$4ab-2-5b=0$ have no solution $a,b$ are natural numbers .
Show that $4ab-2-5b=0$ has no solution .
By contradiction : $b(4a-5)=2$
So $4a-5=${1,2} which gives $a\in\mathbb{Q}$ ( contradiction) 
So the equation has no solution !
Is my proof correct ? 

The essence of your proof is correct, but you've written it down in an awful fashion. Contrary to popular belief, mathematics isn't just writing down some formulae and implications. Every proof must contain some explaining words.
In this case, a proper proof could look like this:
Assume there is an integer solution $a, b$ to the equation. Then the equations implies $b(4a - 5) = 2$. Since $b$ is a nonnegative integer, $4a - 5$ must also be a nonnegative integer [since their product would be nonpositive otherwise]. We also note that $4a - 5$ is a divisor of $2$; But since $2$ is a prime, it only has two nonnegative divisors, namely $1$ and $2$. Neither $4a - 5 = 1$ nor $4a - 5 = 2$ has a solution over the nonnegative integers, which is a contradiction to our assumption.
