Question

if K is an irrational number prove that 2K is also an irrational number​

Answer 1

Answer:

2|a and 2|b

Step-by-step explanation:

Now, assume 2k−−√ is rational. Then, 2k−−√=ab where a,b∈Z, b not equal 0 and a,b have no common factors.

2k−−√=ab⟹2k=a2b2⟹(b2)(2k)=a2⟹2|a2⟹2|a, since 2 is prime⟹∃c∈Z such that a=2c

Then, (2k)(b2)=a2⟹(2k)(b2)=4c2⟹kb2=2c2⟹2|kb2⟹2|b2⟹2|b, since 2 is prime.

So, 2|a and 2|b , a contradiction.