I'm trying to solve:
$$\frac{\sqrt 6}{\sqrt7} \cdot \sqrt {21}$$
I assumed this would breakdown to:
$$\sqrt{\frac{6}{7}\cdot\frac{21}{1}}$$
The exercise says I should break it down to:
$$\sqrt 6\cdot\sqrt\frac{21}{7}$$
What they didn't do is offer any explanation as to why the whole number 21 suddenly becomes 3 and $\frac{6}{7}$ suddenly becomes the whole number 6.
If someone could explain how this works I'd greatly appreciate it.
On
Hint: $$ \sqrt{\dfrac{6}{7}\times 21}=\sqrt{\dfrac{6}{7}\times 7\times 3}=\sqrt{6\times\dfrac{1}{7}\times 7 \times 3} $$ and $\dfrac{1}{7}\times 7 =1$.
On
Cancel before multiplying! Thus: $$ \frac{\sqrt 6}{\sqrt7} \cdot \sqrt {21} = \sqrt6 \cdot \sqrt{\frac{21}7}\text{ and } \underbrace{\frac{21} 7 = 3}_{\text{canceling the 7}}. $$ Then you have $$ \sqrt 6 \cdot \sqrt 3 = \sqrt 2\cdot\overbrace{\sqrt 3\cdot\sqrt 3}^{\text{This is 3.}} = 3\sqrt 2. $$
The whole number $21$ is not suddenly becoming $3$, nor is $\frac{6}{7}$ becoming $6$. What is actually happening is that we can re-order our terms to change the product $\frac{6}{7}\cdot 21$ into $6 \cdot \frac{21}{7}$. The reason why we are able to do this is due to the fact that our numbers are associative under multiplication. See? $$\frac{6}{7} \cdot 21 = \frac{6\cdot 21}{7} = 6 \cdot \frac{21}{7} =6\cdot 3$$