Let us assume that {tex}\\sqrt2{/tex} is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q {tex}\\ne{/tex} 0{tex}\\sqrt2{/tex} = p/qHere p and q are coprime numbers and q {tex}\\ne{/tex}\xa00Solving{tex}\\sqrt2{/tex}\xa0= p/qOn squaring both the side we get,{tex}\\Rightarrow{/tex}\xa02 = (p/q)2{tex}\\Rightarrow{/tex}\xa02q2\xa0= p2\xa0.. (1)p2/2 = q2So 2 divides p and p is a multiple of 2.{tex}\\Rightarrow{/tex}\xa0p = 2m{tex}\\Rightarrow{/tex}\xa0p2\xa0= 4m2\xa0..(2)From equations (1) and (2), we get,2q2\xa0= 4m2{tex}\\Rightarrow{/tex}\xa0q2\xa0= 2m2{tex}\\Rightarrow{/tex}\xa0q2\xa0is a multiple of 2{tex}\\Rightarrow{/tex}\xa0q is a multiple of 2Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number{tex}\\sqrt2{/tex}\xa0is an irrational number.
Let us assume that ✓2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q 0 = p/qHere p and q are coprime numbers and q 0Solving = p/qOn squaring both the side we get, 2 = (p/q)2 2q2 = p2 .. (1)p2/2 = q2So 2 divides p and p is a multiple of 2. p = 2m p2 = 4m2 ..(2)From equations (1) and (2), we get,2q2 = 4m2 q2 = 2m2 q2 is a multiple of 2 q is a multiple of 2Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number Hence it is proved that ✓2 is irrational number