Question
Prove that $\sqrt { 3 }$ is irrational.
✓
Answer
Let us assume $\sqrt{ } 3$ be a rational, then as every rational can be represented in the form $p / q$
where $q \neq 0$ Let $\sqrt{ } 3=p / q$ where $p, q$ have no common factor.
Now squaring on both sides we get $3=p^2 / q^2 $
$\Rightarrow 3 \times q^2=p^2$ Which means 3 divides $p^2$ which implies $3$ divides $p$
Hence we can write $p=3 \times k$, where $k$ is some constant.
This gives $3 \times q^2=9 \times k^2 q^2=3 \times k^2$ Which means $3$ divides $q^2$ which implies 3 divides $q$.
$3$ divides $p$ and $q$ which means $3$ is a common factor for $p$ and $q .$
And this is a contradiction for our assumption that $p$ and $q$ have no common factor...
Hence we can say our assumption that $\sqrt{ } 3$ is rational is wrong...
And therefore $\sqrt{ } 3$ is an irrational...
where $q \neq 0$ Let $\sqrt{ } 3=p / q$ where $p, q$ have no common factor.
Now squaring on both sides we get $3=p^2 / q^2 $
$\Rightarrow 3 \times q^2=p^2$ Which means 3 divides $p^2$ which implies $3$ divides $p$
Hence we can write $p=3 \times k$, where $k$ is some constant.
This gives $3 \times q^2=9 \times k^2 q^2=3 \times k^2$ Which means $3$ divides $q^2$ which implies 3 divides $q$.
$3$ divides $p$ and $q$ which means $3$ is a common factor for $p$ and $q .$
And this is a contradiction for our assumption that $p$ and $q$ have no common factor...
Hence we can say our assumption that $\sqrt{ } 3$ is rational is wrong...
And therefore $\sqrt{ } 3$ is an irrational...
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