Question

Prove that √3 is irrational​

Answer 1

`\textsf{Proof that }\sqrt3\textsf{ is irrational by contradiction method.}\\`

`\textsf{Let }\sqrt3\textsf{ be a rational number. Then }\sqrt3\sf{ =(p)/(q),}\textsf{ where p and q are integers and q}\\e0.`

`\sf\\\sqrt3=(p)/(q)\\\\\implies \sqrt3 \ q=p\\\\\textsf{Squaring both sides,}\\p^2=3q^2..............(1)\\`

`\sf\\\textsf{As q is an integer, }q^2\textsf{ is also an integer. Therefore, }p^2/3\textsf{ is also an integer.}`

`\sf\\\textsf{That is, since }p^2=3q^2,\textsf{which is multiple of 3, means p itself must be a multiple}\\\textsf{of 3 such as p = 3n.}`

`\sf\\\textsf{Now we have }p^2=(3n)^2=9n^2.....(2)\\\textsf{From (1) and (2),}\\9n^2=3q^2\\\implies 3n^2=q^2\\\textsf{This means, q is also multipe of 3, contradicting the fact that p and q had no}\\\textsf{common factors.}\\\textsf{Hence,}\ \sqrt3\textsf{ is an irrational number.}`