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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.

Chinnapparaj R
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Carlos Bacca
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  • You need first to find the roots of the trionym, otherwise you cannot find the desired interval for $x$. – dmtri Nov 07 '18 at 15:07

1 Answers1

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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?

5xum
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  • Thanks, well we used the contrapositive to reverse the original statement. So is it the case that i cannot make this step in the proof as this step itself needs to be proved. thanks – Carlos Bacca Nov 07 '18 at 15:19