Rationalizing Denominators

[King Friday]

Why are we able to reduce the 2/6 while ignoring the radical expression next to it?

 

[fresh_42]

It becomes clearer if you write the entire equation with multiplications only. Can you do this by using [imath] \dfrac{1}{3}=3^{-1} [/imath] or is this new to you?

 

[Dr.Peterson]

Fill in more steps: [math]\frac{2\sqrt{2}}{6}=\frac{2}{6}\frac{\sqrt{2}}{1}=\frac{1}{3}\frac{\sqrt{2}}{1}=\frac{\sqrt{2}}{3}[/math]
It's also useful to be able to see [imath]\frac{a\sqrt{2}}{b}[/imath] as [imath]\frac{a}{b}\sqrt{2}[/imath]: the "top floor" of a fraction is "on the same level as the ground outside". with the bottom being the "basement". So that's an alternative way to see this: [math]\frac{2\sqrt{2}}{6}=\frac{2}{6}\sqrt{2}=\frac{1}{3}\sqrt{2}=\frac{\sqrt{2}}{3}[/math]

 

[King Friday]

It becomes clearer if you write the entire equation with multiplications only. Can you do this by using [imath] \dfrac{1}{3}=3^{-1} [/imath] or is this new to you?

If going through this 52 years ago as a High School Sophomore makes it new, then yes I'm new to it.

 

[King Friday]

Fill in more steps: [math]\frac{2\sqrt{2}}{6}=\frac{2}{6}\frac{\sqrt{2}}{1}=\frac{1}{3}\frac{\sqrt{2}}{1}=\frac{\sqrt{2}}{3}[/math]
It's also useful to be able to see [imath]\frac{a\sqrt{2}}{b}[/imath] as [imath]\frac{a}{b}\sqrt{2}[/imath]: the "top floor" of a fraction is "on the same level as the ground outside". with the bottom being the "basement". So that's an alternative way to see this: [math]\frac{2\sqrt{2}}{6}=\frac{2}{6}\sqrt{2}=\frac{1}{3}\sqrt{2}=\frac{\sqrt{2}}{3}[/math]

Yes, that makes a conceptual difference!