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Class 10MathematicsReal NumbersProof of irrationality of √2, √3, √5Hard
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In the proof of (\sqrt{5}), if (p=5k) and (q=5r) are obtained, which initial condition breaks?

Answer and explanation

Correct answer: (p) and (q) are coprime

Step 1: (p=5k) and (q=5r) show factor (5) in both (p) and (q). Step 2: So they cannot be coprime. Step 3: This breaks the initial lowest-form condition.

Related tags

Sqrt5 ProofCoprime ConditionHardClass 10Proof Of Irrationality Of √2√3√5Proof Of Irrationality Of 2 3 5Real NumbersMathematics
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10 more MCQs
  1. 1Which assumption is taken first to prove that (\sqrt{2}) is irrational?
  2. 2If (\sqrt{2}=\frac{p}{q}), what is obtained after squaring both sides?
  3. 3From the equation (p^2=2q^2), what do we learn about (p^2)?
  4. 4If (p^2) is even, which conclusion about (p) is correct?
  5. 5If (p) is even, in which form can (p) be written?
  6. 6What conclusion is obtained by putting (p=2k) in (p^2=2q^2)?
  7. 7What is the final contradiction in the proof of (\sqrt{2})?
  8. 8What is the correct final conclusion of the proof of (\sqrt{2})?
  9. 9If (\sqrt{3}) is assumed rational, in which form is it written?
  10. 10If (\sqrt{3}=\frac{p}{q}), which equation is obtained after squaring?
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