prove that √2 is an irrational number.
Answers
Answer:
Given √2
To prove, √2 is an irrational number.
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 side 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.
Answer:
The square root of a number is the number that gets multiplied to itself to give the original number. The square root of 2 is represented as √2. The actual value of √2 is undetermined. The decimal expansion of √2 is infinite because it is non-terminating and non-repeating. Any number that has a non-terminating and non-repeating decimal expansion is always an irrational number. So, √2 is an irrational number.
Step-by-step explanation:
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 side 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.