Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

xartaan

  • 3 years ago

Got a question about a simple proof. Prove: If \[\frac{ x }{ x-1 } \le 2\] then x < 1 or x >= to 2. I approached this by multiplying both sides by (x-1), and factoring out x, which simplified to 2 <= x. I feel like this satisfies the hypothesis, but I feel like something is missing. I guess I just don't see how this is PROOF. Is it just that simple and I am missing the fundamental connection, or am I just way off base?

  • This Question is Closed
  1. Jemurray3
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    You have to differentiate between the case where x<1 and x>1 when you multiply by x-1, because one case preserves the sign of the inequality and the other reverses it.

  2. Jemurray3
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    My proof would go something like: First, note that x cannot equal 1, as the expression is then undefined. Let us assume x >1. Then we may multiply both sides of the inequality by (x-1) which yields \[x \leq 2x-2 \] which can be simplified to \[x \geq 2\] On the other hand, lets assume x < 1. Then when we multiply we must reverse the sign of the inequality, yielding \[x \geq 2x-2 \] which simplifies to \[x \leq 1 \implies x<1 \text{ since }x\neq1\] Therefore, the condition \[\frac{x}{x-1} \leq 2 \implies x<1 \text{ or }x \geq 2\]

  3. xartaan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks a ton, I had a feeling something was missing!

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy