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My confusion comes from the fact that addition can be interpreted both as combining two sets, or as extending a lengthm, see this article:https://en.wikipedia.org/wiki/Addition.

Why can addition be interpreted as combining two sets, yet can also be interpreted as extending a length? How can the same thing be interpreted in two different ways, with two very different meanings? How will we know which interpretation is being used, when see problems and situations involving addition?

By the way, can I try to make the explanation as simple as possible, I'm still a beginner, so I don't understand more difficult math terms.

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  • $\begingroup$ I don't see the issue. Addition is an operation, defined on a domain... Depending on the domain, the "interpretation" might be different. You can add integers, complex numbers, matrices... It is more difficult to see addition of matrices as extending a length. $\endgroup$ – Martigan Jun 14 '18 at 9:44
  • $\begingroup$ I just don't get how the same thing can be interpreted as two very different ways. This is important because each interpretation only makes sense in some situations. For instance, the adding a length analogy only makes sense when using it in some situations; as you said, it doesn't make sense with matrices. So if they are both the same thing, how come they only work in some situations? How does your "domain" response solve this issue? $\endgroup$ – Ethan Chan Jun 14 '18 at 9:52
  • $\begingroup$ In the wikipedia example, each element of a set just counts as "one element", i.e., all elements are equivalent, and you're just adding the cardinality of the sets. So addition of sets (as wikipedia illustrates it) is just extending the set's cardinality. And that's pretty much analogous to extending length. Generally speaking, any valid notion of addition just has to satisy some axioms, as given by your wikipedia page, and you can call any "rule of combination" that satisfies those axioms "addition", interpretation notwithstanding. ... $\endgroup$ – John Forkosh Jun 14 '18 at 9:55
  • $\begingroup$ Thanks, but can you try to simplify that? I'm still a beginner, so I don't understand terms like cardinality, etc. $\endgroup$ – Ethan Chan Jun 14 '18 at 9:55
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    $\begingroup$ One way of understanding the difference beteen the two interpretation (even if, in fact, the second is a specific case of the first), is to consider that in fact addition is different in different cases. So in fact we are not talking about the same thing. Yes it looks the same, for convenience on one hand, and also because the "simple" addition used for reals is very often used in a way for the more complex additions (such as complex numbers, matrices...) $\endgroup$ – Martigan Jun 14 '18 at 11:01
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It's all part of a pattern.

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Lets start with two sets, $A$ and $B$. Let $A \setminus B$ represent the set of all elements of $A$ that are not also in set $B$. Define $B \setminus A$ similarly. Then we know that

\begin{align} A &= (A \setminus B) \ \cup (A \cap B) \\ B &= (B \setminus A) \ \cup (A \cap B) \\ A \cup B &= (A \setminus B) \ \cup (A \cap B) \ \cup (B \setminus A) \end{align}

Notice first that the three sets $A \setminus B$, $A \cap B$, and $B \setminus A$ have no points in common. They are pairwise disjoint. Now notice that this is true without saying anything at all about the context of the sets $A$ and $B$. So now lets add some context.

AREA

We can think about $A$ and $B$ as representing two-dimensional regions. Like circles, triangles, quadrilateral, and so on. If we let $\alpha(X)$ represent the area of the object X, then it is not unreasonable to require that $\alpha$ at least has the following properties.

\begin{align} \alpha(A) &= \alpha(A \setminus B) + \alpha(A \cap B) \\ \alpha(B) &= \alpha(B \setminus A) + \alpha(A \cap B) \\ \alpha(A \cup B) &= \alpha (A \setminus B) + \alpha(A \cap B) + \alpha(B \setminus A) \end{align}

PROBABILITY

We can think about $A$ and $B$ as representing events with probabilities. If we let $P(X)$ represent the probability of the event X, then it is not unreasonable to require that $P$ has the following properties.

\begin{align} P(A) &= P(A \ \text{and not} \ B) + P(A \ \text{and} \ B) \\ P(B) &= P(B \ \text{and not} \ A) + P(A \ \text{and} \ B) \\ P(A \ \text{or} \ B) &= P(A \ \text{and not} \ B) + P(A \ \text{and} \ B) + P(B \ \text{and not} \ A) \end{align}

Notice that it follows that $$P(A \ \text{or} \ B) = P(A) + P(B) - P(A \ \text{and} \ B)$$

COUNTING

We can think about $A$ and $B$ as representing finite sets. If we let $\#(X)$ represent the number of elements in the set X, then we would expect

\begin{align} \#(A) &= \#(A \setminus B) + \#(A \cap B) \\ \#(B) &= \#(B \setminus A) + \#(A \cap B) \\ \#(A \cup B) &= \#(A \setminus B) + \#(A \cap B) + \#(B \setminus A) \end{align}

If we suppose that $A$ and $B$ have no points in common, then we would expect to have $\#(A \cap B) = 0, \ \#(A \setminus B) = \#(A)$, and $\#(B \setminus A) = \#(B)$. It would follow that $\#(A \cup B) = \#(A) + \#(B)$.

LENGTH

Let $a$ and $b$ be real numbers. For $a \le b$, define the intervals $(a,b), [a,b), (a,b]$, and $(a,b)$ as usual and define the length function, $L$ as follows $L(a,b) = L[a,b) = L(a,b] = L(a,b) = b-a$.

Then, for $a < b < c$, $L[a,b] + L[b,c] = (b-a) + (c-b) = c-a = L[a,c]$.

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To at least address the "how" questions... :)

Consider adding apples to a cart and keeping track of two quantities: the number of apples in the cart and the weight of the loaded cart. Which interpretation of addition is being used in each case, and the difference between the interpretations, becomes very clear when we think about what happens to each quantity when we add a particularly small (or a particularly large) apple to the cart.

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  • $\begingroup$ Thanks, but my question was, how can the same thing have two different interpretations, when these two interpretations have very different meanings and applications? If both interpretations mean addition (ie: they mean the same thing), why can't the be used in the same situation? Why do they have very different uses and meanings? $\endgroup$ – Ethan Chan Jun 14 '18 at 10:02
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Why can addition be interpreted as combining two sets, yet can also be interpreted as extending a length?

Because one characterizes addition by certain properties and those two examples fulfill those properties.

E.g. let $+$ be a map from a set $M$ to itself such that

  • $(A)$: For all $x, y, z \in M$ we have $(x+y)+z = x+(y+z)$.
  • $(N)$: There is an element $n \in M$ such that for all $x \in M$ we have $n+x=x+n$.
  • $(I)$: For all $x \in M$ there is an element $y \in M$ such that $x+y=y+x=n$.

In algebra such a structure is called an additive group and $+$ is an addition. It is associative, has neutral element and there are inverse elements. For convenience one writes $n$ as $0$ and $y$ as $-x$.

The example with extending length can be fit to the above, e.g for the set of translations.

The example with combining sets will not fulfill to the (I) property, only to $(A)$ and $(N)$, with the empty set as neutral element.

Alas the characterization via those group axioms is not yet specific enough to distinguish addition and multiplication.

E.g. let $*$ be a map from a set $M$ to itself such that

  • $(A)$: For all $x, y, z \in M$ we have $(x*y)*z = x*(y*z)$.
  • $(N)$: There is an element $n \in M$ such that for all $x \in M$ we have $n*x=x*n$.
  • $(I)$: For all $x \in M$ there is an element $y \in M$ such that $x*y=y*x=n$.

In algebra such a structure is called a multiplicative group and $*$ is a multiplication. It is associative, has neutral element and there are inverse elements. For convenience one writes $n$ as $1$ and $y$ as $x^{-1}$.

Structural this is the same, and depending on the context one picks the one which feels more natural.

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