Proof by contradiction for simple inequality

[tomwilliam2]

Homework Statement

I'm trying to show that if $a \approx 1$, then
$$-1 \leq \frac{1-a}{a} \leq 1$$

I've started off trying a contradiction, i.e. suppose
$$ \frac{|1-a|}{a} > 1$$

either i)
$$\frac{1-a}{a} < -1$$
then multiply by a and add a to show
$$1 < 0$$
which is clearly false,

or ii)
$$1 < \frac{1-a}{a}$$
Which by the same reckoning leads me to
$$a < \frac{1}{2}$$

Which seems inconsistent with the original statement that $a \approx 1$.
It's hardly a rock solid proof though, is it?
Is there a better way of doing this?
Thanks

 

[HallsofIvy]

Yes, it is inconsistent! This is a proof by contradiction, remember? Your result that a< 1/2 contradicts the fact that a is close to 1.

 

[tomwilliam2]

Thanks HallsofIvy. So you think that is enough to prove the original statement then?
The reason I thought it was a bit wooly, was that $a \approx 1$ is unclear regarding exactly how close to 1 it is!
Also, given that my final conclusion limits $a$ to less than 1/2, I'm unclear as to why they chose 1 as the bound in the original question. They could have chosen a range of other numbers (i.e $-0.5 < (1-a)/a < 0.5$) and normally these things work out neater.
But if you think it's a sound proof, that's all that counts.
Thanks

 

[HallsofIvy]

No, your conclusion is NOT that a is less than 1/2! Your conclusion is that assuming that (1- a)/a> 1 leads to "a< 1/2" which contradicts the given fact that "$a\approx. 1$".

 

[tomwilliam2]

Sorry, I didn't make myself clear:
Yes, I am concluding that assuming (1-a)/a > 1 leads to a < 1/2. But my problem is with the second point. Does that necessarily contradict the fact that $a \approx 1$? That doesn't seem mathematically rigorous to me (although I'm only a beginner). Imagine, for example, that $a$ was a measure of the distance on a galactic scale...something like a=1/4 would be fairly close to 1, on that scale wouldn't it? Doesn't the "approximately equal to" have some more precise definition?
Thanks

 

[CompuChip]

You can make it more precise by defining "If $a \approx 1$ then X" to mean "There is an $\epsilon > 0$ such that X for all $a$ satisfying $|a - 1| < \epsilon$."

Then $a < 1/2$ contradicts $a \approx 1$ because for any candidate $\epsilon$, you can find a value $a$ such that $1/2 < a < 1$ for which $|a - 1| < \epsilon$ but not $a < 1/2$.

 

[tomwilliam2]

Thanks CompuChip, that tidies it up nicely.