Question

If $\sqrt{2}$ is irrational then state whether 2 $\sqrt{2}$ is rational or irrational.

Answer 1

Step-by-step explanation:

Let us assume on the contrary that 2 is a rational number. Then, there exist positive integers a and b such that

2=ba where, a and b, are co-prime i.e. their HCF is 1

⇒(2)2=(ba)2 

⇒2=b2a2 

⇒2b2=a2 

⇒2∣a2[∵2∣2b2 and 2b2=a2] 

⇒2∣a...(i) 

⇒a=2c for some integer c

⇒a2=4c2 

⇒2b2=4c2[∵2b2=a2] 

⇒b2=2c2 

⇒2∣b

Answer 2

Step-by-step explanation:

Proof:

Let us assume that √2 is a rational number.

So it can be expressed in the form p/q where p, q are co-prime integers and q≠0

√2 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√2 = p/q

On squaring both the sides we get,

=>2 = (p/q)²

=> 2q² = p²……………………………..(1)

p²/2 = q²

So 2 divides p and p is a multiple of 2.

⇒ p = 2m

⇒ p² = 4m² ………………………………..(2)

From equations (1) and (2), we get,

2q² = 4m²

⇒ q² = 2m²

⇒ q² is a multiple of 2

⇒ q is a multiple of 2

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√2 is an irrational number.

2√2 is also irrational

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