Let $x$ be any real number such that $x^2-3x+2<0$ then $1<x<2$.
In my class we proved this by the contrapositive.
I am wondering if this is necessary, why can't we just say:
Equivalent to $(x-2)(x-1)< 0$
This is equivalent to
$(x-2)<0 \;\text{and}\; (x-1)>0$ or $(x-2)>0 \;\text{and}\; (x-1)<0$
Which implies that $x$ must be in that range.. so basically why can't we just solve the inequality as usual to show that statement is true. Why should we go the long way around? Thanks.
How do you show that
$$a\cdot b < 0$$ is equivalent to $$(a<0 \text{ and } b>0) \text{ or }(a<0 \text{ and } b>0)?$$
without using the contrapositive?