- #1
bobbarker
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Hey all, I'm new here so I'm a little noobish at the formatting capabilities of PF. Trying my best though! :P
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].
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.
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.
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.