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