Question

prove that ✓2 is an irrational number ??






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Answer 1

Answer:

To prove that √2 is an irrational number, we will use the contradiction method.

Let us assume that √2 is a rational number with p and q as co-prime integers and q ≠ 0

⇒ √2 = p/q

On squaring both sides we get,

⇒ 2q2 = p2

⇒ p2 is an even number that divides q2. Therefore, p is an even number that divides q.

Let p = 2x where x is a whole number.

By substituting this value of p in 2q2 = p2, we get

⇒ 2q2 = (2x)2

⇒ 2q2 = 4x2

⇒ q2 = 2x2

⇒ q2 is an even number that divides x2. Therefore, q is an even number that divides x.

Since p and q both are even numbers with 2 as a common multiple which means that p and q are not co-prime numbers as their HCF is 2.

Since p and q both are even numbers with 2 as a common multiple which means that p and q are not co-prime numbers as their HCF is 2. This leads to the contradiction that root 2 is a rational number in the form of p/q with p and q both co-prime numbers and q ≠ 0.

Answer 2

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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 ...(1)

p²= 2q²

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.