Answer
shown below
Work Step by Step
To prove that\[\sqrt{2}\]is an irrational number.
Let\[\sqrt{2}\]is a rational number of the form \[a/b\], where a and b are integers and\[b\ne 0\].
It is written as:
\[\sqrt{2}=\frac{a}{b}\].
Square both the side of the above equation and simplify,
\[\begin{align}
& {{\left( \sqrt{2} \right)}^{2}}={{\left( \frac{a}{b} \right)}^{2}} \\
& 2=\frac{{{a}^{2}}}{{{b}^{2}}}
\end{align}\]
Multiply both the side with\[{{b}^{2}}\]:
\[\begin{align}
& 2\times {{b}^{2}}=\frac{{{a}^{2}}}{{{b}^{2}}}\times {{b}^{2}} \\
& 2{{b}^{2}}={{a}^{2}}
\end{align}\]
Now\[2{{b}^{2}}\]is divisible by 2, this implies that \[{{a}^{2}}\]is also divisible by 2.
It implies \[a\]is divisible by 2.
Therefore,
Let\[a=2c\]
Square both the sides of the equation:
\[\begin{align}
& {{\left( a \right)}^{2}}={{\left( 2c \right)}^{2}} \\
& {{a}^{2}}=4{{c}^{2}}
\end{align}\]
As it is calculated above\[{{a}^{2}}=2{{b}^{2}}\], this implies that:
\[\begin{align}
& 2{{b}^{2}}=4{{c}^{2}} \\
& {{b}^{2}}=2{{c}^{2}}
\end{align}\]
Therefore, \[2{{c}^{2}}\]is divisible by 2.
This implies that \[{{b}^{2}}\]is divisible by 2, which implies that b is divisible by 2.
Thus, a and b are divisible by 2.
This is a contraction that \[\sqrt{2}\]is a rational number, because it cannot be written as\[a/b\].
Hence, the number \[\sqrt{2}\]is an irrational.
