# Why can we use inspection for solving equation with multiple unknowns?

In our algebra class, our teacher often does the following:

$a + b\sqrt{2} = 5 + 3\sqrt{2} \implies \;\text{(by inspection)}\; a=5, b = 3$

I asked her why we can make this statement. She was unable to provide a satisfactory answer. So I tried proving it myself.

$a + b\sqrt{2} = x + y\sqrt{2}$. We are required to prove that $a = x$, and $b = y$. Manipulating the equation, we get $\sqrt{2}(b - y) = x - a$, or $\sqrt{2} = \frac{x-a}{b-y}$. Expanding this, we get $\sqrt{2} = \frac{x}{b-y} + \frac{a}{b-y}$. I tried various other transformations, but nothing seemed to yield a result.

• (+1) for not accepting the "inspection" claim of your teacher without further arguments! – azimut Aug 9 '13 at 11:59

We want to show that the only rational solutions $a,b,x,y$ of $$a + b\sqrt{2} = x + y\sqrt{2}$$ are given by $a = x$ and $b = y$.

(Note that if you allow for real values of $a,b,x,y$, then for example $a = x + \sqrt{2}$, $b = y - 1$ would be a solution, too.)

If $y = b$, then obviously $a = x$.

Now assume $y \neq b$. Then $$a + b\sqrt{2} = x + y\sqrt{2}\\ \implies a-x = (y-b)\sqrt{2}\\ \implies (a-x)^2 = 2(y-b)^2\\ \overset{y - b \neq 0}{\implies} 2 = \left(\frac{a-x}{y-b}\right)^{\!2}.$$ So $2$ is the square of a rational number. This contradicts the fact that $\sqrt{2}$ is irrational.

By the above argument, we have shown that the only rational solution of $a + b\sqrt{2} = 0$ is $a = b = 0$. In terms of linear algebra, this property is formulated as "$1$ and $\sqrt{2}$ are $\mathbb Q$-linearly independent". From this it follows that any representation $a + b\sqrt{2}$ with $a,b\in\mathbb Q$ uniquely determines $a$ and $b$.

• But, what if $a = x, b = y$? The equation would become, $2 = \left(\frac{a-a}{b-b}\right)^2$, which makes no sense at all. – Gerard Aug 9 '13 at 12:00
• @AnanayAgarwal: The case $b = y$ is treated separately. In this place I'm assuming that $b\neq y$. – azimut Aug 9 '13 at 12:01
• Why do you know $(a-x)/(y-b)$ is rational? Are you make that assumption on the coefficients? – Clayton Aug 9 '13 at 12:02
• You are prooving there's only one solution made up of rationals, but what if $x,y$ are irrational? – MyUserIsThis Aug 9 '13 at 12:04
• @clayton I assume that $a,b,x,y$ are rational. Otherwise, you get more solutions (I added a comment on this), so I guess this assumption was also made by the OP. – azimut Aug 9 '13 at 12:06

HINT:

Assuming $a,b,x,y$ are rationals $\sqrt2(b-y)=x-a$ rational which is only possible if $b-y=0$

As for $b-y\ne0,\sqrt 2=\frac{x-a}{y-b}$ which is rational

• Oh! I get it now. My teacher didn't tell me that this could be only possible if the numbers in question were rationals. That's why I thought there was a certain ambiguity to the method. – Gerard Aug 9 '13 at 11:58
• @AnanayAgarwal, please find the edited answer and related math.stackexchange.com/questions/457382/… – lab bhattacharjee Aug 9 '13 at 12:02