Prove that underroot3 is a irrational no

1.	Prove that underroot3 is a irrational no
Answer» Let √3 be rational no. Then it exist in the form p/q where q≠0√3=p/q. (here slash means upon) (on squaring both side)(√3)²= (p/q)²3=p²/q²q²=p²/3. -------(1)From (1) we noticed that p² is divisible by 3 therefore p is also divisible by 3 -----(2)Now put p=3r in (1)Therefore q²= (3r)²/3So, q²= 9r²/3 q²= 3r² (÷3) q²/3= r² OR r²= q²/3-----(3)From (3) we noticed that q² is divisible by 3 therefore q is also divisible by 3 ------(4)So by (2) & (4) it contradicts our suppose that √3 is rational but it is irrational because both p and q are divisible by 3Hence proved

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