Why does everyone say that the faster you move through space, the slower you move through time, when that's not the case? Suppose, I'm on earth and my brother is moving away from earth at a constant speed, $v=0.8c$. Now, if 5 seconds $(t_0)$ pass for me, the amount of time that will pass for my brother according to me will be $t$:
$$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$t=\frac{5}{\sqrt{1-(0.8)^2}}$$
$$t=8.33s$$
So, if $5$ decades pass for me, $8.33$ decades will pass for my brother. He will experience rapid aging according to me. So, why is everyone saying time will go slower for him when the case is the exact opposite?
 A: Your question is as such completely unrelated to your title. However, that it seems related to the body of your question to you is at the heart of your misconception. One simply moves at $1$ second per second through time. It is a meaningless thing to ask how fast one moves through time, so it is true that it is not the case that the faster you move through space, the slower you move through time but it's not because you instead move faster through time. It's a meaningless assertion either way.
What you want to ask is whether a clock $A$, as seen by a given observer $O$, ticks at a slower or faster rate than their own clock $B$ when clock $A$ is moving faster or slower w.r.t. the said observer $O$. Now, this is a meaningful question. Given the correction to your calculation indicated in the other answers, the answer to this meaningful question is that a clock $A$ ticks at a slower rate, as observed by $O$, the faster it moves w.r.t. $O$.
However, it is crucial to understand that this does not mean that "time slows down" for clock $A$. That is, in fact, the central tenet of special relativity -- that all inertial observers observe the same physics. So, nothing unusual happens to $A$ from its own reference frame. That is the very axiom on the basis of which we derive the implication (via coupling it to the existence of a finite invariant speed) that when viewed from the reference frame of $B$, $A$ must appear to have slowed down.
A: You have applied the equation incorrectly.
This is because $t$ is the time you observe (the dilated time) on your brothers clock and $t_0$ is the proper time, or the time inside your brother’s frame of reference.
That is, if $5$ seconds elapsed on your brother's clock as measured from your frame, then the elapsed time on his clock in his frame is $t_0$ where $$5=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \\ \rightarrow t_0=5\sqrt{1-0.8^2}=3\ \ \text{seconds}$$
It's interesting to note that since your brother is also observing you to be moving away from him at $v=0.8c$, if $5$ seconds passes for you inside your frame, then your brother will observe your clock to take $$t=\frac{5}{\sqrt{1-(0.8)^2}}=8.3\ \ \text{seconds}$$ This is probably how you meant to apply the equation.
When you observe his clock you will see a dilated time and when he observes your clock, he to will also see a dilated time. So who is correct? The solution to this apparent contradiction "the twin paradox" is addressed  here.
A: Actually, you have used the relativistic formula incorrectly, which is creating the confusion. Let's understand the equation first:
$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Here $\Delta t_0$ is called the proper time, which is the time measured by the observer who is inside the moving frame of reference relative to the other observer. Here the person inside the spaceship measures the proper time.
$\Delta t$ is the dilated time, measured by the observer outside the moving frame of reference; or here, the person on earth. So actually, $\Delta t=5$ should yield correct answer.
A: None of the answers point to the fact that time dilation is not an effect that happens to the clocks themselves. To explain it correctly you need to first identify what are the two spacetime event points between which you are calculating the time interval. If the two events are on the world-line of the brother on board the spaceship, then his time should be considered as the proper time and the earth-bound brother's clock will show the dilated time. For example if the spaceship-brother goes from earth to some distant star then his starting point and finishing point, both are on his worldliness. Starting the journey and finishing it are the two events. If by the spaceship clock the journey took say 5 yrs, the earth clock will show the corresponding time interval dilated by some amount depending on the velocity of the ship. It is 5 yrs times the Lorentz factor if you ignore the initial acceleration and final deceleration of the ship. If you care to include these two non-uniform motions , no big deal, you just need to use the time dilation formula with integration. Look up the 1st chapter of Landau Lifshitz Vol 3.
You may look up a few videos I made on this, here are the links:
https://youtu.be/KEeEhP89SZs
https://youtu.be/8iybUbKDTLs
