1
$\begingroup$

For example we have:

How does it equal to:

I know that

But in that shifting they have wroten that it equals to

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    The shifting you are referring to is nothing more than another way of re-expressing 1 - $\sqrt{5}$ – 2017-02-03
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    Multiply $\frac{1-\sqrt 5}{2}$ by $\frac{1+\sqrt 5}{1+\sqrt 5}$ to get a difference of squares in the numerator. – 2017-02-03
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    Basically rationalize your numerator is what amWhy is saying. – 2017-02-03
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    Thank you both for helping :) – 2017-02-04

2 Answers 2

3

It is not a rule that numerator and denominator can change places. $\frac{a-b\sqrt{c}}{d}\not=\pm\frac{d}{a+b\sqrt{c}}$ but a coincidence.The correct method to solve this type of questions is to multiply top and bottom by the conjugate in this case $\frac{1+\sqrt{5}}{1+\sqrt{5}}$ $$\frac{1-\sqrt{5}}{2}\cdot\frac{1+\sqrt{5}}{1+\sqrt{5}}=\frac{1^2-5}{2(1+\sqrt{5})}=-\frac{4}{2(1+\sqrt{5})}=-\frac{2}{1+\sqrt{5}}$$

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    Thanks so much for the help :) – 2017-02-04
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    I wouldn't call it a _coincidence_, I'd call it a _method_ for dealing with $a+b\sqrt{c}$ in the denominator. – 2017-02-04
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    @EthanBolker I guess I should've worded my answer better,I was talking that in general $\frac{a-b\sqrt{c}}{d}\not=\pm\frac{d}{a+b\sqrt{c}}$ that numerator and denominator don't change places. – 2017-02-04
1

$$ \dfrac{A}{B} =\dfrac{C}{D}$$

can shift places to

$$ \dfrac{D}{B} =\dfrac{C}{A}$$

or

$$ \dfrac{A}{C} =\dfrac{B}{D}$$

all because cross products taken of numerator/denominator on opposite sides of $=$ sign

$$ {A}{B} ={C}{D}$$

remain quite the same.