'Prove by contradiction that √5 is irrational.'
Proof: Assume that √5 is rational i.e. √5=p/q, where p,q∈Z.
Then √5q=p
Now for p to be an integer, √5q must be an integer, i.e q=√5, 2√5, 3√5, ...
⟹ q must be irrational for p to be an integer.
⟹ p,q∉Z.
Contradiction. Therefore √5 is irrational. #
Answer
Your proof is not correct because you have to prove that q=k√5 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=m2⇒5|m2⇒5|m because 5 is a prime number.
Also because 5|m we have that m2=25s2
Thus 5n2=25m2⇒n2=5m2⇒5|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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