How can it be that Doppler Factor $k=\frac{1}{k}$?

Question

I am currently studying special relativity mainly using this paper which uses spacetime diagrams to derive the formulas for the properties of special relativity (Time dilation, length contraction etc.).

In chapter II, the "Doppler k-Factor" is derived as $$k=\sqrt{\frac{1+\beta}{1-\beta}} \tag{9}$$ where $\beta=\frac{V}{c}$ ($V$ is the speed of which an inertial frame of reference $K'$ is receding from a "stationary" frame $K$.
(I use the same tags for equations here as used in the linked paper to avoid confusion).

Earlier in the paper, $k$ has been defined as $$k=\frac{cT_2}{cT'}=\frac{cT'}{cT_1}\tag{7}$$

You can get from here to euquation $9$ by multiplying the two "versions" of $k$ and then taking the square root.
$cT_2$, $cT_1$ and $cT'$ are different "events" that happen when a ligght ray is sent from $K$ to $K'$ and reflected from $K'$ back to $K$:

(Illustration from the linked paper)

Where it gets confusing for me is chapter II,I where the Doppler Factor is used to derive the formula for time dilation. It is said that

$$\frac{cT_2}{cT'}=\frac{(1+\beta)cT}{cT'}=\frac{1}{k}\tag{10}$$

The first step makes sense since $cT_2$ is simply replaced by $(1+\beta)cT$ which has been shown to be the same earlier:

$$cT_2=cT+VT=(1+\beta)cT\tag{4}$$

But how is $\frac{cT_2}{cT'}$ equal to $\frac{1}{k}$? As we saw in equation $7$, $k=\frac{cT_2}{cT'}$, which means that $$k=\frac{1}{k}$$

Obviously, this cannot be true. So what am I missing here? How does one get to $\frac{cT_2}{cT'}=\frac{1}{k}$?

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Please note that I (unfortunately) do not have physics class in school, so I cannot ask any teacher. I apologize if the question/answer is trivial.

Answer 1

Eq.10 is incorrect.

Observe that: $$\frac{cT_2}{cT'}=\frac{(1+\beta)cT}{cT'}=\frac{1}{k}\tag{10}$$ implies $$ \begin{align} cT &=\frac{1}{(1+\beta)k}cT'\\ &=\frac{1}{(1+\beta)\sqrt{\frac{1+\beta}{1-\beta}}}cT'\\ &=\sqrt{\frac{1-\beta}{(1+\beta)^3}}cT' \neq \frac{1}{\sqrt{1-\beta^2}}cT'\\ \end{align} $$

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So, instead $$\frac{cT_2}{cT'}=\frac{(1+\beta)cT}{cT'}=k\tag{10-corrected}$$ implies $$ \begin{align} cT &=\frac{k}{(1+\beta)}cT'\\ &=\frac{1}{(1+\beta)}\sqrt{\frac{1+\beta}{1-\beta}}cT'\\ &=\frac{1}{\sqrt{1+\beta}}\frac{1}{\sqrt{1-\beta}}cT' = \frac{1}{\sqrt{1-\beta^2}}cT'\\ \end{align} $$

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In my opinion, you should use another reference.

It is disappointing that the reference you provided
provides very few references.
In particular, there should be some reference to Hermann Bondi's $k$-calculus, such as
Bondi, Relativity and Common Sense,
https://archive.org/details/RelativityCommonSense