Intuitively, why are two integers $(a,b)$ and $(c,d)$ equal iff $a+d=c+b$? I am going off the construction of integers from natural numbers as stated in this wikipedia article: https://en.m.wikipedia.org/wiki/Integer#construction
I have heard that the first integer in the question would mean $a-b$ and the second one $c-d$, so if they were equal, then we have $a-b=c-d \iff a+d=c+b$, but this just seems like circular reasoning to me, as we have yet to prove that this property holds true for integers.
I tried to make my own "intuitive" definition in the following way. Let every integer be a string of "L"-s and "R"-s, where the integer starts from a fixed point $0$ and "L" means you have to move left, while "R" means you have to move right. In the end, we get an ordered pair $(a,b)$ that shows that we moved $a$ times on the left and $b$ times on the right. Also, moving first on the right and then on the left and vice versa is the same as not moving at all, so "L" and "R" "cancel" each other.
E.g. "RLRRRLLR" would mean we have to move on the right 5 times and on the left 3 times, which is the same as moving on the rigth 2 times and on the left 0 times. Note that how many times we move on either side is always a natural number, so we aren't using any sort of circular reasoning for our "definition". In other words, following the example, we know that $(5,3)=(2,0)$. The same would have been true if the instructions where reversed, i.e. $(3,5)=(0,2)$.
Saying that $(a,b)=(c,d) \iff a-b=c-d$ simply does not work, because $a-b$ is not always a natural number, e.g. in the second example. Yet, I can't seem to grasp intuitively why $(a,b)=(c,d) \iff a+d=c+b$, which would hold true in both examples.
Any help is much appreciated!
 A: First of all, let's represent this construction of the integers accurately.
The construction in Wikipedia never says that $(a,b) = (c,d) \iff a+d = c+b$.
Throughout the construction, pairs such as $(1,3)$ and $(2,4)$ are regarded as distinct objects. But we define a particular equivalence relation on pairs of natural numbers by the statement $$(a,b) \sim (c,d) \iff a+d = c+b,$$
where $(a,b) \sim (c,d)$ means "$(a,b)$ is equivalent to $(c,d)$."
To be completely rigorous, at this point we would have to prove that $\sim$ is in fact an equivalence relation, although the article does not mention that a proof is required.
The article then introduces the notation $[(a,b)]$ to represent the equivalence class containing $(a,b),$ that is, by definition
$$ [(a,b)] = \{ (x,y) \mid (x,y) \sim (a,b) \}. $$
The construction then has to define various arithmetic operations on these equivalence classes. Some tedious work would be required to show that these definitions are self-consistent, which the article skips over by saying this "is easily verified".
The mechanics of proof have to work this way: start with known facts about objects that are already well-defined, then combine them in logically sound ways.
Intuition does not need to work that way.
There are various ways this intuition might work.
For example, we could suppose that some as-yet-not-rigorously-defined mathematical objects exist, then we could list the properties our intuition says they should have, and we could try to rearrange those properties into something that can be modeled by objects that are already rigorously defined.
So we have an intuition that we should be able to form any integer by the difference of two integers, $a - b,$ in particular we can always find a way to do it such that $a$ and $b$ both are non-negative. This gives us a representation of any integer as a pair of non-negative integers $(a,b)$.
We have an intuitive idea that the non-negative integers ought to correspond somehow to the natural numbers, so a pair of natural numbers $(a,b)$ should represent the same integer as the pair of non-negative integers $(a',b')$ (where $a'$ is the integer that corresponds to $a$ and $b'$ is the integer that corresponds to $b$).
We want the integers to be such that $(a',b')$ and $(c',d')$ represent the same integer if and only if $a' - b' = c' - d',$
and we also want the integers to support arithmetic facts such as that
$x - y = z \iff x = y + z.$
We use those facts to decide that we need
$a' - b' = c' - d'$ to be equivalent to $a' + d' = b' + c',$
and we now have an equivalence property that we can map back to natural numbers,
$$(a,b) \sim (c,d) \iff a+d = c+b.$$
But none of these steps based on our intuition are proof. All we have at this point is an idea about a definition that might work as a definition of integers based on natural numbers. We still have to go through all the tedious steps of proving that this formula defines an equivalence class, defining and proving the consistency of all the arithmetic operations on integers, and so forth.
In short, the intuition that $(a,b)$ means $a - b$ is only a way to develop an informed guess about how the integers could be defined from the natural numbers.
And circular reasoning is an acceptable way to make guesses,
provided that we follow up by using correct (non-circular) reasoning to show that the guesses are correct.

Your intuition about strings of letters L and R representing "move left" and "move right" is another good intuition,
although I would prefer that when we count the Ls and Rs in a pair we list the number of Rs first so that the result corresponds to motion along the integer number line represented in the usual way (with positive numbers on the right).
In other words, let $(a,b)$ be a representative of the number you reach on the integer number line by starting at $0$, moving $a$ places to the right and $b$ places to the left.
Then we say $(a,b) \sim (c,d)$ if the two sets of right and left motions arrive at the same point, and we define $[(a,b)]$ as the equivalence class of all sets of motions that arrive at the same point as $(a,b).$
Now suppose $(a,b) \sim (c,d),$ and let's suppose we arrange the strings of motions for each of these representations so that first we do all the left motions of $(a,b)$ and put down a temporary marker to remember where that got us to and we do all the right motions of $(c,d)$ and put down a temporary marker to remember where that got us to.
Then we do the rightward motions if $(a,b)$ starting at its marker and the leftward motions of $(c,d)$ starting at its marker.
In the end, both of these sets of motions should end up at the same place on the number line, which is the integer represented by both $(a,b)$ and $(c,d).$
Then the marker for $(a,b)$ will be $b$ places to the left of $0$ and the marker for $(c,d)$ will be $c$ places to the right of $0$. The distance between those markers is
$b + c$. In order for $(a,b)$ and $(c,d)$ both to end up at the same point on the number line, moving $a$ places to the right of the first marker and $d$ places to the left of the second marker must bring us to the same point.
That is, the distance between markers must be $a + d.$
But the distance between markers is the same no matter how we calculate it,
so $a + d = b + c.$
A: The wikipedia article defines $(a,b)=(c,d)$ when $a+d=b+c$.  There is nothing to prove since it's the definition.  Now, the pair $(a,b)$ represents the integer $a-b$.  We're not actually subtracting, but defining something using addition that acts the same way as negative numbers would.
A: We define equality between the integers $(a,b)\in\mathbb{Z}$ and $(c,d)\in\mathbb{Z}$ iff they satisfy $a + d = b + c$ where $a,b,c,d,\in\mathbb{N}$ and the operation of addition $+$ is the one which was defined within the naturals. As to your question, it is not circular, but it is based on the underlying idea of difference. Precisely speaking, once we do not have the difference defined yet, we use its additive version to provide a valid construction.
The same thing happens when we define equality between rational numbers, which are pairs of integers where the second entry is not zero. To be accurate, let $(a,b)\in\mathbb{Q}$ and $(c,d)\in\mathbb{Q}$. We define equality between such rationals iff they satisfy the relation $ad = bc$ where the multiplication is the one defined within the integers. That is because we do not have the definition of division yet (so we make use of its multiplicative version instead).
Hopefully this helps!
