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
Problem
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.
Getting Started
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.
Student Solutions
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.