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If (p^2=3q^2) and (p=3k), why will (q) be divisible by (3)?

Answer and explanation

Correct answer: Because substitution gives \(q^2=3k^2\), so \(q^2\) is divisible by 3

Given \(p=3k\), substitute it into \(p^2=3q^2\): \((3k)^2=3q^2\), so \(9k^2=3q^2\). Dividing by 3 gives \(q^2=3k^2\); hence \(q^2\) is divisible by 3. Since 3 is prime, if the square of an integer is divisible by 3, then the integer itself is divisible by 3. Therefore, \(q\) is divisible by 3. Exam tip: for a prime \(r\), \(r\mid n^2\) implies \(r\mid n\).

Related tags

Number SystemsIrrationality ProofSquare Root 3Prime DivisibilityContradiction MethodProof Of Irrationality Of Square Root 2 And Square Root 3MathematicsClass 9 Mcq

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