How do you proof that for a,b,cinRR, a/b=b/c=c/a<=>a=b=c?

Dec 19, 2015

The reverse direction is trivial, assuming we add the condition that $a = b = c \ne 0$ as then $\frac{a}{b} = \frac{b}{c} = \frac{c}{a} = 1$.

For the forward direction, note first that all of $a , b , c$ must be of the same sign, as otherwise one of $\frac{a}{b} , \frac{b}{c} , \frac{c}{a}$ would differ in sign from the others.

Now, there are two possible cases in which the claim would not hold. In the first case, one differs from the other two. Without loss of generality, suppose $a = b$ and $b \ne c$. Then $\frac{a}{b} = 1 \ne \frac{b}{c}$, a contradiction.

In the second case, they all differ. Then, without loss of generality, suppose $a > b > c$. Then $\frac{a}{b} > 1$ and $\frac{c}{a} < 1$ meaning $\frac{a}{b} \ne \frac{c}{a}$, a contradiction.

Thus the only way to avoid a contradiction is if $a = b = c$

Dec 20, 2015

Here's an alternative proof not using proof by contradiction...

Explanation:

Let me try to prove this without using contradiction:

First, if $a = b = c$ then $\frac{a}{b} = \frac{b}{c} = \frac{c}{a} = 1$.

To show in the other direction, suppose $\frac{a}{b} = \frac{b}{c} = \frac{c}{a} = k$ for some constant $k$.

Then $a = b k$, $b = c k$ and $c = a k$

So we find:

$a = b k = c {k}^{2} = a {k}^{3}$

Dividing both ends by $a$ we get:

${k}^{3} = 1$

Since we are told that $a , b , c \in \mathbb{R}$, we must have $k \in \mathbb{R}$, so $k = \sqrt[3]{1} = 1$ is the Real cube root of $1$.

So $a = b k = b = c k = c$

$\textcolor{w h i t e}{}$
Notice my comment above about the Real cube root. If we were not told that $a , b , c \in \mathbb{R}$, then $k = \omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ (the primitive Complex cube root of $1$) would also work and we would not get $a = b = c$.

For example, $a = 1$, $b = \omega$, $c = {\omega}^{2}$ would satisfy
$\frac{a}{b} = \frac{b}{c} = \frac{c}{a}$ but not $a = b = c$