Alg2.1.20

Which one of the two numbers $\sqrt{3}+2, 3+\sqrt{2}\,$ is greater?

Taking the first number and rationalising, we get

$\sqrt{3}+2=\frac{(\sqrt{3}+2)(\sqrt{3}-2)}{\sqrt{3}-2}\,$

This is equal to

$\sqrt{3}+2=\frac{3-4}{\sqrt{3}-2}=\frac{-1}{\sqrt{3}-2}\,$

Taking the second number and rationalising,

$3+\sqrt{2}=\frac{(3+\sqrt{2})(3-\sqrt{2})}{3-\sqrt{2}}\,$

$3+\sqrt{2}=\frac{7}{3-\sqrt{2}}\,$

After rationalising, the second denominator is greater than first denominator. But the second numerator is greater.

Hence the inequality

$(3+\sqrt{2})>(\sqrt{3}+2)\,$