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\frac{\sqrt{2}\times 2}{\sqrt{3}+1}
Divide \sqrt{2} by \frac{\sqrt{3}+1}{2} by multiplying \sqrt{2} by the reciprocal of \frac{\sqrt{3}+1}{2}.
\frac{\sqrt{2}\times 2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}
Rationalize the denominator of \frac{\sqrt{2}\times 2}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.
\frac{\sqrt{2}\times 2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}
Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\sqrt{2}\times 2\left(\sqrt{3}-1\right)}{3-1}
Square \sqrt{3}. Square 1.
\frac{\sqrt{2}\times 2\left(\sqrt{3}-1\right)}{2}
Subtract 1 from 3 to get 2.
\frac{2\sqrt{2}\sqrt{3}-\sqrt{2}\times 2}{2}
Use the distributive property to multiply \sqrt{2}\times 2 by \sqrt{3}-1.
\frac{2\sqrt{6}-\sqrt{2}\times 2}{2}
To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.
\frac{2\sqrt{6}-2\sqrt{2}}{2}
Multiply -1 and 2 to get -2.
\sqrt{6}-\sqrt{2}
Divide each term of 2\sqrt{6}-2\sqrt{2} by 2 to get \sqrt{6}-\sqrt{2}.