prove that √2 is an irrational number ???
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To prove that √2 is an irrational number, we will use the contradiction method. ⇒ p2 is an even number that divides q2. Therefore, p is an even number that divides q. ... 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.
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√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.
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