Question

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?

Answer 1

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.

Answer 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
\[\frac{x}{x-1} \leq 2 \implies x<1 \text{ or }x \geq 2\]

Answer 3

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