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Given some compound inequality like $2x < x^2 < 3x, x \in \mathbb{Z}^+$, what is the reason why the common factor $x$ cannot be divided out? I haven't taken a general algebra course yet, but my hunch would be that the integers not being closed under division wouldn't imply a common factor $x$ couldn't be divided out given that we know $2, 3, x \in \mathbb{Z}^+$ and $\frac{2x}{x}=2, \frac{3x}{x}=3$, and $\frac{x^2}{x}=x$.

The context of the original problem is to prove that no $x$ could exist that satisfies the inequality. I understand his intended solution to the proof, my question is only about why this particular approach is invalid. My professor asserts that dividing away the $x$ is not allowed because our domain is the integers, but clearly dividing by $x$ in this context keeps us in the positive integers. I'd like to better understand what I'm missing.

Thank you!

EDIT: My original post didn't specify that x was a positive integer, several contributors made notes about the possibility of $x = 0$.

Reece McMillin
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    The only things I can think of are, $x$ may be zero, in which case you can never divide by $x$, or $x$ may be negative, in which case division by $x$ swaps the inequalities. But those can be dealt with in separate cases. – Akiva Weinberger Oct 27 '21 at 17:21
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    Assuming $x>0$, $~a,b,x\in\Bbb Z$, it is always true that $ax<bx$ implies $a<b$. This follows because $\Bbb Z\subseteq\Bbb Q$ and $\Bbb Q$ is an ordered field. Perhaps you want a proof that stays within $\Bbb Z$, but that can be done. That lemma is justification for dividing in inequalities. – Akiva Weinberger Oct 27 '21 at 17:24
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    Don't you see a problem in posting a question and then, after getting an answer, to change it to something else to which that answer does not apply? – José Carlos Santos Oct 27 '21 at 17:54
  • @JoséCarlosSantos Sorry, I made a typo and didn't realize until after the fact, hence my edit note. Would you rather I delete the question along with the other answers? Yours was still relevant and helpful to my understanding. – Reece McMillin Oct 27 '21 at 18:09

2 Answers2

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You cannot divide by $0$. And, if $x$ is negative, the you can divide by $x$… but you must reverse the inequalities then, getting $2>x>3$.

José Carlos Santos
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You are given that $x$ is positive integer , so there is no harm in dividing it by zero.

In that case your answer simply becomes

$2<x<3$

However if the question would have mentioned that $x$ is a non-negative integer, than you must split your question into two cases $\implies$ $x=0$ or $x>0$

The former one gives $0<0<0$ which is false and second one gives $2<x<3$ which is also false since $x$ is an integer.

Lalit Tolani
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