Math, asked by ujjavallking, 2 months ago

prove that √5 is a rational number​

Answers

Answered by malothulutherking
1

Answer:

let root5 is irrational. so root 5 = p/q .p and q are rational . Gcd of p and q is 1

Step-by-step explanation:

q ×root 5 =p

q × root 5-p=0

s.o.b.s.

5q^2+p^2-2pqroot5=0

2pqroot5=5q^2+p^2

root5=5q^2+p^2 /2pq

lhs is irrational and rhs is rational this is not possible

hence our assumption is wrong.

so root5 is irrational number

Answered by muzamilahmad051762
0

Answer:

if p is a prime no and divides a² then p divides a,where a greater than 0

Step-by-step explanation:

if possible,suppose that√5 be a rational no.

~ we can find 2 co- prime no's a and b,b is not equal to 0

such that a/ b = √5

a = b√5→ 1

squaring on both sides

a²= (b√5)²

a² = b²5

5 divides a² → x

therefore,a= 5c → 2

put value of a in equation 1 we get

5c = b√5

squaring on both sides

5c² = (b√5)²

25c² = b²5

dividing 5 on both sides

5c² = b²

5 divides b²,

therefore,5 divides b²→ xx

from x and xx we get

5 is a factor of a as well as b

so,a and b are not co- primes

therefore our supposition is wrong

hence√5 is a irrational no.

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