Question

Prove that root 5 is irrational.​

Answer 1

Answer:

let assume that √5 is irrational

so it can be expressed as in the form of p/q

where p, q are co-prime integers q≠0

√5= p/q

squaring on both sides

5= p²/q²

5q²=p²——————(1)

p²/5= q²

so 5 divided p

p is a multiple of 5

p = 5m

p² =25m²——————(2)

from equation(1) and (2) we get

5q²=25m²

q²=5m²

q² is a multiple of 5

q is a multiple of 5

Hence p , q have a common factor 5 . they are co primes therefore p/q is not a rational number.

√5 is an irrational number

Hence proved

Answer 2

We have to prove √5 is irrational

let us assume the opposite, i.e., √5 is irrational

√5 can be written as a/b

whereas a and b ( b ≠ 0) are co-prime ( no common factors other than 1 ) .

So, √5 = a/b

√5b = a

(squaring both sides)

(√5b² = a²

5b² = a²

a/5 = b²

So, 5 divides a

[ By theorem - if a prime number , and p divides a² , then p divides a , where a is a positive number]

So , 5 shell divide a also

Hence , we can say

a/5 = c where c is some integers

so , a = 5c

Now we know that

5b² = a²

putting a = 5c

5b² = (5c)²

5b²= 25c²

5b² = 25c²

b² = 1/5 × 25c

b² = 5c²

b²/5 = c²

Hence , 5 divides b²

[ by theorem : if p is prime , and p divides a², then p divides a , where a is a positive number]

So ,5 divides b also

By (1) and (2)

5 divided both a and b

So , a & b have a factor 5

therefore, a & b are not co-prime.

Hence , our assumption is wrong

therefore , by contrediction

√5 is irrational