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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?
 one year ago
 one year 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?
 one year ago
 one year ago

This Question is Closed

Jemurray3Best ResponseYou've already chosen the best response.2
You 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.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.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 (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 \[\frac{x}{x1} \leq 2 \implies x<1 \text{ or }x \geq 2\]
 one year ago

xartaanBest ResponseYou've already chosen the best response.0
Thanks a ton, I had a feeling something was missing!
 one year ago
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