Show that root 2 is an …

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/q
Here p and q are coprime numbers and q {tex}\ne{/tex} 0
Solving
{tex}\sqrt2{/tex} = p/q
On squaring both the side we get,
{tex}\Rightarrow{/tex} 2 = (p/q)²
{tex}\Rightarrow{/tex} 2q² = p² .. (1)
p²/2 = q²
So 2 divides p and p is a multiple of 2.
{tex}\Rightarrow{/tex} p = 2m
{tex}\Rightarrow{/tex} p² = 4m² ..(2)
From equations (1) and (2), we get,
2q² = 4m²
{tex}\Rightarrow{/tex} q² = 2m²
{tex}\Rightarrow{/tex} q² is a multiple of 2
{tex}\Rightarrow{/tex} q is a multiple of 2
Hence, 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} is 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/q Here p and q are coprime numbers and q 0 Solving = p/q On squaring both the side we get, 2 = (p/q)2 2q2 = p2 .. (1) p2/2 = q2 So 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 2 Hence, 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