Prove that $$\sqrt{5}-\sqrt{3}$$ is not a rational number.
Let $$\sqrt 5-\sqrt 3 $$ be a rational number of form $$\frac{a}{b}$$ ,where $$b\neq 0$$
Squaring on both sides
$$(\sqrt 5- \sqrt 3)^2=(\dfrac{a}{b})^2$$
$$(\sqrt 5)^2+(\sqrt 3)^2- 2(\sqrt 5)(\sqrt 3)=\dfrac{a^2}{b^2}$$
$$ 5+3+2\sqrt 15=\dfrac{a^2}{b^2}$$
$$8+2\sqrt 15=\dfrac{a^2}{b^2}$$
$$2\sqrt 15=\dfrac{a^2}{b^2}-8$$
$$\sqrt 15=\dfrac{a^2-8b^2}{2b^2}$$
since $$\sqrt 15$$ is irrational ,$$\dfrac{a^2-8b^2}{2b^2}$$ is rational
Since LHS $$\neq$$ RHS, contradiction arises,
Therefore $$\sqrt 5-\sqrt 3$$ is irrational
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