If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|?

Question

If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|

Is that true for all real values? If so, can someone point me to a proof?

(EDIT: I made a mistake in my context. When comparing "closeness", I should have put 'a' in the denominator. Then I'm pretty sure |a - b| < |a - c| implies |1 - $\frac{b}{a}$| < |1 - $\frac{c}{a}$|. However, I'm going to let the question remain, as it might be interesting to see what happens when a is put to the numerator.)

I used to think so but I may have come up with a counter-example. It's very confusing though so I want to first provide some background. Also, I don't know all the official terminology for these things, which is why the background might help.

The idea is how to measure the "closeness" of an approximation to some true value. There are two ways of doing it, absolute difference and relative ratio.

So if our true value is 2.5, and we want to test approximations of 2.25 and 2.75, we could say that 2.5 - 2.25 is 0.25 and 2.75 - 2.5 is also 0.25, so they are equally close.

But we could also try looking at the ratios. $\frac{2.5}{2.25}$ = 1.111..., and $\frac{2.5}{2.75}$ = 0.9090909... So this time, the second number is closer, when viewing it from the perspective of proportions.

Another concrete example. Our true value is 3, and our test values are 2.9 and 3.05. This time 3.05 is closer in both ways. In other words, the rule holds true:

|3 - 3.05| < |3 - 2.9|, then |1 - $\frac{3}{3.05}$| < |1 - $\frac{3}{2.9}$|

It seems impossible to ever violate that rule. If there is a smaller absolute distance between two numbers, then there should be a smaller percentage difference as well.

Now for my supposed counter-example.

a = $\sqrt{2}$, b = $\frac{7}{5}$, c = $\frac{10}{7}$

|a - b| = 0.014 213 562...

|a - c| = 0.014 357 866..., which implies b is closer.

|1 - $\frac{a}{b}$| = 0.010 152 545...

|1 - $\frac{a}{c}$| = 0.010 050 506..., which implies c is closer.

How is it possible? Does it have something to do with the coincidence that $\frac{b}{a}$ = $\frac{a}{c}$ due to square root "normalization"? By that I mean:

$\displaystyle \frac{\frac{7}{5}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{7*\sqrt{2}}{5*2} = \frac{\sqrt{2}}{\frac{10}{7}}$

If so I can't fathom how. I feel like I might be missing something obvious.

Btw if someone can clarify what the official terminology is, that would be a big help. I might be able to search and find answers on my own if I just knew the proper names of these things.

Answer 1

Hint: You are given that $ | a - b | < |a - c|$, and are asking if

$$ \frac { | a - b | } { |b| } < \frac{ |a-c|} { |c| }. $$

An obvious place to start would be if $ |b| < |c|$, then this statement might not follow.

This suggests we should look for a counterexample where $|b| < |c|$, so take a really small $b$ and a really large $c$.

Answer 2

Let $a=1$, $b=10^{-3}$, $c=2$

$$|1-10^{-3}|<|1-2|$$

but not

$$ |1-10^3|<|1-1/2|.$$

Answer 3

If $\left|c\right| \leq \left|b\right|$ we have

$$\frac1{\left|b\right|} \leq \frac1{\left|c\right|}$$

which can be combined with $\left|a - b\right| < \left|a - c\right|$ to derive $\left|1-\frac{a}{b}\right| < \left|1 - \frac{a}{c}\right|$