1

I am trying to prove that there is no positive integer x such that $2x < x^2 < 3x$. I started by assuming that this statement is true. I then subtracted 3x from each part of the inequality to get $-x < x^2-3x < 0$, for every positive integer $x$. This is a contradiction since there is an x such that $x^2-3x > 0$. Is this the proper way to use proof by contradiction or is my reasoning wrong?

Mike Pierce
  • 18,938
user3699546
  • 559
  • 4
    Divide by $x$. – André Nicolas Mar 27 '15 at 01:14
  • 2
    Try placing $$$ before and after each of your expressions in your post. e.g. you can write $$$ 2x < x^2 < 3x $$$ which will result into $ 2x < x^2 < 3x $. Then you already learned your first $\LaTeX$ rules. – Pedro Mar 27 '15 at 01:14
  • Well done! Second tip for future posts: for longer exponents you need to use { and }. For instance x^23 is in LaTeX $$$ x^{23} $$$, which gives smoothly $ x^{23} $. – Pedro Mar 27 '15 at 01:19
  • If you want to show that there is no positive integer such that ... , the counter assumption is that there is such a positive integer. Not, that it is true for all positive integers. As for a hint how to proceed: as Andre Nicolas said you can just divide by $x$ to get a contradiction. – T. Eskin Mar 27 '15 at 01:24

1 Answers1

1

As mentioned in the comments, there is an easier way. Note that since $x$ is non-zero, we can divide by $x$ in each part of the inequality. So then if there does exists some positive integer $x$ such that $2x < x^2 < 3x$, then for that same $x$ we would have $2 < x < 3$. But we know that no positive integer satisfies this last inequality because $3$ is the successor of $2$. This is our contradiction.

Mike Pierce
  • 18,938