Question
Prove that $\sqrt{3}+\sqrt{2}$ is irrational.
✓
Answer
Let $\sqrt{3}$ is a rational number.
So, two integers $a$ and $b$ can be found so that $\sqrt{3}=\frac{a}{b}$Assume that a and are co$-$prime.
$\Rightarrow a=\sqrt{3} b$
Squaring both the sides,
$\Rightarrow a^2=3 b^2$
So, $a^2$ is divisible by $3$ and it can be said that a is divisible by $3 .$
Let $a^2=3 c$, where $c$ is an integer.
$a^2=3 b^2$
$\Rightarrow(3 c)^2=3 b^2 \Rightarrow b^2=3 c^2$
So, $b^2$ is divisible by $3$ and it can be said that $b$ is divisible by $3 .$
So, two integers $a$ and $b$ can be found so that $\sqrt{3}=\frac{a}{b}$Assume that a and are co$-$prime.
$\Rightarrow a=\sqrt{3} b$
Squaring both the sides,
$\Rightarrow a^2=3 b^2$
So, $a^2$ is divisible by $3$ and it can be said that a is divisible by $3 .$
Let $a^2=3 c$, where $c$ is an integer.
$a^2=3 b^2$
$\Rightarrow(3 c)^2=3 b^2 \Rightarrow b^2=3 c^2$
So, $b^2$ is divisible by $3$ and it can be said that $b$ is divisible by $3 .$
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