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xartaan
 2 years ago
Got a question about a simple proof.
Prove: If \[\frac{ x }{ x1 } \le 2\] then x < 1 or x >= to 2.
I approached this by multiplying both sides by (x1), 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?
xartaan
 2 years ago
Got a question about a simple proof. Prove: If \[\frac{ x }{ x1 } \le 2\] then x < 1 or x >= to 2. I approached this by multiplying both sides by (x1), 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?

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Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.2You have to differentiate between the case where x<1 and x>1 when you multiply by x1, because one case preserves the sign of the inequality and the other reverses it.

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.2My 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 (x1) which yields \[x \leq 2x2 \] 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 \[x \geq 2x2 \] which simplifies to \[x \leq 1 \implies x<1 \text{ since }x\neq1\] Therefore, the condition \[\frac{x}{x1} \leq 2 \implies x<1 \text{ or }x \geq 2\]

xartaan
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks a ton, I had a feeling something was missing!
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