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SamuelALDEN Group Title

The square root of two is a rational number.

  • 2 years ago
  • 2 years ago

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  1. UnkleRhaukus Group Title
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    False

    • 2 years ago
  2. SamuelALDEN Group Title
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    True or false?

    • 2 years ago
  3. Boblovesmath Group Title
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    Sorry I meant it is an irrational number. So it is false.

    • 2 years ago
  4. SamuelALDEN Group Title
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    Thank you!

    • 2 years ago
  5. UnkleRhaukus Group Title
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    prove by contradiction , assuming the square root of two is rational , it can be written as a ratio of natural numbers that have no common factors p,q say \[√2=p/q\] \[2=p^2/q^2\]\[2q^2=p^2\] this implies \(p^2\) is even if \(p^2\) is even then \(p\) is even if \(p \) is even it can be written as two times another natural number \(r\) \[p=2r\] \[2q^2=p^2=(2r)^2=4r^2\]\[q^2=2r^2\] which similarly implies that \(q\) is even With \(p,q\) both even the natural numbers necessarily have a common factor of two , which contradicts the statement of rationality, \[\sqrt2\not\in\text{rational numbers}\] \[\square\]

    • 2 years ago
  6. Boblovesmath Group Title
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    So in order to be rational the none of them should be even or at least one of them should be even and the other not even in order not to have a common factor of 2? Can you please explain this situation to me if you may?

    • one year ago
  7. UnkleRhaukus Group Title
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    if the numer is rational there must be a way to express it as a ratio of (relatively prime) natural numbers, (ie no common factors)

    • one year ago
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