Question

Prove that 2 is an irrational number

Answer 1

Heya!

here's your answer :
$\sqrt{2}$
Let us assume that √2 is rational. So, we can find integers p and q (≠ 0) such that √2 = a/b
Suppose p and q have a common factor other than

1.Then, we divide by the common factor to get √2 =a/b where a and b are coprime.
So, b√2 = a.
Squaring on both sides, we get
2b2 = a2
Therefore, 2 divides a2.
Now, by Theorem which states that Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer,
 2 divides a2.
So, we can write a = 2c for some integer c
Substituting for a, we get 2b2 = 4c2 ,i.e. b2 = 2c2 .
This means that 2 divides b2, and so 2 divides b (again using the above Theorem with p = 2). Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that √2 is irrational.

Answer 2

Answer:

Let √2 be a rational number

Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get

p²= 2q² ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p² [since, 2q²=p²]

⇒ 2 is a factor of p

Let p =2 m for all m ( where m is a positive integer)

Squaring both sides, we get

p²= 4 m² ...(2)

From (1) and (2), we get

2q² = 4m² ⇒ q²= 2m²

Clearly, 2 is a factor of 2m²

⇒ 2 is a factor of q² [since, q² = 2m²]

⇒ 2 is a factor of q

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.