# Is this argument valid for a proof of a double inequality

I have solved the following double inequality $1\leq x < x^{2}$ as follows:

$1\leq x$ is already solved

$x< x^{2}$ using a table of sings gives me $(-\infty,0) \cup (1,\infty)$

together the solution is $(1,\infty)$.

Now my question: is the following argument correct?

$1\leq x$ is already solved

For the second inequality $x< x^{2}$ I use from the first one that $1\leq x$ so I can divide both sides of the inequality (I divide by a number different from zero and a positive number which preserves the inequality) and I get $1 < x$.

Again together the solution is $(1,\infty)$.

It is valid to use the first inequality as a condition to solve the second one?

• Yes, it is absolutely valid. – Win Vineeth Feb 17 '16 at 5:52
• However $x=1$ is a solution. – fosho Feb 17 '16 at 6:13
• My bad, it should be strict inequality $x < x^{2}$ – Novato Feb 17 '16 at 6:15