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This question already has an answer here:

$a,b,c,~$and $d$ are rational numbers. $b>0$ and $d>0$ the $\sqrt{b}$ and the $\sqrt{d}$ are both irrational.

if $a+\sqrt{b}=c+\sqrt{d}$

show that $a = c$ and $b = d$.

I know that a=c and b=d intuitively, but I'm not sure how to prove it.

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marked as duplicate by egreg, Ross Millikan, Gerry Myerson, user61527, user122283 May 24 '14 at 0:02

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  • $\begingroup$ What's the data and what's the question?? $\endgroup$ – DonAntonio May 23 '14 at 23:31
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$a-c=\sqrt(d)-\sqrt(b)$ If this is true, then that means $\sqrt(d)-\sqrt(b)$ is rational because rational numbers are closed under subtraction. The only way for this to be true is that $\sqrt(d)-\sqrt(b)=0$.

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    $\begingroup$ I don't think it's obvious that $\sqrt{d}-\sqrt{b}=0,~$ in order for it to be rational. For example $x=1+\sqrt{2}$ and $y=\sqrt{2}$ are both irrational, but $x-y=1$. $\endgroup$ – Peter Woolfitt May 23 '14 at 23:28
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    $\begingroup$ This can be fixed by noticing that $$\sqrt d + \sqrt b = \frac{d^2 - b^2}{a - c}$$ is also rational; now sum some things. $\endgroup$ – user61527 May 23 '14 at 23:31
  • $\begingroup$ @T.Bongers Thanks! $\endgroup$ – Peter Woolfitt May 23 '14 at 23:33
  • $\begingroup$ @T.Bongers (d-b)/(a-c) not (d^2-b^2)/(a-c) $\endgroup$ – user152341 May 23 '14 at 23:45
  • $\begingroup$ @user152341 Yes, you're quite right. $\endgroup$ – user61527 May 23 '14 at 23:47

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