kanishka2002
Let root 2 be a rational no. therefor root2=p/q where p and q be co primes and q is not = to 0 i.e.q root2=p squaring both sides 2q square=p square............1 i.e.2 divides p sq. 2 divides p therefore p=2c for any +ve integer c put p=2c in eq.1 2q sq.=4c sq. i.e. q sq.=4c sq./2 q sq.=2c sq. i.e.2 divides q 2 divides p and q both i.e. 2 is common factor of p and q but p and q are co primes hence contradiction root 2 is irrational
Answers & Comments
therefor root2=p/q where p and q be co primes and q is not = to 0
i.e.q root2=p
squaring both sides
2q square=p square............1
i.e.2 divides p sq.
2 divides p
therefore p=2c for any +ve integer c
put p=2c in eq.1
2q sq.=4c sq.
i.e. q sq.=4c sq./2
q sq.=2c sq.
i.e.2 divides q
2 divides p and q both
i.e. 2 is common factor of p and q
but p and q are co primes
hence contradiction
root 2 is irrational
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