anonymous
  • anonymous
The square root of two is a rational number.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
False
anonymous
  • anonymous
True or false?
anonymous
  • anonymous
Sorry I meant it is an irrational number. So it is false.

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anonymous
  • anonymous
Thank you!
UnkleRhaukus
  • UnkleRhaukus
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, \[\sqrt2\not\in\text{rational numbers}\] \[\square\]
anonymous
  • anonymous
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?
UnkleRhaukus
  • UnkleRhaukus
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)

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