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### What does a 4-dimensional space look like?

We often hear people in physics and mathematics talking about 4-dimensional, 5-dimensional spaces, etc... so what are those spaces? how do we picture these high dimensional spaces?

We often agree that a line is 1-dimensional, a plane is 2-dimensional and a space is 3-dimensional. This is because if we fix a coordinate system 0xyz then any point in the space is determined by its coordinate (x,y,z).

When we talk about the 4th dimension, we immediately relate it to the time line. A point in the space is no longer a point (x,y,z) but it becomes a point (x,y,z,t). So a fixed point in the 3-dimensional space will correspond to infinite many points in the 4-dimension, its coordinate (x,y,z,t) changes as the time variable t changes. It means that a point (x,y,z) of today is different from a point (x,y,z) of yesterday, even though x, y, z are unchanged, but t has changed from t = a moment of yesterday to t = a moment of today.

So if we think of the 4th dimension as the time line then what should we think of the 5th dimension? It seems that we are stuck. Even if we elude ourselves and imagine it as an imaginary line towards heaven above then what do we made out the rumor by those theoretical physicists spreading around that our universe is 11-dimensional?

So, when talking about dimension, what do mathematicians mean? When a mathematician talking about 4-dimensional, 5-dimensional, or 11-dimensional spaces, he actually means the level of freedom in those spaces. A dimension measures how many "free variables" that space can accommodate. A dimension is the number of "free variables" that need to be used to describe that space.

Let us look at some examples.

The first example is a circle. A circle of radius 1 has the formula $x^2 + y^2 = 1$. This circle is in a 2-dimensional plane and is described by two variables $x$ and $y$. But clearly these two variables are not free variables. If we assign a value to the variable $x$ then $y$ is no longer "free", it must take the value $\pm \sqrt{1 - x^2}$. So, even though the circle is on a 2-dimensional plane, it is a 1-dimensional object. You do not have to take my word for it. You may just need to ask a tiny little virus living on that circle, she'll tell ya.

- "Hello tiny little virus, please tell me, what does your house look like?"
- "My house? Oh, I live on a straight line" - little Miss Virus answers.
 on the circle, a little Miss Virus will feel like she is living on a straight line

The second example is the plane $x+y+z=1$. This plane is obviously a 2-dimensional plane. However, it is  inside a 3-dimensional space and it is specified by a formula with three variables $x$, $y$, $z$. But these variables are not independent variables. If we assign $x$ and $y$ to some values then variable $z$ is no longer free and must take the value $1-x-y$. So we have two free variables $x$ and $y$ and the third variable $z$ depends on them.
 2-dimensional plane x+y+z=1 inside 3-dimensional space

When solving problems, sometimes it is simpler if we can find a way to reduce the dependency between variables. It is best if we can make all variables become independent variables. But if we cannot do that, then the less number of relations between them the better. Let us look at the following problem.

Problem: Prove that if
$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$$
then
$$\left( \frac{a+b+c+d+e+f}{b+d+f} \right)^3 = \frac{a^3 + b^3 + c^3 + d^3 + e^3 + f^3}{b^3 + d^3 + f^3} + \frac{acf + ceb + ead}{bdf} + \frac{a}{b} + \frac{c}{d} + \frac{e}{f} .$$

Analyzing: First of all, we can see that this is an equality proving problem. The simplest way (but may not be an efficient way) to solve this kind of problem is to expand it all out and then using the given relationship between variables to show that both sides of the equation are the same. We have 6 variables $a$, $b$, $c$, $d$, $e$ and $f$, and they are related by the following equations
$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$$

In the context of our story about dimension, this problem is not about 6-dimensional space even we have 6 variables. These six variables have overlapping relationships, for example, $ad = bc$, $cf = de$, etc...

So how do we eliminate these inter-relationships? The trick is to create a new variable $k$ as follows
$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f} = k$$

With this new variable $k$, the problem becomes 4-dimensional with four "free variables" $b$, $d$, $f$ and $k$. Other variables are now dependent on these free variables, that is $a = bk$, $c = dk$, $e = fk$.

Solution: Let
$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f} = k,$$
then, $a = kb$, $c = kd$ and $e = kf$.

Therefore,
$$\frac{a+b+c+d+e+f}{b+d+f} = \frac{kb+b+kd+d+kf+f}{b+d+f} = k+1,$$

$$\frac{a^3 + b^3 + c^3 + d^3 + e^3 + f^3}{b^3 + d^3 + f^3} = \frac{k^3 b^3 + b^3 + k^3 d^3 + d^3 + k^3 f^3 + f^3}{b^3 + d^3 + f ^3} = k^3 + 1,$$

$$\frac{acf + ceb + ead}{bdf} = \frac{kb ~kd ~f + kd ~kf ~b + kf ~kb ~d}{bdf} = 3 k^2,$$

$$\frac{a}{b} + \frac{c}{d} + \frac{e}{f} = 3k .$$

So the equation that we need to prove becomes
$$(k+1)^3 = k^3 + 1 + 3 k^2 + 3 k,$$
this is a familiar equality, and so the problem is solved.

Homework: Prove that if
$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$$
then
$$\frac{a^7 + c^7 + e^7}{b^7 + d^7 + f^7} = \left( \frac{a+c+e}{b+d+f} \right)^7$$