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Prove that√5 is irrational |
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Answer» Let us consider that √5 is a “rational number”.We were told that the rational numbers will be in the “form” of \\frac {p}{q}form Where “p, q” are integers.So, \\sqrt { 5 } = \\frac {p}{q}p = \\sqrt { 5 } \\times qwe know that \'p\' is a “rational number”. So 5 \\times q should be normal as it is equal to pBut it did not happens with √5 because it is “not an integer”Therefore, p ≠ √5qThis denies that √5 is an “irrational number”So, our consideration is false and √5 is an “irrational number”. Let us assume√5 is irrational number.So, √5 is in the form of p/q where p and q are some Integers where q≠0.Since ,We divide numerator and denominator by common factor to get a and b where a and b are co -prime.So , √5=a/b √5b = aSquaring both side,(√5b)^2 =a^2 Prove that√5 is irrational |
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