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BoblovesmathBest ResponseYou've already chosen the best response.0
Sorry I meant it is an irrational number. So it is false.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
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, \[\sqrt2\not\in\text{rational numbers}\] \[\square\]
 one year ago

BoblovesmathBest ResponseYou've already chosen the best response.0
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

UnkleRhaukusBest ResponseYou've already chosen the best response.1
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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