sqrt{3}$

And we are done.

$-3 = -2\sqrt{3xy}$ and $2=x+y$.

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here's something kind of interesting that I've never noticed before:

If you have $\sqrt{q}-\sqrt{p} = r$ where $p,q,r$ are rational then you have $\sqrt q = \sqrt p + r$ and $q = p + r^2 + 2r\sqrt{p}$ so either $r=0$ and $p=q$ or both $q$ and $p$ are rational squares.

So $-3+2\sqrt 3= -2\sqrt{3xy} + (x+y)\sqrt 3$ mean

$-3 = (-2 -2\sqrt{xy} + x+y)\sqrt 3$

$-\sqrt 3 = -2 -\sqrt{4xy} +x+y$ means that either $4xy = 3$ or both $3$ and $4xy$ are rational squares. So $xy = \frac 34$

From there we have $-3+2\sqrt 3=-2\sqrt{\frac 94} + (x+y)\sqrt 3$ and

$-3 + 2\sqrt 3=-3 + (x+y)\sqrt 3$ and so $x+y = 2$.

This let's us go further with $x,y = \frac 12, \frac 34$.

Answer 4

Since the OP has not specified their level, nor from which book this is taken, I feel inclined to post this:

Suppose $x+y \neq 2$. Then, $$2\sqrt{3xy}-3 = (x+y-2)\sqrt3 \in \Bbb Q(\sqrt3)$$where $\Bbb Q(\sqrt3)$ is the smallest field containing the RHS. Then, if $\Bbb F$ is the smallest field containing the LHS, we must have $\Bbb Q(\sqrt3) = \Bbb F$. Then the LHS is of the form $a+b\sqrt3, a,b \in \Bbb Q$, and upon observation, $a=0$. So, if we divide by $\sqrt 3$, we must end up with rational values for both LHS and RHS. That is $$0 \neq 2\sqrt{xy} - \sqrt{3} \in \Bbb Q$$ Squaring, $2\sqrt{3xy} \in \Bbb Q \implies 2\sqrt{3xy} - 3 \in \Bbb Q$, contradiction!

Hence, $x+y = 2$, which also implies $2\sqrt{3xy} = 3$.

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Another approach: Assume $\sqrt{3xy}$ is irrational. Then $$-3 = \sqrt{3}(x+y - 2 - 2\sqrt{xy}) \iff -\sqrt{3} =x+y-2-2\sqrt{xy}$$ This must imply $-2\sqrt{xy} = -\sqrt 3 \iff \sqrt{3xy} = 3/2 \in \Bbb Q$, contradiction.

Hence, $\sqrt{3xy}$ is rational, then comparing yields $x+y=2$. Done!

Answer 5

It is a special case of a linear independence criteria for square roots. Said simply for this special case: $ $ if $\color{#0a0}{\sqrt c,\,\sqrt d}\,$ are irrational then the only way that their rational linear combination $\,\color{#c00}{e} = \color{#0a0}{a\sqrt c+ b \sqrt d}\,$ can be rational is when $\,\color{#c00}{e = 0},\,$ i.e. when the radicands are $\,\Bbb Q$-equivalent $\,\sqrt c = q\sqrt d\,$ for some $\,q\in \Bbb Q,\,$ so then $\,\color{#c00}e = (aq+b)\sqrt d\in \Bbb Q\!\iff\! \,aq+b=0\!\iff\! \color{#c00}e = 0,\,$ e.g. $\,2\sqrt {45}-3\sqrt{20} = 0\,$ by $\sqrt{45}/\sqrt{20} = 3/2,\,$ i.e. the sum can be rational only in the "trivial" case that the radicals can be rewritten to be rational multiples $\,b_i \sqrt d\,$ of the same radical $\,\sqrt d,\,$ so their sum is rational $\!\iff\! b_1+b_2=0.\,$ Below is a simple proof (high-school level).

$\bbox[1px,border:1px solid #c00]{\bbox[8px,border:1px solid #c00]{{\bf Theorem}\ \ \ \color{#0a0}{a\sqrt c+ b \sqrt d} = \color{#c00}e\in\Bbb Q\,\Rightarrow\, \color{#0a0}b\color{#c00}{e\!=\!0},\ {\rm if}\ \,\sqrt d\not\in\Bbb Q}}\ \ {\rm for}\,\ \color{0a0}{a,b,c,d}\in\Bbb Q$

Proof $\,\ (a\sqrt c = e-b\sqrt d)^2\Rightarrow a^2c = e^2\!+\!b^2d-2be\sqrt d,\,$ so $\,be\neq 0\Rightarrow\sqrt d\in\Bbb Q\Rightarrow\!\Leftarrow $

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So,$ $ in OP: $\,\ \color{#0a0}{2\sqrt{3xy}+b\sqrt{3}}=\color{#c00}3\in\Bbb Q\,\Rightarrow\, \color{#0a0}b\:\!\color{#c00}3 = 0,\,$ so $\,0=\color{#0a0}b\, (= 2\!-\!x\!-\!y\,$ in OP), as sought.

Remark $ $ The theorem can be generalized to linear combinations of any number of square roots. Then it says roughly that if such a sum is rational then the square-roots must be multiplicatively dependent over $\Bbb Q$, i.e. some product of them is rational, see the proof in this answer (Besicovic's theorem). These results are greatly clarified when one exploits the innate symmetries by studying field / Galois theory, in particular that of radical (Kummer) extensions - see the papers cited there.