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√2 is irrational prove |
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Answer» Let us assume that √2 is rational no. √2=p/q. (q is not equal to 0) √2q=p. (S.B.S) 2q^2=p^2Since 2 divides p square ,So 2 also divides pAlso. p=2m. (M is some integer) p^2=4m^2. {Sbs} 2q^2=4m^2 q^2=2m^2Since 2 divides q square so 2 also divides q-+But we know that rational numbers have only one common factor that is oneSo this contradicts our fact that under root 2 is rationalSo our exemption may be wrong,Hence √2 is rational number Given √2 is irrational number.Let √2 = a / b wher a,b are integers b ≠ 0we also suppose that a / b is written in the simplest formNow √2 = a / b ⇒ 2 = a2 / b2\xa0⇒ 2b2 = a2∴ 2b2 is divisible by 2⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2 = 4c2 ⇒ 2b2 = 4c2 ⇒\xa0b2 = 2c2∴ 2c2 is divisible by 2∴ b2 is divisible by 2∴ b is divisible by 2∴a are b are divisible by 2 .this contradicts our supposition that a/b is written in the simplest formHence our supposition is wrong∴ √2 is irrational number. |
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