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Given this simple equation:

$$ a + b\sqrt{5} = 7\sqrt{5} - 7 $$

I can really easily say that $a = -7$ and $b = 7$, because:

$ a + b\sqrt{5} = 7\sqrt{5} - 7 \\ b\sqrt{5} + a = 7\sqrt{5} - 7 \Rightarrow b = 7 \land a = -7 $

But, what if I want to show it with step-by-step passages?

I know I would just need to find out that $\,\,\,a + b = 0 \,\,\,$ and then I would say that $\,\,a = -b \,\,$ and substitute $\,a\,$ in the original equation:

$ a + b\sqrt{5} = 7\sqrt{5} - 7 = -b + b\sqrt{5} \\ b(\sqrt{5} - 1) = 7\sqrt{5} - 7 \\ b = \frac{7\sqrt{5} - 7}{\sqrt{5} - 1} \\ b = \frac{7\sqrt{5} - 7}{\sqrt{5} - 1} \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = 7 \\ a = -b = -7 $

But starting from:

$a + b\sqrt{5} = 7\sqrt{5} - 7$

I have an irrational radical which is multiplied by $b$, and if I put the equation into a system:

$$ \left\{ \begin{array}{c} a + b\sqrt{5} = 7\sqrt{5} - 7 \\ a = 7\sqrt{5} - 7 -b\sqrt{5}\\ \end{array} \right. $$ $$ \left\{ \begin{array}{c} 7\sqrt{5} - 7 -b\sqrt{5} + b\sqrt{5} = 7\sqrt{5} - 7 \\ 7\sqrt{5} - 7 = 7\sqrt{5} - 7 \\ \end{array} \right. $$

This is where it finishes. So the question is: is there a way to get rid of a radical which is a factor of only one addend, like in $a + b\sqrt{5}$ and obtain $a + b$? What are the passages?

Thanks for the attention.

user3019105
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2 Answers2

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You can transform to $$ a+7=(7-b)\sqrt 5$$ and - unless $b=7$ - to $$ \sqrt 7=\frac{a+7}{7-b}.$$ If $a,b$ are rational this claims that $\sqrt 7$ is rational, which is absurd. Hence division by $7-b$ was not allowed after all, i.e., $b=7$ and then $a+7=0$, i.e., $a=-7$.

Hagen von Eitzen
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  • Sorry, where did you get that $\sqrt{7}$? Did you mean $\sqrt{5} = \frac{a + 7}{7 - b}$. Also, giving $\sqrt{5} = \frac{a + 7}{7 - b}$ you correctly deduce that if $a$ and $b$ are rational, there equation doesn't make sense because \sqrt{5} is not rational, but at the end how do you deduce that $b = 7$, I didn't understand the last reasoning you made when you say Hence division by 7−b was not allowed after all, i.e., b=7 Also as the two values for a and b are -7 and 7, we would obtain $\sqrt{5} = \frac{-7 + 7}{7 - 7} = \sqrt{5} = \frac{0}{0}$, how should we interpret this result? – user3019105 Jun 11 '15 at 22:37
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First of all you are solving the equation for rational $a,b$ as otherwise this equation has an infinite set of solutions. Secondly you should know that rationals and irrationals are closed sets under multiplication by rational numbers. Therefore you can consider the irrational parts of the equation and the rational parts separately to solve for $a$ and $b$.

RowanS
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  • rationals and irrationals are closed sets under multiplication by rational numbers, you mean if I multiply a rational by a rational I get another rational but if I multiply an irrational by a rational I will get an irrational, right? – user3019105 Jun 11 '15 at 18:58
  • You have understood correctly – RowanS Jun 11 '15 at 19:01
  • The irrationals are not closed under multiplication: for example $\sqrt{2} \cdot \sqrt{2} = 2$. However, the rationals are closed under multuplication. – Ashkay Jun 11 '15 at 19:02
  • @Ashkay Yes, but as RowanS says, irrationals are closed sets under multiplication by rational numbers, not by irrationals , therefore $\sqrt{2} ⋅ x = y ,,,,, : x \in \mathbb{Q} ,,,,, \implies y \in \mathbb{I} ,,,,, \forall x \in \mathbb{Q}$ – user3019105 Jun 11 '15 at 19:11
  • @RowanS All right, but what do you mean with Therefore you can consider the irrational parts of the equation and the rational parts separately to solve for a and b.? – user3019105 Jun 11 '15 at 19:13
  • the rational part of the equation can only affect the rational part and the irrational part can only affect the rational part. So we can write rational( $a+b\sqrt{5}$) = a and similarly rational( $7+7\sqrt{5}$) = 7 – RowanS Jun 11 '15 at 19:17
  • Oops, totally misread that. My bad. – Ashkay Jun 11 '15 at 19:17
  • @RowanS I am sorry for answering only now, still I didn't get it, could you make another example, please? – user3019105 Jun 11 '15 at 23:01
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    Are you familiar with complex numbers? It's a similar idea so a+bi =7+7i you can consider the real parts and get real(a+bi) = a and real (7+7i) = 7 so a=7 but instead of doing this with complex numbers we are doing this with rational and irrational numbers because similarly to how complex numbers are closed under multiplication by rational numbers irrational numbers are closed under multiplication by rational numbers. – RowanS Jun 11 '15 at 23:12
  • Now I understood, the only thing is that in this case we are leveraging the fact that irrationals are closed under addition by rational numbers too, therefore the rational part of the addition $a + b\sqrt{5}$ is rational $a$ and so on... Is that correct? – user3019105 Jun 12 '15 at 06:47
  • Yes that is correct – RowanS Jun 12 '15 at 08:30
  • Thank you! It makes sense now! – user3019105 Jun 12 '15 at 12:54