Can someone explain why this line of reasoning concerning ordered relations is right?

Choose δ ∈ (0,1) such that 1 - δ < x < 1 implies 3/M < 2x² - 3x + 1 < 0; i.e., M/3 > 1/(2x² - 3x + 1). Notice that 0 < x < 1 also implies 2 < x + 2 < 3. It follows that f(x) = (x+2)/(2x² - 3x + 1) < M for all 1 - δ < x < 1.

I do not understand the last statement because it seems to say that if 0 < a < b and d < c < 0, then ad < bc. Is this true?

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Originally Posted by lm1988

Choose δ ∈ (0,1) such that 1 - δ < x < 1 implies 3/M < 2x² - 3x + 1 < 0; i.e., M/3 > 1/(2x² - 3x + 1). Notice that 0 < x < 1 also implies 2 < x + 2 < 3. It follows that f(x) = (x+2)/(2x² - 3x + 1) < M for all 1 - δ < x < 1.
I do not understand the last statement because it seems to say that if 0 < a < b and d < c < 0, then ad < bc. Is this true?

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Personally, I think that it is unreasonable for you to ask for explanation starting in the middle of someone's else proof.

Why not simply state the actual question and ask for help?

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Originally Posted by Plato

Personally, I think that it is unreasonable for you to ask for explanation starting in the middle of someone's else proof.

Why not simply state the actual question and ask for help?

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There is no question. I am just asking what the line of reasoning is as in the original post. The only additional information that you need to know is that M < 0 and is in R. This is actually how the proof begins.

This is what I am trying to prove:
http://latex.codecogs.com/gif.latex?...-3x+1}=-\infty

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Originally Posted by lm1988

There is no question. I am just asking what the line of reasoning is as in the original post. The only additional information that you need to know is that M < 0 and is in R. This is actually how the proof begins.

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Again, without knowing the reason for that argument, it impossible to comment on the argument itself.

Here is the best guess I have: if
$\displaystyle \begin{align*}\frac{3}{M} &< f <0\\ \frac{3}{fM} &> 1\\\frac{1}{fM} &>\frac{1}{3}\\ \frac{1}{f}&<\frac{M}{3} \end{align*}$.

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Originally Posted by Plato

Again, without knowing the reason for that argument, it impossible to comment on the argument itself.

Here is the best guess I have: if
$\displaystyle \begin{align*}\frac{3}{M} &< f <0\\ \frac{3}{fM} &> 1\\\frac{1}{fM} &>\frac{1}{3}\\ \frac{1}{f}&<\frac{M}{3} \end{align*}$.

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The proof was provided in a textbook, so I am assuming that the author intended for the reader to understand the reason for that argument.