Question Regarding Proof Using Quadratic Formula

Question

I'm learning proof techniques. I'm following this PDF at: http://persweb.wabash.edu/facstaff/turnerw/Writing/proofs.pdf

On the Theorem $4$ example (pg. 7), I am confused!

I see that we assume $x \gt 0$ by using the negative (or minus) version of the quadratic formula.

But, there are three things I don't understand.

1. Why is $\frac{2a - 2b} { 2a}$ simplified into $1 - (\frac{b}{a})$? When I simplify this equation, I get $\frac{(a-b)}{a}$.

2. Why is the value of b important, or is it not? $b = a(1 − x)$

3. And where it's stated at the bottom that a divides b, is that referring to the equation $x = 1 - (\frac{b}{a})$?

Thank you for the help.

Update: Thank you to Eevee for the answer. I still have one more question which is -- why do we choose the negative (or minus) quadratic formula for when x > 0? I had sort of inferred that we did because otherwise x would be negative. But I'm not sure.

Update: Thank you Sten for the answer. On that note, was it assumed that I would have worked the quadratic formula out both ways? I don't see anything that infers that I would choose the negative (or minus) version of the formula.

Answer 1

I see that we assume x > 0 by using the negative (or minus) version of the quadratic formula.

No, we assume $x$ is a positive integer from the very beginning. The goal is a proof by contrapositive:

• What we wish to prove: *"If $a$ does not divide $b$ then $ax^2 + bx + b - a = 0$ has no positive integer solution $x$."
• The equivalent contrapositive: If $x$ is a positive integer solution to $ax^2 + bx + b - a = 0$, then $a$ divides $b$.

Notice that, since we're proving the latter, we want to start with the hypothesis of $x$ being a positive integer solution to the quadratic.

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Why is 2a - 2b / 2a simplified into 1 - (b/a)? When I simplify this equation, I get (a-b)/a.

Split up the fraction:

$$\frac{a-b}{a} = \frac a a - \frac b a = 1 - \frac b a$$

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Why is the value of b important, or is it not? b = a(1 − x)

The definition of "$p$ divides $q$" is that "there exists an integer $k$ such that $q = pk$." If the goal is to show that $a$ divides $b$, then $1-x$ is that integer (since the proof also assumes $a,b,x$ are integers).

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And where it's stated at the bottom that a divides b, is that referring to the equation x = 1 - (b/a)?

It refers, more precisely, to the fact that $b = a(1-x)$, i.e. $b$ is an integer multiple of $a$.