Need Help Using Proof by Contraposition > Let $x$ be a real number. Prove that if $x^3+5x+1\le0$, then $x<0.$ My solution: Suppose that $x\ge0$, what should I do next?

Well, if $x\ge 0,$ then $x^3+5x+1\ge 1>0$. So, by contraposition we have shown the statement. More generally, the idea is that a statement $p\implies q$ is equivalent to the statement $\
eg q\implies \
eg p$.

Here, we assume $\
eg q$. That is, that $x$ is not less than $0$. That's the same thing as saying that $x$ is greater than or equal to $0$. $x\ge 0$.

Using this, we'd like to show the negation of $x^3+5x+1\le 0$, $\
eg p$. That is, we'd like to show that assuming that $x\ge 0$, $x^3+5x+1>0$.