Rational and irrational numbers proof

  • Thread starter 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

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.
 

Answers and Replies

  • #2
jgens
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If [itex]\sqrt{c} + \sqrt{d}[/itex] is rational, then clearly its square must also be rational. So suppose [itex]\sqrt{c} + \sqrt{d} = a[/itex] where [itex]a \in \mathbb{Q}[/itex] and consider [itex]\sqrt{c} = a - \sqrt{d}[/itex]. Can you use the first bit of information to derive a contradiction now?
 
  • #3
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If [itex]\sqrt{c} + \sqrt{d}[/itex] is rational, then clearly its square must also be rational. So suppose [itex]\sqrt{c} + \sqrt{d} = a[/itex] where [itex]a \in \mathbb{Q}[/itex] and consider [itex]\sqrt{c} = a - \sqrt{d}[/itex]. 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 [itex]e\in \mathbb{Q} = \sqrt{d}+ \sqrt{b}[/itex] )

So since [itex]\sqrt{b}+\sqrt{d}[/itex] is rational, [itex](\sqrt{b}+\sqrt{d})^{2}[/itex] is rational, or [itex]d+b+2 \sqrt{db}[/itex] is rational, or [itex]d+b+2 \sqrt{d} \sqrt{b}[/itex]

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

But 2 is rational and [itex]\sqrt{b}[/itex] is irrational so [itex]e - d/e[/itex] 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?
 
  • #4
jgens
Gold Member
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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 [itex]\sqrt{c} + \sqrt{d} = e[/itex] where [itex]e \in \mathbb{Q}[/itex] and consider [itex]\sqrt{c} = e - \sqrt{d}[/itex]. Clearly [itex]c = e^2 - 2e\sqrt{d} + d = (e^2 + d) - 2e\sqrt{d}[/itex]. Since the first term is rational and the second term necessarily irrational, their sum must also be irrational, contradicting the fact that [itex]c \in \mathbb{Q}[/itex].

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.
 

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