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I was doing math for school and got to something that really confused me. With having the rule $\frac{2}{4} = \frac{4}{8}$ (or some simular fraction equation) in mind, I got to the following confusing equation: $$\frac{\sqrt 3}{3} = \frac{(\sqrt{3})^2}{3^2} = \frac{3}{9} = \frac{1}{3}$$

This is really confusing me since $\frac{\sqrt{3}}{3}$ can never equal $\frac{1}{3}$. Where am I going wrong?

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It is incorrect that $\frac{\sqrt3}3=\frac{\sqrt3^2}{3^2}$. A simple way of seeing this is because you can multiply both the top and the bottom of a fraction by the same number, but in this case, you are multiplying $\sqrt3$ to the top, and $3$ to the bottom, but $\sqrt3\ne3$, so they are different numbers. What you proved, instead, is that $\big(\frac{\sqrt3}3\big)^2=\frac13$, which is true.

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$$\frac{\sqrt 3}{3} \ne \frac{(\sqrt{3})^2}{3^2}$$

In general $\dfrac a b \ne \dfrac{a^2}{b^2}$. You can multiply both the numerator and the denominator by the same number, but not by different numbers.

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