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 $3=p^2 / q^2$
$\Longrightarrow 3 \times q^2=p^2$
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...
Now squaring on both sides we get $3=p^2 / q^2$
$\Longrightarrow 3 \times q^2=p^2$
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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