Tuesday, January 29, 2019

discrete mathematics - Is this a valid proof for sqrt5 being irrational?




'Prove by contradiction that 5 is irrational.'



Proof: Assume that 5 is rational i.e. 5=p/q, where p,qZ.



Then 5q=p



Now for p to be an integer, 5q must be an integer, i.e q=5, 25, 35, ...



q must be irrational for p to be an integer.




p,qZ.



Contradiction. Therefore 5 is irrational. #


Answer



Your proof is not correct because you have to prove that q=k5 is irrational for every integer k by proving that 5 is irrational.



Note that if we multiply a non-zero integer and an irrational the product will be an irrational number.(You can prove this for practise)



So you assume what you want to prove.




Here it is a valid proof which i hope it will help you:




Assume that 5=mn where (m,n)=1.



We assume that (m,n)=1 because if its not then we can cancel a priori every common factor of the numerator and denominator until we remain with a fraction sl in its lowest terms namely (s,l)=1.



Thus 5n2=m25|m25|m because 5 is a prime number.




Also because 5|m we have that m2=25s2



Thus 5n2=25m2n2=5m25|n



Now we have a contradiction because 5|n and 5|m and we assumed that (n,m)=1



So 5 is an irrational.



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