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Prove that `sqrt(2)`is an irrational number.

Answer» Let statement `p:sqrt(2)` is an irrational number.
Let if possible p is false. `therefore sqrt(2)` is a rational number.
`rArr.sqrt(2)=(a)/(b)` where a and b are prime integers and `bne0`.
`rArr 2=(a^2)/(b^2)`
`rArr a^2=2b^2`
`rArra^2` is divisible by 2.
`rArra` is divisible by 2.
Let `a=2c` where c is an interger.
`rArr a^2=4c^2`
`rArr 2b^2=4c^2" "(because a^2=2b^2)`
`rArr b^2=2c^2`
`rArrb^2` is divisible by 2.
`rArr` b is divisible by 2 .
Now a and b both are divisible by 2.
which contradiction our assumption that a and b are prime.
`therefore sqrt(2)` is rational number is false.
`rArr sqrt(2)` is an irrational number.


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