Question
Show that $(\sqrt{3}+\sqrt{5})^2$ is an irrational number.
✓
Answer
Let $(\sqrt{3}+\sqrt{5})^2$ is a rational number.
$\Rightarrow(\sqrt{3}+\sqrt{5})^2=\frac{p}{q}$
Where $p, q$ are co$-$prime
Using $(a+b)^2=a^2+b^2+2 a b$ we get,
$(\sqrt{3})^2+(\sqrt{5})+2 \sqrt{3} \sqrt{5}=\frac{p}{q}$
$\Rightarrow 3+5+2 \sqrt{15}=\frac{p}{q}$
$\Rightarrow 8+2 \sqrt{15}=\frac{p}{q}$
$\Rightarrow 2 \sqrt{15}=\frac{p}{q}-8$
$\Rightarrow \sqrt{15}=\frac{1}{2}\left(\frac{p}{q}-8\right)$
$\Rightarrow \sqrt{15}=\left(\frac{p}{2 q}-4\right)$
The $\text{RHS}$ is the difference of two rational numbers.
Therefore $\text{LHS}$ will also be rational.
But we know that $\sqrt{15}$ is irrational.
So our assumption is wrong.
Hence, $(\sqrt{3}+\sqrt{5})^2$ is an irrational number.
$\Rightarrow(\sqrt{3}+\sqrt{5})^2=\frac{p}{q}$
Where $p, q$ are co$-$prime
Using $(a+b)^2=a^2+b^2+2 a b$ we get,
$(\sqrt{3})^2+(\sqrt{5})+2 \sqrt{3} \sqrt{5}=\frac{p}{q}$
$\Rightarrow 3+5+2 \sqrt{15}=\frac{p}{q}$
$\Rightarrow 8+2 \sqrt{15}=\frac{p}{q}$
$\Rightarrow 2 \sqrt{15}=\frac{p}{q}-8$
$\Rightarrow \sqrt{15}=\frac{1}{2}\left(\frac{p}{q}-8\right)$
$\Rightarrow \sqrt{15}=\left(\frac{p}{2 q}-4\right)$
The $\text{RHS}$ is the difference of two rational numbers.
Therefore $\text{LHS}$ will also be rational.
But we know that $\sqrt{15}$ is irrational.
So our assumption is wrong.
Hence, $(\sqrt{3}+\sqrt{5})^2$ is an irrational number.
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