Question

prove that √2 is an irrational number ??? ​

Answer 1

Answer:

Lets 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.

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.

Thus, √2 is an irrational number by the contradiction method.

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Answer 2

Step-by-step explanation:

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