Is the expression (sqrt{2} - sqrt{3})(sqrt{3} | vedclass

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(B) Given expression: $(\sqrt{2} - \sqrt{3})(\sqrt{3} + \sqrt{2})$This is in the form of $(a - b)(a + b) = a^2 - b^2$.Here,$a = \sqrt{2}$ and $b = \sq

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(B) Given expression: $(\sqrt{2} - \sqrt{3})(\sqrt{3} + \sqrt{2})$This is in the form of $(a - b)(a + b) = a^2 - b^2$.Here,$a = \sqrt{2}$ and $b = \sqrt{3}$ is not quite right,let us rewrite it as $(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})$.Using the identity $(a - b)(a + b) = a^2 - b^2$:$(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$.Since $-1$ is an integer,it is a rational number.Therefore,the expression is not an irrational number.

Is the expression $(\sqrt{2} - \sqrt{3})(\sqrt{3} + \sqrt{2})$ an irrational number?

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