prove that underroot 3+underroot7 is an irrational number
Answers
Step-by-step explanation:
Sol: let us assume that √7 be rational. then it must in the form of p / q [q ≠ 0] [p and q are co-prime] √7 = p / q => √7 x q = p squaring on both sides => 7q2= p2 ------> (1) p2 is divisible by 7 p is divisible by 7 p = 7c [c is a positive integer] [squaring on both sides ] p2 = 49 c2 --------- > (2) subsitute p2 in equ (1) we get 7q2 = 49 c2 q2 = 7c2 => q is divisble by 7 thus q and p have a common factor 7. there is a contradiction as our assumsion p & q are co prime but it has a common factor. so that √7 is an irrational.
Hii friend,
If possible , let 3✓7 be rational Number. Then 3✓7 is rational , 1/3 is rational.
=> (1/3 × 3✓7) is rational . [ Because product of two rationals is rational]
=> ✓7 is rational.
This contradicts the fact that ✓7 is irrational.
This contradiction arises by assuming that 3✓7 is rational.
Hence,
3✓7 is irrational...... PROVED......
HOPE IT WILL HELP YOU.... :-)