Hi, i have just a few questions on how Rationalizing the Denominator works.
First if you had to rationalise 2/radical: 5 why does it become 2 and radical 5 over 5. Why is not 10/25?
The basic idea is that \(\displaystyle \sqrt{a}(\sqrt{a})= (\sqrt{a})^2= a\). So to "get rid of" a square root
you multiply by that same square root, Also I hope you understand that, in order not to change the actual value of a fraction you have to multiply both numerator and denominator by the same thing.
To "get rid of" \(\displaystyle \sqrt{5}\) in the denominator, multiply
both numerator and denominator by \(\displaystyle \sqrt{5}\). \(\displaystyle \frac{2}{\sqrt{5}}= \frac{2}{\sqrt{5}}\frac{\sqrt{5}}{\sqrt{5}}= \frac{2\sqrt{5}}{(\sqrt{5})^2}= \frac{2\sqrt{5}}{5}\)
I am puzzled as to why you would think it should be \(\displaystyle \frac{10}{25}\)! For the numerator to go from 2 to 10 you must have multiplied by 5. But multiplying the denominator by 5 does not give 25, it gives \(\displaystyle 5\sqrt{5}\). In fact, it should be easy to see that \(\displaystyle \frac{10}{25}= \frac{2\cdot 5}{5\cdot 5}= \frac{2}{5}\), not \(\displaystyle \frac{2}{\sqrt{5}}\). Or are you simply not clear on what "\(\displaystyle \sqrt{}\)" means?
Second question is 4/3 and radical 7,
I think you mean \(\displaystyle \frac{4}{3\sqrt{7}}\)?
why do we multiply 3 and radical 7 to get 21?
We
don't. We multiply both numerator and denominator by \(\displaystyle \sqrt{7}\) to get
\(\displaystyle \frac{4}{3\sqrt{7}}\frac{\sqrt{7}}{\sqrt{7}}= \frac{4\sqrt{7}}{3\sqrt{7}\cdot \sqrt{7}}= \frac{4\sqrt{7}}{3(\sqrt{7})^2}= \frac{4\sqrt{7}}{3\cdot 7}= \frac{4\sqrt{7}}{21}\)
I hope i've explained it well...