How can I show that Mathematics is not all about numbers? My goal is to be able to show that many concepts in Mathematics do not necessarily need numbers, and they can be formalized for sets that do not involve numbers. To explain what I mean, suppose I define the following set of objects:
$$
G = \{phone,volleyball,shoe,water \ bottle\}
$$
I want to know what questions I need to answer so that I can be able to treat set $G$ the same as the set of real numbers $\mathbb{R}$. More precisely, I want to be able to perform the basic operations (addition, subtraction, etc), I want to be able to define functions from the set $G$ to another arbitrary set of objects, and I want to be able to differentiate and integrate these functions.
However, as I have no formal background in Mathematics, I'm interested in resources that can help me do this. Honestly, I am not even sure where to start, so any help would be greatly appreciated.

EDIT
For future reference, based on the comments, I found a free book called Abstract Algebra: Theory and Applications by Thomas W. Judson that I think is helpful.
 A: Let do a little Frame challenging here:

I want to know what questions I need to answer so that I can be able to treat set $G$ the same as the set of real numbers $\Bbb R$.

You can't. One characteristic of $\Bbb R$ is that it has infinitely many elements. Many of the things you do with $\Bbb R$ require infinitely many elements to do. But $G$ has only four elements. You can set up a lot of things on $G$ that look like similar operations on $\Bbb R$, but only if you do not look closely. No matter what you define on $G$, it will not be long before you discover differences in behavior from $\Bbb R$.

Okay, so we abandon the $G$ you defined, and instead pick one with as many elements in it as $\Bbb R$. (This is infinite - indeed a bigger infinity than even the rationals have, so it cannot be physical items. But it doesn't have to be a subset of $\Bbb R$ or $\Bbb C$ or some other acknowledged set of numbers.) Can we define operations on $G$ so that it acts just like $\Bbb R$?
Yes, in theory. Because it is infinite, how one goes about that is going to depend exactly how the set is picked. But we'll just assume that part can be worked out. What you need to define is:

*

*An element "$0$" of $G$ and an operation "$+$" on $G\times G$ with values in $G$ which has the properties

*

*(Additive Identity) For all $x \in G, 0 + x = x$

*(Additive Inverse)  For all $x \in G$ there exists a $y \in G$ with $x + y = 0$

*(Commutivity of Addition) For all $x, y \in G, x + y = y + x$

*(Associativity of Addition) For all $, y, z \in G, (x + y) + z = x + (y + z)$



*An element "$1$" of $G$, $1 \ne 0$, and an operation "$\cdot$" on $G \times G$ with values in $G$ which has the properties:

*

*(Multiplicative Identity) For all $x \in G, 1 \cdot x = x$

*(Multiplicative Inverse)  For all $x \in G, x \ne 0$, there exists a $y \in G$ with $x \cdot y = 1$

*(Commutivity of Multiplication) For all $x, y \in G, x \cdot y = y \cdot x$

*(Associativity of Multiplication) For all $, y, z \in G, (x \cdot y) \cdot z = x \cdot (y \cdot z)$

*(Distributivity of Multiplication over Addition) For all $x, y, z \in G, x\cdot(y + z) = (x\cdot y) + (x\cdot z)$



*A binary relation $\le$ between elements of $G$ satisfying

*

*(Reflexivity) For all $x\in G, x \le x$

*(Antisymmetry) For all $x,y \in G$, if $x \le y$ and $y \le x$, then $x = y$.

*(Transitivity) For all $x, y ,z \in G$, if $x \le y$ and $y \le z$, then $x \le z$.

*(Linearity) For all $x, y \in G$, either $x\le y$ or $y \le x$.

*If $x, y, z \in G$ and $x \le y$, $x + z \le y + z$.

*If $x, y, z \in G, z > 0$ and $x \le y$, then $x \cdot z \le y \cdot z$.

*(Completeness) If $A, B \subseteq G, A \ne \varnothing, B \ne \varnothing, A\cup B = G, A \cap B = \varnothing$ and for all $a \in A, b \in B, a \le b$, then there is some $x \in G$ such that if $y < x$, then $y \in A$ and if $y > x$, then $y \in B$.



(That is, if $G$ is divided into two non-empty sets $A, B$ such that all elements of $A$ are $<$ all elements of $B$, there is a point $x \in G$ where $A$ and $B$ meet. Either $A = \{r\mid r < x\}, B = \{r\mid r \ge x\}$ or $A = \{r\mid r \le x\}, B = \{r\mid r > x\}$. This property is what differentiates $\Bbb R$ from $\Bbb Q$. The sets $\{r\in \Bbb Q\mid r < \sqrt 2\}$ and $\{r\in \Bbb Q\mid r > \sqrt 2\}$ divide $\Bbb Q$ with no point in the middle.)
You get all of that done, and you have a set $G$ that can be used in every way exactly like $\Bbb R$, no matter what the elements of $G$ actually are. So you've shown that Math doesn't require numbers, right? Wrong.
This is where the frame challenge comes in. No matter what the elements of $G$ were. Now they are numbers. Defining those operations turned them into numbers. What makes $\Bbb R$ to be $\Bbb R$ is not the elements of the set. Instead, it is exactly those properties listed. It can be shown that given any two sets that meet all of those properties, you can match up the elements of one to the other in such a way that every property is perfectly preserved by the match.
This is how $\Bbb R$ is unique. You can define the same operations and properties on other sets, but all you get is something that is effectively just $\Bbb R$ again. The underlying objects do not matter.
