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I have to find (a/b), b=(1-(√3/3)), a= (1+ (1/√3))

I know I can just punch this into my calculator. However, my teacher says we need to know how to add these and subtract these but just as said that, class was over. How can I add and subtract roots with regular integers?

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$$\frac{a}{b}=\frac{1+\frac{1}{\sqrt 3}}{1-\frac{\sqrt 3}{3}}=\frac{1+\frac{\sqrt 3}{3 }}{1-\frac{\sqrt 3}{3}}=\frac{\frac{3+\sqrt 3}{3}}{\frac{3-\sqrt 3}{3}}=\frac{3+\sqrt 3}{3-\sqrt 3}=\frac{3+\sqrt 3}{3-\sqrt 3}\cdot 1$$ $$=\frac{3+\sqrt 3}{3-\sqrt 3}\cdot \frac{3+\sqrt 3}{3+\sqrt 3}=\frac{12+6\sqrt 3}{6}=2+\sqrt 3$$

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In order to find the ratio $\frac{a}{b}$, first express a and b as fractions. $$a = 1+ \frac{\sqrt{3}}{3} = \frac{3+\sqrt{3}}{3}$$ $$b = 1 - \frac{\sqrt{3}}{3} = \frac{3-\sqrt{3}}{3}$$

Now, we take the ratio of a and b. $$\frac{a}{b} = \frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}} = \frac{3+\sqrt{3}}{3-\sqrt{3}}$$

To get rid of the square root at the denominator, multiply by the conjugate. $$\frac{a}{b} = \frac{3+\sqrt{3}}{3-\sqrt{3}} \times \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{9+6\sqrt{3} + 3}{9-3} = \frac{12 + 6\sqrt{3}}{6}$$

Then you can simplify.

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Note that $$b = 1-\sqrt3/3 = 1-1/\sqrt3$$ Hence, $$\dfrac{b}{a} = \dfrac{1-1/\sqrt3}{1+1/\sqrt3} = \dfrac{1-1/\sqrt3}{1+1/\sqrt3} \cdot \dfrac{1-1/\sqrt3}{1-1/\sqrt3} = \dfrac{1+1/3-2/\sqrt3}{1-1/3} = \dfrac{\dfrac43-\dfrac2{\sqrt3}}{2/3} = 2 - \sqrt3$$ Similarly, $$\dfrac{a}b = 2+\sqrt3$$

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