I saw this video on Lorentz transformation and needed to refresh my memory a bit.
If
$$\frac{t}{t'}= \sqrt{1-\frac{v^2}{c^2}}$$
and
$$\gamma = \frac {t'}t $$
How do you make this equal ?
$$\gamma = \frac 1{\sqrt{1-\frac {v^2}{c^2}}}$$
Most of you will probably say "Oh If you can't do this you're way to ahead of yourself". You don't need to say that since I am aware of that and just took this as an example.
On
So we have an expression for $t/t'$, $$ \frac{t}{t'} = \sqrt{1-v^2/c^2},$$
but we need the reciprocal, $t'/t$, in order to know what $\gamma$ is. To do this we will write the right hand side of our equation as a fraction. This can be done by writing it "over one".
$$ \frac{\color{red}{t}}{\color{blue}{t'}} = \frac{\color{red}{\sqrt{1-v^2/c^2}}}{\color{blue}{1}}.$$
Now we can find the reciprocal by just flipping both sides upside down.
$$ \frac{\color{blue}{t'}}{\color{red}{t}} = \frac{\color{blue}{1}}{\color{red}{\sqrt{1-v^2/c^2}}},$$
Now that we know what $t'/t$ is we can replace it with $\gamma$ (because we know that $\gamma=t'/t$). This replacement gives us the desired equation.
$$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}}.$$
Let $\frac{x}{y} = z$ then:
$\frac{x}{y}*y = z*y$
$x = z*y$
$1 = z*\frac{y}{x}$
$\frac{1}{z} = \frac{y}{x}$
Basically $\frac{1}{\frac{x}{y}} = \frac{y}{x}$ is such a basic identity we tend to forget that it's not immediately obvious to everyone.