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## Homework Statement

Let [tex]a, b, c, d \in Q[/tex], where [tex]\sqrt{b}[/tex] and [tex]\sqrt{d}[/tex] exist and are irrational.

If [tex]a + \sqrt{b} = c + \sqrt{d}[/tex], prove that [tex]a = c[/tex] and [tex]b = d[/tex].

## Homework Equations

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

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

[tex]a - c = \sqrt{d} - \sqrt{b} [/tex]

Then I multiply by the conjugate of [tex]\sqrt{d} - \sqrt{b}[/tex] to get

[tex](a - c)\times(\sqrt{d}+\sqrt{b}) = d - b[/tex]

By the contrapositive to a different result we proved in class, since (a-c) and (d-b) are rational, either a-c = 0 or [tex]\sqrt{d}+\sqrt{b}[/tex] 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 [tex]\sqrt{d}+\sqrt{b}[/tex] can be rational). Is this the wrong direction to be going in?

Thanks.