[tex]\huge{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\fcolorbox{red}{black}{\red{Answer}}}}}}}}}}}}}}}}}}}[/tex] Let us assume√5 is rational. √5=[tex]\frac{p}{q}[/tex] [p and q are co-prime] p=√5q …(1) p²=5q [Squaring both the sides] [tex]\frac{p²}{5}=q²[/tex]…(2) p² divides 5, p also divides 5. p=5m [m is any integer] From equation 1, √5q=5m q=[tex]\frac{5m}{√5}[/tex] q=√5m q²=5m² [Squaring both the sides] 5 divides both p and q. But p and q are co-primes. It means our assumption is wrong. √5 is irrational _______________________________________ Reply
[tex]\huge{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\fcolorbox{red}{black}{\red{Answer}}}}}}}}}}}}}}}}}}}[/tex]
Let us assume√5 is rational.
√5=[tex]\frac{p}{q}[/tex] [p and q are co-prime]
p=√5q …(1)
p²=5q [Squaring both the sides]
[tex]\frac{p²}{5}=q²[/tex]…(2)
p² divides 5, p also divides 5.
p=5m [m is any integer]
From equation 1,
√5q=5m
q=[tex]\frac{5m}{√5}[/tex]
q=√5m
q²=5m² [Squaring both the sides]
5 divides both p and q.
But p and q are co-primes.
It means our assumption is wrong.
√5 is irrational
_______________________________________