Question

prove √2 is irrational no.

Answer 1

hello!

HERE IS THE SOLUTION BY THE CONTRADICTION METHOD:

if possible, let √2 be not an irrational.

> √2 is rational

√2 can be written as p/q (where p&q are integers and q not equal to 0)

on squaring both sides:

(√2)^2 = (p/q)^2

2 = p^2 / q^2

Thus, p square = 2q square. ....... (1)

since 2 is a factor of p square
therefore, 2 is also a factor of p. ..... (2)

let p = 2m, where m is any +ve integer

on putting p = 2m, in eqn (1)

(2m)^2 = 2q^2

4m^2 / 2 = q^2

2m^2 = q^2

since 2 is a factor of q square
therefore, 2 is also a factor of q .... (3)

from (2) and (3) we have:

2 is a factor of both P and q

this contradicts that HCF (p,q) = 1

> so the assumption that√2 is a rational number is wrong.

Therefore, √2 is an irrational number.


hope it helps ....

Answer 2

If possible let √2 be rational and let and simplest form p/q
then p and q is a integers no common factor other then ,1and q not equal 0
Now, √2=p/q

2=p²/q² (no squaring both side)
2q²=p²
2divide p² (·.·2divideds2q²)

2divides p(…2is prime and dividesq²=2divided q)
let a=2c for some integer c .

puting p=2c in (¡) we get
2q²=4c² == q²=2c²
2dividedq² ( ·.·2divides2c²)
2dividedq. (·.·2is prime and divide q²=== 2divides q)
thus 2is a common factor of p and q.
the contradiction arises by assuming that √2 is rational
Hence √2is irrational.