The square root of two is a rational number.

Question

unklerhaukus · Answer

False

anonymous · Answer

True or false?

anonymous · Answer

Sorry I meant it is an irrational number. So it is false.

anonymous · Answer

Thank you!

unklerhaukus · Answer

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 · Answer

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 · Answer

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)