Rational Roots
Given that a, b and c are natural numbers show that if sqrt a+sqrt
b is rational then it is a natural number. Extend this to 3
variables.
Age
16 to 18
Challenge level
Suppose that a and b are natural numbers. If $\sqrt{a} + \sqrt{b}$ is rational then show that it is a natural number. Show that, indeed both $\sqrt{a}$ and $\sqrt{b}$ are then integers.
Suppose that a, b and c are natural numbers. If $\sqrt{a} + \sqrt{b} + \sqrt{c}$ is rational then show that it is a natural number. Moreover show that $\sqrt{a}$ , $\sqrt{b}$ and $\sqrt{c}$ are then integers.
In this problem you are given that $a$, $b$ and $c$ are natural numbers. You have to show that if $\sqrt{a}+\sqrt{b}$ is rational then it is a natural number.
You could use the fact that if $\sqrt{a}+\sqrt{b}$ is rational then so is its square which means that $\sqrt ab $ is also rational. Knowing this the next step is to use $$\sqrt{a}(\sqrt{a}+\sqrt{b}) = a+\sqrt{ab}$$ to show that $\sqrt a$ is rational and to do likewise for $b$.
This is all you need because it has been proved that if $\sqrt a$ is rational then $a$ must be a square number.
Try to apply this method and then to extend it to three variables for the last part.
David sent in this solution, using the hints we gave you.
$\sqrt{a}+\sqrt{b}$ rational
$\Rightarrow (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{a b}$ rational
$\Rightarrow 2\sqrt{a b}$ rational
$\Rightarrow \sqrt{a b}$ rational
$\Rightarrow a+\sqrt{a b}$ rational
i.e., $\sqrt{a}(\sqrt{a}+\sqrt{b})$ rational
$\Rightarrow \sqrt{a}$ rational (and so $a$ is a square)
$\Rightarrow \sqrt{b}$ is also rational and hence $b$ is a square.
For the second part:
$\sqrt{a}+\sqrt{b}+\sqrt{c}$ rational
$\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ rational
$\Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ rational
$(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})^2=a b+b c+c a+2\sqrt{a b
c}(\sqrt{a}+\sqrt{b}+ \sqrt{c})$
so $\sqrt{a b c}$ is also rational
$\Rightarrow \sqrt{a}(\sqrt{a b}+\sqrt{b c}+\sqrt{c
a})-\sqrt{a b c}=a(\sqrt{b} +\sqrt{c})$ is rational
so $\sqrt{b}+\sqrt{c}$ is rational and $\sqrt{a}$ is rational,
and use the above.