Rational and irrational numbers proof

[bobbarker]

Hey all, I'm new here so I'm a little noobish at the formatting capabilities of PF. Trying my best though! :P

Homework Statement

Let $$a, b, c, d \in Q$$, where $$\sqrt{b}$$ and $$\sqrt{d}$$ exist and are irrational.

If $$a + \sqrt{b} = c + \sqrt{d}$$, prove that $$a = c$$ and $$b = d$$.

Homework Equations

A number is rational iff it can be expressed as $$m/n$$, where $$m,n \in Z$$ and $$n \neq 0$$.

A rational number added to a rational number is a rational number.

A rational number added to an irrational number is an irrational number.

A rational number multiplied by a rational number is a rational number.

A rational number multiplied by an irrational number is an irrational number.

The Attempt at a Solution

I reordered things so that

$$a - c = \sqrt{d} - \sqrt{b}$$
Then I multiply by the conjugate of $$\sqrt{d} - \sqrt{b}$$ to get
$$(a - c)\times(\sqrt{d}+\sqrt{b}) = d - b$$
By the contrapositive to a different result we proved in class, since (a-c) and (d-b) are rational, either a-c = 0 or $$\sqrt{d}+\sqrt{b}$$ is rational.

Now if the first case is true the result is proved, but I'm struggling with the second half of this (proving or disproving whether $$\sqrt{d}+\sqrt{b}$$ can be rational). Is this the wrong direction to be going in?

Thanks.

 

[jgens]

If $\sqrt{c} + \sqrt{d}$ is rational, then clearly its square must also be rational. So suppose $\sqrt{c} + \sqrt{d} = a$ where $a \in \mathbb{Q}$ and consider $\sqrt{c} = a - \sqrt{d}$. Can you use the first bit of information to derive a contradiction now?

 

[bobbarker]

If $\sqrt{c} + \sqrt{d}$ is rational, then clearly its square must also be rational. So suppose $\sqrt{c} + \sqrt{d} = a$ where $a \in \mathbb{Q}$ and consider $\sqrt{c} = a - \sqrt{d}$. Can you use the first bit of information to derive a contradiction now?

Let me run this by you... I played with it for a few minutes.

(I put the problem's variables in place of yours, and defined a new quantity $e\in \mathbb{Q} = \sqrt{d}+ \sqrt{b}$ )

So since $\sqrt{b}+\sqrt{d}$ is rational, $(\sqrt{b}+\sqrt{d})^{2}$ is rational, or $d+b+2 \sqrt{db}$ is rational, or $d+b+2 \sqrt{d} \sqrt{b}$

Since $\sqrt{d} = e - \sqrt{b}$, we can write
$d+b+2 \sqrt{d} \sqrt{b} = e^{2}$
$d+b+2 (e - \sqrt{b}) \sqrt{b} = e^{2}$
$d + b + 2e \sqrt{b} - b = e^{2}$
$2 \sqrt{b} = e - d/e$

But 2 is rational and $\sqrt{b}$ is irrational so $e - d/e$ must be irrational... a contradiction since d is rational and we made the assumption that e and therefore e^2 was indeed a rational number.

Is this the right idea?

 

[jgens]

That was the general idea that I was trying to get at; however, you can save yourself a little work by considering another square . . .

Suppose that $\sqrt{c} + \sqrt{d} = e$ where $e \in \mathbb{Q}$ and consider $\sqrt{c} = e - \sqrt{d}$. Clearly $c = e^2 - 2e\sqrt{d} + d = (e^2 + d) - 2e\sqrt{d}$. Since the first term is rational and the second term necessarily irrational, their sum must also be irrational, contradicting the fact that $c \in \mathbb{Q}$.

This is the exact proof that I was thinking of but it amounts to the same thing as yours, so it doesn't really matter.