Question

prove that 3√5 is irrational​

Answer 1

Step-by-step explanation:

( first we have to prove $\sqrt{5}$ is rational )

assume that $\sqrt{5$ is rational .

then \sqrt{5 = p/q , where q ≠ 0 , p and q are integers and HCF (p,q) = 1

⇒ q√5=p

on squaring both sides ,

5q²=p²

ie , p²=5q² ---{1}

here ,  5 is a prime no. and 5 divides p² . we know the theorem if p is a prime & p divides  a² , then p divides a also .   ∴  5 divides p also  

p = 5 × a , where a is an integer

p² = (5a)²

p²= 25a²

substituting {1} in p

5q² = 25a²

q²= (5a)² ---{2}

here 5 is a prime number and 5 divides q² .

by the theorem if p is a prime and p divides a ² then p divides a also  , we got 5 divides q also .

∴ we've 5 divides p and 5 divides q ⇒ HCF (p,q) ≠ 1 , which is a contradiction to our assumption that HCF (p,q) .   ∴ our assumption is wrong that  "√5 is a rational "  is wrong .

                                      ∴  √5 is irrational

assume that  3√5 is rational

then 3√5 = p/q , where q ≠ 0 , p & q are integers , HCF ( p,q) = 1

√5=p/q x 3

√5 = 3p / q

p & q are integers and 3 is also an integer . so 3p/q is also an integer

 ∴ 3p/q is a rational no .

but √5 is an irrational ( proved ) , which is contradiction to the fact that √5 is irrational

 ∴ our assumption is wrong

 ∴  3√5 is irrational

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Answer 2

Answer:

I hope it helps u dear friend ^_^